V7/usr/lib/learn/C.a
eL0 #next
0.1a 10
L0.1a U #print
This script is loosely based on the material in
"The C Programming Language", by Brian Kernighan
and Dennis Ritchie (Prentice-Hall, 1978).
You may find it helpful to have that at hand.
Some of the lessons will indicate
the section that they are based on.
Do you have a copy at hand? Answer yes or no.
#copyin
#user
#uncopyin
#match yes
Fine.
#fail
It might help, but it's not critical.
#log
#next
1.1a 10
L1.1a #print
(Section 1.1)
The way you compile a C program is to say
cc name.c
where name.c is the name of the file the program
is on. Here is a short C program that prints
out a line containing "hello". Type it in and
compile it. Call it anything you want but
don't rename the output.
When done, type "ready".
main()
{
printf("hello\n");
}
#once #create Ref
hello
#user
a.out >x
#cmp x Ref
#log
#next
1.1b 10
L1.1b +#print
(Section 1.1)
Now write a C program that prints two lines,
the first of which says "hello" and the second
"goodbye". Don't forget those \n delimiters.
Compile and test it. When satisfied,
type "ready".
#once #create Ref
hello
goodbye
#user
a.out >test
#cmp test Ref
#succeed
Here is one possible solution to compare against yours.
main()
{
printf("hello\n");
printf("goodbye\n");
}
You could also combine the two messages into one
call to printf, like
printf("hello\ngoodbye\n");
but this is harder to read at a glance.
#log
#next
1.1c 10
tL1.1c #print
(Section 1.1)
The program in Ref.c has an error in it.
Find it, fix it, and run it.
Then type ready.
#once #create Ref.c
main()
{
printf("hello\");
}
#once #create Ref
hello
#user
a.out >x
#cmp x Ref
#log
#next
1.1d 10
fL1.1d #print
(Section 1.1)
Write a program which prints these four lines,
exactly as shown:
A tab is \t
A backspace is \b
A quote is \"
A backslash is \\
Compile it, test it, then type ready.
#once #create Ref
A tab is \t
A backspace is \b
A quote is \"
A backslash is \\
#user
a.out >x
#cmp x Ref
#succeed
One solution:
main()
{
printf("A tab is \\t\n");
printf("A backspace is \\b\n");
printf("A quote is \\\"\n");
printf("A backslash is \\\\\n");
}
#fail
Watch out for those backslashes.
#log
#next
1.1e 10
L1.1e #print
What will be printed by the following printf statement?
printf("\"#@\"");
Type "answer XXX", where XXX is the set of characters
that will be printed.
#copyin
#user
#uncopyin
#match "#@"
#log
#next
1.1f 10
tL1.1f #print
(Section 1.2)
Write a program to print
the value of the character 'X' in
octal. Compile it and run it.
Then type ready.
#user
a.out >test
grep 130 test >/dev/null
#succeed
A possible solution:
main()
{
printf("%o\n", 'X');
}
#log
#next
2.1a 10
L10 G w#print
The lessons from this point on are from the old C script.
They have not been cleaned up as much as they could, so some
are not very illuminating, and the code they illustrate
is not always the best. You're welcome to try them nonetheless,
but be warned.
If you want to proceed, type yes;
otherwise, type bye.
#copyin
#user
#uncopyin
#match yes
#next
11.1a 10
11.2a 5
sL11.1a
P #print
With your 'cc' command you can give the name of
an object file to be loaded with your program.
For example
cc x.c y.o
will load the previously compiled program 'y' along with
the program 'x' to be compiled now.
The file "getnum.o" contains a subroutine "getnum" which
reads an integer and returns its value.
Write a program which reads a number and decides
whether or not it is a multiple of 23. If so print
"yes" and otherwise print "no".
Compile and test; then type "ready".
#once #create Ref1
23000
#once #create Ref2
23001
#once cp %s/getnum.o .
#user
a.out <Ref1 >z1
a.out <Ref2 >z2
grep yes z1 >/dev/null && grep no z2 >/dev/null
#succeed
/* One way: */
main() {
if (getnum()%23 == 0)
printf("yes\n");
else
printf("no\n");
}
#log
#next
12.1a 10
L11.2a #print
With your 'cc' command you can give the name of
an object file to be loaded with your program.
For example
cc x.c y.o
will load the previously compiled program 'y' along with
the program 'x' to be compiled now.
There is a file in this directory named "getnum.o"
that contains a subroutine "getnum" that will read digits
from the standard input, convert them to binary, and
return an integer value.
Write a program which reads an integer and prints
it back in octal. Compile and test as usual.
#once #create Ref
254
#once cp %s/getnum.o .
#user
a.out <Ref >test
grep 376 test >/dev/null
#succeed
/* One way: */
main() {
printf("%o\n", getnum());
}
#log
#next
11.1a 10
L12.1a
P C#print
Write a program which reads two numbers and
prints the larger one in decimal. Use the same
"getnum" subroutine. Compile, test and type
"ready" as usual.
#once #create Ref1
14039 89
#once #create Ref2
20022 23001
#once cp %s/getnum.o .
#user
a.out <Ref1 >x1
a.out <Ref2 >x2
grep 14039 x1 >/dev/null && grep 23001 x2 >/dev/null
#succeed
/* One way: */
main() {
int n1, n2;
n1 = getnum();
n2 = getnum();
printf("%d\n", n1 > n2 ? n1 : n2);
}
/* You could also use something like
if (n1 > n2)
printf("%d\n", n1);
else
printf("%d\n", n2);
*/
#log
#next
12.1b 10
rL12.1b P '#print
The function getnum actually returns -1 when it
encounters end of file. (The source is in getnum.c
if you're interested.)
Write, compile and run a program that
reads numbers one per line with getnum
and, for each, prints:
small if the number is >0 and <=100
big if the number is >100 and <=1000
huge if the number is >1000.
Type "ready" when you're done.
#once cp %s/getnum.o .
#once cp %s/getnum.c .
#once #create Ref
1001
1000
999
101
100
1
#once #create Ref1
huge
big
big
big
small
small
#user
a.out <Ref >test
#cmp Ref1 test
#succeed
/* One way:*/
main() {
int n;
while ((n = getnum()) >= 0)
if (n > 0 && n <= 100)
printf("small\n");
else if (n > 100 && n <= 1000)
printf("big\n");
else if (n > 1000)
printf("huge\n");
}
/* Notice that in principle n could be negative,
so we need the last case to say
else if (n > 1000)
instead of just falling into it with a bare
else
Also it's a good idea to indent the else-if's
exactly the way they are here; otherwise
you'll lose track of what's going on.
**/
#log
#next
13.1a 10
nL13.1a P #print
Write a program which reads
its input and counts the number of
characters and the number of spaces
(where a space is either a blank or
a tab or a newline). Print both numbers.
Compile, test, and type "ready".
#once #create Ref
hoboken harrison newark roseville avenue grove street
east orange brick church orange highland avenue
mountain station south orange maplewood millburn short hills
summit chatham madison convent station morristown
new providence murray hill berkeley heights
gillette stirling millington lyons basking ridge
bernardsville far hills peapack gladstone
#user
a.out <Ref >x1
a.out <Ref >x2
grep 348 x1 >/dev/null && grep 45 x2 >/dev/null
#succeed
#include <stdio.h>
/* One way: */
main() {
int nchar, nspace;
char c;
nchar = nspace = 0;
while ((c = getchar()) != EOF) {
nchar++;
if (c == ' ' || c == '\t' || c == '\n')
nspace++;
}
printf("spaces = %d, chars = %d\n", nspace, nchar);
}
#log
#next
14.1a 10
14.2a 5
L14.1a P #print
Using the familar "getnum.o" routine
write a program that reads numbers one per line and determines
for each if it is prime. Print "prime"
for a prime number and "composite" for a non-prime number.
Compile, test, and type "ready".
#once #create Ref
10039
17947
#once #create Ref1
prime
composite
#once cp %s/getnum.o .
#user
a.out <Ref >x1
#cmp x1 Ref1
#succeed
/* A slow but sure prime-tester */
main()
{
int p, i, comp;
while ((p = getnum()) >= 0) {
comp = 0;
for (i = 2; i*i <= p; i++)
if (p%i == 0) {
comp = 1;
break;
}
if (comp)
printf("composite\n");
else
printf("prime\n");
}
}
#log
#next
15.1a 10
L14.2a 8F #print
Using the "getnum" routine on "getnum.o", write a program
that reads a list of positive numbers and prints their sum. Stop reading
numbers when "getnum" returns a negative or zero value.
Compile and test your program; then type "ready".
#once #create Ref
5 43 293 400 75 832 903 33
#once cp %s/getnum.o .
#user
a.out <Ref >xxx
grep 2584 xxx >/dev/null
#succeed
/* Read numbers and count */
main()
{
int s, n;
s = 0;
while ((n=getnum()) > 0)
s += n;
printf("Sum is %d\n", s);
}
#log
#next
14.2b 5
15.1a 10
L14.2b #print
Write a program which counts the number of five letter
words in its input (define a word as anything between
blanks, tabs or newlines). Compile and run it, then type "ready".
Note that all that is wanted is the total number of
five letter words - nothing was said about distinct
words. Just count the number of times exactly five
characters appear between spaces.
#once #create Ref
This is a passage of text which contains
exactly twelve words of five letters.
Words may appear at the start or at the final
part of a line. Other words show up in
the middle. Avoid counting seven or eight letters
but every five must be noted.
#user
a.out <Ref >xxx
grep 12 xxx >/dev/null
#succeed
/* one way to count five letter words */
#include <stdio.h>
main()
{
int since, wdnum, c;
since = 0;
while ((c=getchar()) != EOF) {
if (c == ' ' || c == '\t' || c == '\n') {
if (since == 5)
wdnum++;
since = 0;
}
else
since++;
}
printf("%d\n", wdnum);
}
#log
#next
15.1a 10
L15.1a P #print
Write a program that reads in lines one at a time,
and prints them out if their length (including
the newline) is odd.
You can use the function getline if you like; the object
file is in getline.o.
Compile and run it, then type "ready".
#once #create Ref1
this line contains an odd number of letters!
this line, however, contains an even number of letters!
#once #create Ref2
this line contains an odd number of letters!
#once cp %s/getline.o .
#user
a.out <Ref1 >x1
#cmp x1 Ref2
#succeed
/* It's certainly easiest with getline: */
#include <stdio.h>
main()
{
char line[500];
int n;
while ((n = getline(line, 500)) > 0)
if (n % 2 == 1)
printf("%s", line);
}
#log
#next
15.1b 10
L15.1b AF ]#print
Write a program that reads in lines, and prints each out
in reverse order (except that the newline should be
at the end).
Thus the line
cat food
should come out as
doof tac
Compile it and run it, then type "ready".
#once #create Ref1
This is odd.
This is even.
#once #create Ref2
.ddo si sihT
.neve si sihT
#once cp %s/getline.o .
#user
a.out <Ref1 >x1
#cmp x1 Ref2
#succeed
/* one way to do this */
#include <stdio.h>
main()
{
char line[500];
int n;
while ((n = getline(line, 500)) > 0) {
for (n -= 2; n >= 0; n--)
putchar(line[n]);
putchar('\n');
}
}
#log
#next
17.1a 10
16.2a 5
L16.2a P #print
Write a program which reads a file with lines of up
to 200 characters and shortens them to 60 characters
by throwing away any characters past the first 60.
Compile and test it; then type "ready".
#once #create Ref
hoboken harrison newark roseville avenue grove street
east orange brick church orange highland avenue east orange
mountain station south orange maplewood millburn short hills
summit chatham madison convent station morristown summit cha
new providence murray hill berkeley heights
gillette stirling millingon lyons basking ridgexxxxxxxxxxxxx
bernardsville far hills peapack gladstone
#once #create badin
hoboken harrison newark roseville avenue grove street
east orange brick church orange highland avenue east orange brick church orange highland avenue east orange brick church orange highland avenue
mountain station south orange maplewood millburn short hills
summit chatham madison convent station morristown summit chatham madison convent station morristown summit chatham madison convent station morristown
new providence murray hill berkeley heights
gillette stirling millingon lyons basking ridgexxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
bernardsville far hills peapack gladstone
#user
a.out <badin >xxx
#cmp Ref xxx
#succeed
/* one way to do this */
#include <stdio.h>
main()
{
char line[61];
int c, k;
k = 0;
while ((c = getchar()) != EOF) {
if (c == '\n') {
line[k] = 0;
printf("%s\n", line);
k = 0;
}
else if (k < 60)
line[k++] = c;
}
}
/* Note that this version works regardless of
how long the lines are. If you use getline,
is the same thing true??
*/
#log
#next
16.2b 10
L16.2b FF #print
Write a program which copies all lines containng
the letter 'p' from its input to its output.
Compile and test it; then type "ready".
#once #create Ref
mountain station south orange maplewood millburn short hills
new providence murray hill berkeley heights
bernardsville far hills peapack gladstone
#once #create badin
hoboken harrison newark roseville avenue grove street
east orange brick church orange highland avenue
mountain station south orange maplewood millburn short hills
summit chatham madison convent station morristown
new providence murray hill berkeley heights
gillette stirling millington lyons basking ridge
bernardsville far hills peapack gladstone
#once cp %s/getline.o .
#user
a.out <badin >xxx
#cmp Ref xxx
#succeed
/* a way to find lines with 'p' */
#include <stdio.h>
main()
{
char line[500];
int k;
while (getline(line, 500) > 0)
for (k = 0; line[k] != '\0'; k++)
if (line[k] == 'p') {
printf("%s", line);
break;
}
}
#log
#next
16.2c 5
17.1a 10
sL16.2c P #print
Write a program to read its input and find the
word in it with the most vowels (instances of a,e,i,o, or u).
Print out that word. Compile and test your
program, then type ready.
#once #create Ref
When in the course of human events, it becomes
necessary for one people to dissolve the political bands which have
connected them with another, and to assume among the
powers of the earth the separate and equal station to which
the laws of Nature and of Nature's God entitle them, a decent
respect to the opinions of mankind requires that they should
declare the causes which impel them to the separation.
We hold these truths to be self evident, that all men
are created equal, that they are endowed by their creator
with certain unalienable rights, that among these are life, liberty,
and the pursuit of happiness. That to secure these rights,
governments are instituted among men, deriving their just
powers from the consent of the governed. That whenever
any form of government becomes destructive of these ends,
it is the right of the people to alter or to abolish it, and
to institute new government, laying its foundation on such
principles and organizing its powers in such form, as to them
shall seem most likely to effect their safety and happiness.
#user
a.out <Ref >xxx
grep unalienable xxx >/dev/null
#succeed
/* a way to find a word with lots of vowels */
#include <stdio.h>
main()
{
char bigword[100], thisword[100];
int nvow, maxvow, c, k;
maxvow = k = 0;
while ((c = getchar()) != EOF) {
if (c == '\n' || c == ' ') {
if (nvow > maxvow) {
copy(thisword, bigword, k);
maxvow = nvow;
}
nvow = k = 0;
} else {
thisword[k++] = c;
switch (c) {
case 'a': case 'e': case 'i': case 'o': case 'u':
nvow++;
}
}
}
printf("the word %s had %d vowels\n", bigword, maxvow);
}
copy(a, b, n)
char a[], b[];
{
int i;
for(i = 0; i < n; i++)
b[i] = a[i];
b[i] = 0;
}
#log
#next
17.1a 10
rL17.1a P #print
Write in a program which reads its input and writes it
out line numbered, with a three-digit line number (starting at one)
followed by a single space and then by the original line.
Note: the printf format specification %3d will print a three
digit number.
You might try the function fgets (part of the standard library).
fgets(buf, size, stdin)
reads from the terminal ("stdin") up to size characters
into buf. It returns NULL on end of file.
Compile and test your program; then type "ready".
#once #create Ref
When in the course of human events, it becomes
necessary for one people to dissolve the political bands which have
connected them with another, and to assume among the
powers of the earth the separate and equal station to which
the laws of Nature and of Nature's God entitle them, a decent
respect to the opinions of mankind requires that they should
declare the causes which impel them to the separation.
We hold these truths to be self evident, that all men
are created equal, that they are endowed by their creator
with certain unalienable rights, that among these are life, liberty,
and the pursuit of happiness. That to secure these rights,
governments are instituted among men, deriving their just
powers from the consent of the governed. That whenever
any form of government becomes destructive of these ends,
it is the right of the people to alter or to abolish it, and
to institute new government, laying its foundation on such
principles and organizing its powers in such form, as to them
shall seem most likely to effect their safety and happiness.
#once #create reffed
1 When in the course of human events, it becomes
2 necessary for one people to dissolve the political bands which have
3 connected them with another, and to assume among the
4 powers of the earth the separate and equal station to which
5 the laws of Nature and of Nature's God entitle them, a decent
6 respect to the opinions of mankind requires that they should
7 declare the causes which impel them to the separation.
8 We hold these truths to be self evident, that all men
9 are created equal, that they are endowed by their creator
10 with certain unalienable rights, that among these are life, liberty,
11 and the pursuit of happiness. That to secure these rights,
12 governments are instituted among men, deriving their just
13 powers from the consent of the governed. That whenever
14 any form of government becomes destructive of these ends,
15 it is the right of the people to alter or to abolish it, and
16 to institute new government, laying its foundation on such
17 principles and organizing its powers in such form, as to them
18 shall seem most likely to effect their safety and happiness.
#user
a.out <Ref >xxx
#cmp xxx reffed
#succeed
#include <stdio.h>
main()
{
char s[200];
int k;
k = 0;
while (fgets(s, 200, stdin) != NULL)
printf("%3d %s", ++k, s);
}
#log
#next
18.1a 10
17.1c 5
gL17.1c #print
Print the 20 Fibonacci numbers beginning with 2
(the sequence is 2,3,5,8,... where each number
is the sum of the immediately preceding pair of numbers.
Start with the pair 1,1).
Print each number on a separate line as a five digit
number (remember %3d in printf? %5d does five digits).
Compile and test your program; then type "ready".
#once #create Ref
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
6765
10946
17711
#user
a.out >xxx
#cmp xxx Ref
#succeed
/* one way */
main()
{
int f1, f2, t, count;
f1 = 1;
f2 = 1;
for (count = 0; count < 20; count++) {
t = f1+f2;
f1 = f2;
f2 = t;
printf("%5d\n", t);
}
}
#log
#next
18.1a 10
L18.1a P #print
There is nothing to force you to write a main program
on each file; just as I supplied "getnum" on a separate file,
you can put different routines on different files. Write
a function "length(s)" which if given a character array argument
"s" returns the length of that string.
Put this on a file named "length.c", and leave
the object file in "length.o".
Write the main program for
testing this somewhere else. Type "ready" when you have compiled
and tested your program and are happy with it.
#once #create Ref
This file contains several lines of quite different lengths.
You should get
all of them exactly
r
right.
#once #create answer
60
0
14
19
1
6
#once #create tzaqc.c
main() {
char s[200];
while (gets(s))
printf("%d\n", length(s));
}
#user
cc tzaqc.c length.c
a.out <Ref >value
#cmp value answer
#succeed
/* one way */
length(s)
char *s;
{
int k;
for (k=0; s[k]; k++)
;
return(k);
}
#log
#next
19.1a 10
yL19.1a $ #print
Write a subroutine which counts the number of times it has
been called and returns that count each time. Name it
"count()". Write it on a file named "count.c". Compile
and test it; type "ready" when happy.
#once #create Ref
23080
#once #create tzaqc.c
main() {
int i;
for (i = 0; i < 23079; i++)
count();
printf("%d\n", count());
}
#user
cc tzaqc.c count.o
a.out >value
#cmp value Ref
#succeed
/* one way */
count()
{
static int n = 0;
return(++n);
}
#log
#next
30.1a 10
20.1a 5
L2.1a #print
(Section 1.2)
The file Ref.c contains a copy of
a program to convert Fahrenheit to
Celsius. Modify it to print this
heading at the top:
Fahrenheit-Celsius Conversion
F: C:
Type ready when you're satisfied.
#once #create Ref
Fahrenheit-Celsius Conversion
F: C:
0 -17.8
20 -6.7
40 4.4
60 15.6
80 26.7
100 37.8
120 48.9
140 60.0
160 71.1
180 82.2
200 93.3
220 104.4
240 115.6
260 126.7
280 137.8
300 148.9
#once #create Ref.c
/* print Fahrenheit-Celsius table
for f = 0, 20, ..., 300 */
main()
{
int lower, upper, step;
float fahr, celsius;
lower = 0; /* lower limit of temperature table */
upper = 300; /* upper limit */
step = 20; /* step size */
fahr = lower;
while (fahr <= upper) {
celsius = (5.0/9.0) * (fahr-32.0);
printf("%4.0f %6.1f\n", fahr, celsius);
fahr = fahr + step;
}
}
#user
a.out >x
#cmp Ref x
#fail
Make sure you get the spacing right.
#log
#next
2.1b 10
7L2.1b #print
(Section 1.2)
Write a C program that prints a number which is
the sum of three numbers: 23, 197, and the product
of 23 and 197. Again, compile and test
it. Please do the computation with the program -
if you do it by hand, I'll give you credit for the
work, but it's just stupid.
#user
a.out >test
grep 4751 test >/dev/null
#succeed
/* Here is one possible solution */
main()
{
printf("%d\n", 23 + 197 + 23*197);
}
#log
#next
2.1c 10
7L2.1c #print
(Section 1.2)
What value is printed by this printf statement?
printf("%.3f", 3.141592654);
Type "answer XXX", where XXX is the value.
#copyin
#user
#uncopyin
#match 3.142
#fail
Remember about rounding?
#log
#next
2.1d 10
L2.1d #print
What value is printed by this printf statement?
printf("%%3.1f", 3.141592654);
Type "answer XXX", where XXX is the value.
#copyin
#user
#uncopyin
#match %3.1f
#fail
Look again - it's %%
#log
#next
2.1e 10
gL2.1e #print
Is there any difference between
printf("%s", s);
and
printf(s);
Answer yes or no.
#copyin
#user
#uncopyin
#match yes
#fail
Consider the string "50% of the answers are wrong."
#log
#next
3.1a 10
L20.1a P 0#print
Write a program to read a list of positive numbers
and sort them into ascending order. Print
the sorted list of numbers one per line
as five digit numbers (%5d in printf).
Stop reading numbers when getnum returns -1.
Compile and test your program; then type "ready".
#once #create Ref
1
3
4
9
11
12
13
14
15
16
17
20
34
71
200
225
250
275
300
4095
7111
16384
#once cp %s/getnum.o .
#once #create input
4 20 3 200 16384 4095 71 11 12 13 14
15 16 17 34 9 7111 300 275 250 225 1
#user
a.out <input >xxx
#cmp xxx Ref
#succeed
main()
{
int n;
int list[1000];
n = getlist(list);
shellsort(list, n);
printlist(list, n);
}
getlist(list)
int list[];
{
int n;
n = 0;
while ((list[n]=getnum()) >= 0)
n++;
return(n);
}
/* this is a shell sort, stripped down to process a list
of integers only. Although you probably don't know
how to write this offhand, you should know where to find
it - it is only marginally more code than a bubble sort
and much faster (n**1.5 vs. n**2) in time. */
shellsort(v, n) /* sort v[0]...v[n-1] into increasing order */
int v[], n;
{
int gap, i, j, temp;
for (gap = n/2; gap > 0; gap /= 2)
for (i = gap; i < n; i++)
for (j=i-gap; j>=0 && v[j]>v[j+gap]; j-=gap) {
temp = v[j];
v[j] = v[j+gap];
v[j+gap] = temp;
}
}
printlist(list, n)
int list[], n;
{
int i;
for(i=0; i<n; i++)
printf("%5d\n",list[i]);
}
/* this is a crummy bubble sort which
would work perfectly well for this
problem but can not be recommended
for large jobs.
sortlist()
{
int i, j, k;
for(i=0; i<n; i++)
for(j=n-1; j>0; j--)
if (list[j-1] > list[j]) {
k = list[j];
list[j] = list[j-1];
list[j-1] = k;
}
}
****/
#log
#next
30.1a 10
L3.1a c#print
(Section 1.3)
The file Ref.c contains a copy of
a program to convert Fahrenheit to
Celsius that prints from 0 to 300
degrees in steps of 20.
Modify it to print from 300 down to 0
in steps of 50. Type ready when you're done.
#once #create Ref
300 148.9
250 121.1
200 93.3
150 65.6
100 37.8
50 10.0
0 -17.8
#once #create Ref.c
/* print Fahrenheit-Celsius table
for f = 0, 20, ..., 300 */
main()
{
int lower, upper, step;
float fahr, celsius;
lower = 0; /* lower limit of temperature table */
upper = 300; /* upper limit */
step = 20; /* step size */
fahr = lower;
while (fahr <= upper) {
celsius = (5.0/9.0) * (fahr-32.0);
printf("%4.0f %6.1f\n", fahr, celsius);
fahr = fahr + step;
}
}
#user
a.out >x
#cmp Ref x
#succeed
Here's our solution:
main() /* Fahrenheit-Celsius 300 ... 0 by 50 */
{
int lower, upper, step;
float fahr;
lower = 0; /* lower limit of temperature table */
upper = 300; /* upper limit */
step = 50; /* step size */
for (fahr = upper; fahr >= lower; fahr = fahr - step)
printf("%4.0f %6.1f\n", fahr, (5.0/9.0) * (fahr-32.0));
}
#log
#next
3.1b 10
rL3.1b #print
What is the value of i after this
___for statement executes?
Type "answer N", where N is the value of i.
for (i = 0; i < 10; i = i + 3)
;
#copyin
#user
#uncopyin
#match 12
#log
#next
4.1a 10
pL30.1a P #print
Write a subroutine named "locn(s,c)" which expects two
arguments: the first is a pointer to characters 's' which
points to a null-terminated string, and the second
is a character 'c' which is to be searched for in the
string 's'. If the character 'c' does not
appear in the string return 0; otherwise return a pointer
to the position of 'c' in the string. Name the program "locn.c";
as usual, compile and test it and then type "ready".
#once #create Ref
0
19
0
25
0
#once #create tzaqc.c
char *alpha "abcdefghijklmnopqrstuvwxyz";
main()
{
extern char *locn();
printf("%d\n", locn(alpha, '+'));
printf("%d\n",locn(alpha, 't')-alpha);
printf("%d\n",locn(alpha, 'a')-alpha);
printf("%d\n",locn(alpha, 'z')-alpha);
printf("%d\n",locn("", 'z'));
}
#user
cc tzaqc.c locn.o
a.out >value
#cmp value Ref
#succeed
/* Try this: */
char *
locn (s, c)
char *s;
{
for( ; *s; s++)
if (*s == c)
return(s);
return(0);
}
#log
#next
31.1a 10
L31.1a P #print
Write a function named "rev(s)" which reverses
the string "s" in place. Name the file that contains
the function "rev.c".
When you're satisfied, type "ready".
#once #create Ref
cbax0987654321
#once #create tzaqc.c
main(){
char *s1, *s2, *s3, *s4;
s1 = "abc";
s2 = "x";
s3 = "";
s4 = "1234567890";
rev(s1);
rev(s2);
rev(s3);
rev(s4);
printf(s1);
printf(s2);
printf(s3);
printf(s4);
printf("\n");
}
#user
cc tzaqc.c rev.o
a.out >value
#cmp value Ref
#succeed
/* One way:*/
rev (s)
char *s;
{
char *p;
int t;
for (p=s; *p; p++)
;
for (p--; p > s; p--, s++) {
t = *p;
*p = *s;
*s = t;
}
}
#log
#next
32.1a 10
L32.1a P #print
Write a program
pair(a,b)
which accepts as arguments two pointers to integers
and swaps the integers if necessary so that the
first argument points to the larger one; that is
int x,y;
x = 9;
y = 15;
pair( &x, &y);
results in x being 15 and y 9. Leave the program
on file "pair.c"; compile, test it, and type "ready".
#once #create tzaqc.c
main()
{
int x,y;
y=200;
x = 0;
pair(&y, &x);
if (x != 0 || y != 200)
return(1);
pair(&x,&y);
if (x != 200 || y != 0)
return(1);
x = 30;
y = 23097;
pair(&x,&y);
if (x != 23097 || y != 30)
return(1);
return(0);
}
#user
cc tzaqc.c pair.o
a.out
#succeed
pair(a, b)
int *a, *b;
{
int t;
if (*a <= *b) {
t = *a;
*a = *b;
*b = t;
}
}
#log
#next
33.1a 10
L33.1a P #print
Write a main program which counts the number of command-line arguments
it has which begin with the letter 'b'. Print the
result in decimal. Compile and test it as usual.
Then type "ready".
#user
a.out abc bcd efg rpq b bbvd >xxx
grep 3 xxx >/dev/null
#succeed
/* a possible solution */
main(argc, argv)
char *argv[];
{
int i, k;
for(i=k=0; i<argc; i++)
if (argv[i][0] == 'b')
k++;
printf("%d\n", k);
}
#log
#next
37.1a 10
L35.1a ' #print
Write a program which prints the number of vowels
(instances of 'a', 'e', 'i', 'o', and 'u')
in the input.
#once #create Ref
hoboken harrison newark roseville avenue grove street
east orange brick church orange highland avenue
mountain station south orange maplewood millburn short hills
summit chatham madison convent station morristown
new providence murray hill berkeley heights
gillette stirling millington lyons basking ridge
bernardsville far hills peapack gladstone
#user
a.out <Ref >xxx
grep 109 xxx >/dev/null
#succeed
/* a possible solution */
#include <stdio.h>
main()
{
int k, c;
k = 0;
while ((c = getchar()) != EOF)
switch (c) {
case 'a': case 'e':case 'i': case 'o': case 'u':
k++;
break;
}
printf("%d\n", k);
}
#log
#next
37.1a 10
L37.1a P (#print
Let's try a recursive function. Write a subroutine
power(x,n)
which computes x to the power n by the following
algorithm:
1. if n is zero return 1.
2. if n is odd return x*power(x,n-1).
3. if n is even return the square of
power(x,n/2).
You may assume than x and n are integers, n>=0.
If n is negative return 0 for an answer.
Put your routine on a file "power.c". Compile
it and test it; then type "ready".
#once #create tzaqc.c
main()
{
if (power(-1,-1) != 0)
return(1);
if (power(-3,2) != 9)
return(1);
if (power(2,12) != 4096)
return(1);
if (power(3,5) != 243)
return(1);
if (power(-5, 5) != -3125)
return(1);
if (power(7,3) != 343)
return(1);
if (power(7,4) != 2401)
return(1);
if (power(3,7) != 2187)
return(1);
if (power(2,10) != 1024)
return(1);
return(0);
}
#user
cc tzaqc.c power.o
a.out
#succeed
/* a possible solution */
power(x, n)
{
int k;
if (n < 0)
return(0);
if (n == 0)
return(1);
if (n%2 == 1)
return(x * power(x, n-1));
k = power(x, n/2);
return(k*k);
}
#log
#next
40.1a 10
L4.1a $ #print
Does the # of a "#define" statement absolutely
have to go in column 1? Answer yes or no.
#copyin
#user
#uncopyin
#match yes
#log
#next
5.1a 10
(L40.1a () ^#print
Write a subroutine
errmess(n)
which looks at its argument and prints
one of the following messages:
n message (follow it by a newline)
1 ?
2 syntax error
3 bad syntax error
4 fatal error
5 I give up.
anything else eh?
Leave the routine on errmess.c, compiled and tested
as usual. Then type "ready".
#once #create Ref
eh?
eh?
I give up.
fatal error
bad syntax error
syntax error
?
#once #create tzaqc.c
main()
{
errmess (23069);
errmess (-2000);
errmess (5);
errmess (4);
errmess (3);
errmess (2);
errmess (1);
}
#user
cc tzaqc.c errmess.o
a.out >xxx
#cmp Ref xxx
#succeed
/* a possible solution */
char *message[] = {
"eh?",
"?",
"syntax error",
"bad syntax error",
"fatal error",
"I give up.",
};
errmess(n)
{
if (n < 0 || n > 5)
n = 0;
printf("%s\n", message[n]);
}
#log
#next
41.1a 10
L41.1a P #print
The problem is to produce a function
bitct(x)
which examines the bits in x, returning a count of
the number of 1-bits. There are various ways of doing
this job: here are two.
(1) a sane way. Shift the word x right 16 times (you are
on UNIX) and check the rightmost bit each time, counting
the number of times it is '1'.
(2) a machine-independent (sort of) way. The logical
bitwise AND of x and x-1 contains one fewer one bit than x itself.
Loop anding x and x-1 until you get zero.
Program either algorithm. Compile and test it. Leave it on
a file bitct.c and type "ready".
#once #create tzaqc.c
main()
{
int x;
x=23069;
if (bitct(x) != goodct(x))
return(1);
x=0;
if (bitct(x) != goodct(x))
return(1);
x=16384;
if (bitct(x) != goodct(x))
return(1);
x = -1;
if (bitct(x) != goodct(x))
return(1);
x= -200;
if (bitct(x) != goodct(x))
return(1);
return(0);
}
goodct(x)
{
int k, i;
for(k=i=0; i<16; i++)
{
k =+ (x&1);
x= x>>1;
}
return(k);
}
#user
cc tzaqc.c bitct.o
a.out
#succeed
/* a possible solution */
bitct(x)
{
int k, i;
for(i=k=0; i<16; i++) {
if (x&1)
k++;
x >>= 1;
}
return(k);
}
/* by the way, if you really care about
this problem a table lookup by whole bytes
is faster */
#log
#next
42.1a 10
L42.1a P #print
Write a function
inprod(a,b,n)
that computes the inner product of two integer vectors
a and b which are n items long. Name the file "inprod.c"
and compile and test it; then type ready.
You may assume that the result and all intermediate
values fit in a 16-bit integer, not usually a safe assumption.
#once #create tzaqc.c
main()
{
int x[100], y[100];
int k;
for(k=0; k<100; k++)
{
x[k] = k%10;
y[k] = (k*k)%3;
}
if (inprod(x,y,100) != xprod(x,y,100)) return(1);
return(0);
}
xprod(x,y,n)
int *x, *y;
{
int k, sum;
for(sum=k=0; k<n; k++)
sum=+ *x++ * *y++;
return(sum);
}
#user
cc tzaqc.c inprod.o
a.out
#succeed
/* one way */
inprod(a, b, n)
int *a, *b;
{
int s;
s = 0;
while (n--)
s += *a++ * *b++;
/* none of the spaces in the line above are necessary but
would you really want to read
s+=*a++**b++;
and try to parse it? Even clearer than what I have,
but slower, would be
for(i=0; i<n; i++)
s += a[i]*b[i];
*/
return(s);
}
#log
#next
43.1a 10
L43.1a P %#print
Write a function
cubrt(x)
that returns the cube root of a floating point number.
Put it on a file named "cubrt.c"; compile and test it,
then type "ready".
(If you don't know how to compute cube roots, try Newton's method).
#once #create reldif.c
double reldif(a,b)
double a,b;
{
double c,d;
if (a==0. && b==0.) return(0.);
c = a>0 ? a : -a;
d = b>0 ? b : -b;
c = c>d ? c : d;
return( (a-b)/c );
}
#once #create tzaqc.c
main()
{
double cubrt();
double reldif();
double a, b, eps;
a = 8.;
b = 2.;
eps = 1e-5;
if (reldif(cubrt(a), b) > eps)
exit(1);
a = 0.;
b = 0.;
if (reldif(cubrt(a), b) > eps)
exit(1);
a = 1e6;
b = 1e2;
if (reldif(cubrt(a), b) > eps)
exit(1);
exit(0);
}
#user
cc tzaqc.c cubrt.o reldif.c
a.out
#succeed
/* one way */
double cubrt(x)
double x;
{
/* Newton's method: x <- x - (x**3-a)/(3*x*x) */
double y, yn, dabs();
y = 0.;
yn = x;
while (dabs(y-yn) > y*1.e-8) {
y = yn;
yn = y - (y*y*y-x)/(3*y*y);
}
return(yn);
}
double dabs(x)
double x;
{
return(x>0 ? x : -x);
}
#log
#next
50.1a 10
43.1b 5
>L43.1b * #print
Write a subroutine myexp(x) which expects a floating
argument x and returns the floating point value
of e to the x. An adequate algorithm
for the purpose is the standard Maclaurin series
x 2 3 4
e = 1 + x + x /2! + x /3! + x /4! + ...
Name your routine myexp(), not exp(). You can test it, then,
by comparing it with the system routine as well as by
comparing it with tables. Leave it on file
myexp.c, and then type "ready".
#once #create reldif.c
double reldif(a,b)
double a,b;
{
double c,d;
if (a==0. && b==0.) return(0.);
c = a>0 ? a : -a;
d = b>0 ? b : -b;
c = c>d ? c : d;
return( (a-b)/c );
}
#once #create tzaqc.c
main()
{
double x,y, log(), myexp(), reldif();
for(x=1.; x<5.; x=+ 0.2)
{
y = myexp(x);
if (reldif(x, log(y)) >1.e-5) return(1);
}
return(0);
}
exp()
{
printf("Cheat! you called the system routine\n");
return(1.2);
}
#user
cc tzaqc.c myexp.o reldif.c
a.out
#succeed
/* one way */
double myexp(x)
double x;
{
double term, sum, dabs();
int n;
term = sum = 1.0;
n = 1;
while (dabs(term) > dabs(sum)/1.e8) {
term = term * x / n++;
sum += term;
}
return(sum);
}
double dabs(x)
double x;
{
return(x>0 ? x : -x);
}
#log
#next
50.1a 10
L5.1a * y#print
(Section 1.5)
Write a program that copies exactly three characters
from its input to its output.
When compiled and tested, type ready.
#once #create Ref
XY
#once #create z1
XY
marks the spot.
#user
a.out <z1 >z2
#cmp z2 Ref
#succeed
/* Here is one possible solution */
main()
{
putchar(getchar());
putchar(getchar());
putchar(getchar());
}
#log
#next
5.1b 10
5.2a 5
eL5.1b * #print
(Section 1.5)
Write a program that will read the first character
from its input and print it out in octal.
Compile it, test it, and then type ready.
#once #create Ref
+
#user
a.out <Ref >test
grep 53 test >/dev/null
#succeed
A possible solution:
main()
{
printf("%o\n", getchar());
}
Remember that you can use a function value almost
any place that you could use a variable like x.
Thus many times there's no need for extra variables.
#log
#next
5.1c 10
5.2b 5
L5.1c + `#print
(Section 1.5)
Write a program which reads one character from
its input; if that character is ? it prints "yes",
otherwise it prints "no".
Compile it, test it, and then type ready.
#once #create Ref1
? is here
#once #create Ref2
no ? at beginning
#user
a.out <Ref1 >test1
a.out <Ref2 >test2
grep yes test1 >/dev/null && grep no test2 >/dev/null
#succeed
This is one possible solution
main()
{
if (getchar() == '?')
printf("yes\n");
else
printf("no\n");
}
The indenting and general formatting of C programs
should be done so you make the structure clear,
like the code above.
#log
#next
5.1d 10
L5.1d L T#print
Write a program that counts the blanks, tabs, and newlines
in its input, and prints the total. Don't forget to
define the value of EOF at the beginning of your program.
The best way is to add
#include <stdio.h>
as the first line of your program.
The must____ be in column 1.
(See page 143 of the C book.)
You may also have to say
cc name.c -lS
to compile the program.
#once #create Ref
This is some junk that
contains
blanks
tabs
and newlines.
#user
a.out <Ref >test1
a.out </dev/null >test2
grep 13 test1 >/dev/null && grep 0 test2 >/dev/null
#succeed
One possible solution:
#include <stdio.h>
main()
{
int n, c;
n = 0;
while ((c = getchar()) != EOF)
if (c == ' ' || c == '\t' || c == '\n')
n++;
printf("%d\n", n);
}
This program won't work on huge files, because an int
isn't big enough.
#log
#next
5.1e 10
5.2e 5
L5.1e , #print
Write a program that counts the number of vowels
in its input (excluding 'y'). Don't forget to define
the value of EOF at the beginning of your program.
#once #create Ref
This line contains some vowels, including
the letter 'y'. It also has a capital letter, I think.
#user
a.out <Ref >test
grep 28 test >/dev/null
#succeed
Here is one solution.
#include <stdio.h>
main()
{
int nv, c;
nv = 0;
while ((c = getchar()) != EOF)
if (c=='a' || c=='e' || c=='i' || c=='o' || c=='u'
|| c=='A' || c=='E' || c=='I' || c=='O' || c=='U')
nv++;
printf("%d\n", nv);
}
#fail
Did you remember capital letters?
#log
#next
5.1f 10
L5.1f - #print
(Section 1.5)
Write a program to replace each tab by the
three-character sequence >, backspace, -,
which prints as ->. The program should also replace
each backspace by the analogous sequence <-.
Compile it, test it, then type ready.
#once #create Ref
This contain s a back space and a tab.
#once #create Ref1
This contain <-s a back <-space and a>-tab.
#user
a.out <Ref >test
#cmp test Ref1
#log
#next
5.1g 10
L5.1g u n#print
Write a program to copy its input to its output,
replacing each string of one or more blanks by
a single blank.
#once #create Ref
This has lines with several blanks
including some in funny places.
#once #create Ref1
#once #create Answer
This has lines with several blanks
including some in funny places.
#user
a.out <Ref >test
a.out <Ref1 >>test
#cmp test Answer
#succeed
One way:
#include <stdio.h>
main()
{
int c;
for (c = getchar(); c != EOF; ) {
putchar(c);
if (c == ' ')
while ((c = getchar()) == ' ')
;
else
c = getchar();
}
}
#log
#next
9.1a 10
L5.2a / #print
Write a program which reads a character from its
input and prints "high" if that character is
larger than 100 in numeric value (decimal) and "low"
if it is less than or equal to 100 in numeric value.
Compile it as usual and then type "ready".
#once #create Ref1
u is a big letter
#once #create Ref2
B is a small letter
#user
a.out <Ref1 >test1
a.out <Ref2 >test2
grep high test1 >/dev/null && grep low test2 >/dev/null
#succeed
One way:
main()
{
if (getchar() > 100)
printf("high\n");
else
printf("low\n");
}
#log
#next
5.1b 10
aL5.2b 0 E#print
(Section 1.5)
Write a program which reads a character from its
input and tests whether that character is larger than
100 in numeric value. If so, read two more
characters, and print the value of the second of them
in octal. Compile and test your program, then type "ready".
#once #create Ref1
u V has value 126
#once #create Ref2
. V should not be processed
#user
a.out <Ref1 >test1
a.out <Ref2 >test2
grep 126 test1 >/dev/null && cmp -s test2 /dev/null
#succeed
One way:
main()
{
if (getchar() > 100) {
getchar();
printf("%o\n", getchar());
}
}
#log
#next
5.1c 10
L5.2e +#print
(Section 1.5)
Write a program which copies its input to its output
doubling each character (i.e. each input character
is written twice on the output). Compile and test
it. Then type ready. Don't forget
#include <stdio.h>
at the beginning of your program, and
cc x.c -lS
to compile it.
#once #create Ref
hoboken harrison newark roseville avenue grove street
east orange brick church orange highland avenue
mountain station south orange maplewood millburn short hills
summit chatham madison convent station morristown
new providence murray hill berkeley heights
gillette stirling millington lyons basking ridge
bernardsville far hills peapack gladstone
#once #create Answer
hhoobbookkeenn hhaarrrriissoonn nneewwaarrkk rroosseevviillllee aavveennuuee ggrroovvee ssttrreeeett
eeaasstt oorraannggee bbrriicckk cchhuurrcchh oorraannggee hhiigghhllaanndd aavveennuuee
mmoouunnttaaiinn ssttaattiioonn ssoouutthh oorraannggee mmaapplleewwoooodd mmiillllbbuurrnn sshhoorrtt hhiillllss
ssuummmmiitt cchhaatthhaamm mmaaddiissoonn ccoonnvveenntt ssttaattiioonn mmoorrrriissttoowwnn
nneeww pprroovviiddeennccee mmuurrrraayy hhiillll bbeerrkkeelleeyy hheeiigghhttss
ggiilllleettttee ssttiirrlliinngg mmiilllliinnggttoonn llyyoonnss bbaasskkiinngg rriiddggee
bbeerrnnaarrddssvviillllee ffaarr hhiillllss ppeeaappaacckk ggllaaddssttoonnee
#user
a.out <Ref >xxx
#cmp Answer xxx
#succeed
/* One way: */
#include <stdio.h>
main()
{
int c;
while ((c = getchar()) != EOF) {
putchar(c);
putchar(c);
}
}
#log
#next
5.1e 5
5.3e 2
vL5.3e 2 {#print
Write a program which copies its input to
its output. Compile and test it. Then type ready.
#once #create Ref
hoboken harrison newark roseville avenue grove street
east orange brick church orange highland avenue
mountain station south orange maplewood millburn short hills
summit chatham madison convent station morristown
new providence murray hill berkeley heights
gillette stirling millington lyons basking ridge
bernardsville far hills peapack gladstone
#user
a.out <Ref >xxx
#cmp Ref xxx
#succeed
/* One way: */
#include <stdio.h>
main() {
int c;
while ((c = getchar()) != EOF)
putchar(c);
}
#log
#next
5.1e 10
oL50.1a P #print
(Everything)
Write a program that reads an input file and
prints out the most frequently appearing word in it.
Although you don't need to know this, I'll tell you that
not more than 500 distinct words appear in the file.
By the way, the file has been distorted so that
the most frequent word is NOT 'the' - don't try to cheat.
Compile and test your program.
#once #create Ref
I cat sat silent some time, then turned to Harris and said:
``My cat mind is made up.''
Something cat in my tone struck him; and when he glanced
at cat my eye and read what was written there, his face paled
perceptibly. cat He hesitated a moment, then said:
``Speak.''
I cat answered, with perfect calmness:
``I cat WILL ASCEND THE RIFFELBERG.''
If cat I had shot my poor friend he could not have fallen from
his cat chair more suddenly. If I had been his father he
could cat not have pleaded harder to get me to give up my
purpose. cat But I turned a deaf ear to all he said. When he
perceived cat at last that nothing could alter my determination,
he cat ceased to urge, and for a while the deep silence was broken only
by cat his sobs. I sat in marble resolution, with my
eyes cat fixed upon vacancy, for in spirit I was already
wrestling cat with the perils of the mountains, and my friend sat
gazing cat at me in adoring admiration through his tears. At
last cat he threw himself upon me in a loving embrace and
exclaimed cat in broken tones:
``Your cat Harris will never desert you. We will die together!''
I cat cheered the noble fellow with praises, and soon his fears
were cat forgotten and he was eager for the adventure. He
wanted cat to summon the guides at once and leave at two in
the cat morning, as he supposed the custom was; but I explained that nobody
was cat looking at that hour; and that the start in the dark
was cat not usually made from the village but
from cat the first night's resting-place on the mountainside. I
said cat we would leave the village at 3 or 4 p.m. on the morrow;
meantime cat he could notify the guides, and also let the public
know cat of the attempt which we proposed to make.
I cat went to bed, but not to sleep. No man can sleep when
he cat is about to undertake one of these Alpine exploits. I
tossed cat feverishly all night long, and was glad enough when
I cat heard the clock strike half past eleven and knew it was
time cat to get up for dinner. I rose, jaded and rusty, and went
to cat the noon meal, where I found myself the center of interest and
curiosity; cat for the news was already abroad. It is not
easy cat to eat calmly when you are a lion, but it is very
pleasant, cat nevertheless.
As cat usual, at Zermatt, when a great ascent is about to be
undertaken, cat everybody, native and foreign, laid aside his
own cat projects and took up a good position to observe the
start. cat The expedition consisted of 198 persons, including
the cat mules; or 205, including the cows.
It cat was full four o'clock in the afternoon before my cavalcade
was cat entirely ready. At that hour it began to move. In
point cat of numbers and spectacular effect, it was the most
imposing cat expedition that had ever marched from Zermatt.
I cat commanded the chief guide to arrange the men and
#user
a.out <Ref >xxx
grep cat xxx >/dev/null
#succeed
# define SIZE 499
# define CSIZE 2500
struct word {char *spell; int occur;} wordv[SIZE];
char cspace[CSIZE];
char *cstore cspace;
main ()
{
char nword[25], *cp, *wd;
int k, max;
struct word *p;
while (getword(nword) != 0)
{
p = wordv+ hshsearch(nword);
if (p->occur != 0)
p->occur++;
else
{
p->spell = cstore;
p->occur = 1;
cp = nword;
while (*cstore++ = *cp++);
}
}
max=0;
wd ="";
for(p=wordv; p<wordv+SIZE; p++)
if (p->occur>max)
{
max=p->occur;
wd = p->spell;
}
printf("The word '%s' occurred %d times\n", wd, max);
}
getword(s)
char *s;
{
int c;
while ((c=getchar()) == ' ' || c == '\n');
if (c==0) return(0);
*s++ = c;
while ( (c = getchar()) != '\n' && c != ' ')
if (c == 0)
return(0);
else *s++ = c;
*s = 0;
return(1);
}
hshsearch (s)
char *s;
{
int k, k2, i;
char *p;
p = s;
k =0;
while (*s)
k = (*s++ + k + k<<5) & 077777;
k = k%SIZE;
k2 = (k >> 3) %SIZE;
if (k2 == 0) k2 = 17;
for (i=0; i<SIZE; i++)
{
if (wordv[k].occur == 0)
return(k);
if (comp(wordv[k].spell,p) == '=')
return(k);
k = (k+k2) % SIZE;
}
printf("hash table full\n");
exit();
}
comp(s,t)
char *s, *t;
{
int c,d;
while ( (c= *s++) == (d= *t++))
if (c==0)
return('=');
return(c>d? '>': '<');
}
#log
#next
L9.1a P J#print
(Section 1.9 -- read 1.6-1.8 too.)
Write a program that removes trailing blanks
and tabs from each line of input, and deletes
entirely blank lines. To make your job easier,
you can use the function getline; its source
is in getline.c.
Type read when you are done.
#once #create Ref
This file contains
some trailing
blanks
and tabs
and some empty lines.
#once #create Ref1
This file contains
some trailing
blanks
and tabs
and some empty lines.
#once cp %s/getline.c .
#user
a.out <Ref >test
#cmp test Ref1
#succeed
No answer yet - sorry.
#log
#next
10 10
getline.c #include <stdio.h>
getline(s, lim) /* get line into s, return length */
char s[];
int lim;
{
int c, i;
i = 0;
while (--lim > 0 && (c=getchar()) != EOF && c != '\n')
s[i++] = c;
if (c == '\n')
s[i++] = c;
s[i] = '\0';
return(i);
}
getline.o w ;5
E
5%
%
@ @m H
%
@ @m H
@ @m@ w ) _getline" ~getline csv s lim L1 | L2 c i L4
L5 P __iob L10000 $ L10001 , __filbuf L6 h L3 x cret getnum.c y #include <stdio.h>
getnum()
{
int c, n;
n = 0;
while ((c=getchar()) >= '0' && c <= '9')
n = n*10 + c - '0';
if (c == EOF)
return(-1);
return(n);
}
getnum.o l w 05
E
5%0 %9 A Wp
Ameu%@ w ) _getnum " ~getnum csv L1 f L2 c n L4
__iob L10000 L10001 & __filbuf L5 N L6 \ L3 b cret