第一篇：C程序設(shè)計(jì)語(yǔ)言(第二版)課后答案
“The C Programming Language”, 2nd edition，Kernighan and Ritchie 《c程序設(shè)計(jì)語(yǔ)言》英文的配套答案，所列頁(yè)碼均為英文版的。1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20 1.21 1.22 1.23 1.24 2.01 2.02 2.03 2.04 2.05 2.06 2.07 2.08 2.09 2.10 3.01 3.02 3.03 3.04 3.05 3.06 4.01 4.02 4.03 4.04 4.05 4.06 4.07 4.08 4.12 4.13 4.14 5.01 5.02 5.03 5.04 5.05 5.06 5.07 5.08 5.09 5.10 5.11 5.13 5.14 6.01 6.03 6.04 6.05 7.01 7.02 7.03 7.06 7.08 7.09 8.01 8.03 8.04 8.06
Answer to Exercise 1-1
Run the “hello, world” program on your system.Experiment with leaving out parts of the program, to see what error messages you get.Murphy's Law dictates that there is no single correct answer to the very first exercise in the book.Oh well.Here's a “hello world” program:
#include
int main(void){
printf(“hello, worldn”);
return 0;}
As you can see, I've added a return statement, because main always returns int, and it's good style to show this explicitly.Answer to Exercise 1-2
Experiment to find out what happens when printf 's argument string contains c, where c is some character not listed above.By 'above', the question is referring to: n(newline)t(tab)b(backspace)“(double quote)(backslash)We have to tread carefully here, because using a non-specified escape sequence invokes undefined behaviour.The following program attempts to demonstrate all the legal escape sequences, not including the ones already shown(except n , which I actually need in the program), and not including hexadecimal and octal escape sequences.#include
int main(void){
printf(”Audible or visual alert.an“);
printf(”Form feed.fn“);
printf(”This escape, r, moves the active position to the initial position of the current line.n“);
printf(”Vertical tab v is tricky, as its behaviour is unspecified under certain conditions.n“);
return 0;}
Answer to Exercise 1-3
Modify the temperature conversion program to print a heading above the table.#include
int main(void){
float fahr, celsius;int lower, upper, step;
lower = 0;
upper = 300;
step = 20;
printf(”F Cnn“);
fahr = lower;
while(fahr <= upper)
{
celsius =(5.0 / 9.0)*(fahrstep;
}
return 0;}
This version uses a for loop:
#include
int main(void){
float fahr, celsius;
int lower, upper, step;
lower = 0;
upper = 300;step = 20;
printf(”C Fnn“);
for(celsius = upper;celsius >= lower;celsius = celsius20)
printf(”%3d %6.1fn“, fahr,(5.0/9.0)*(fahr-32));
return 0;} Answer to Exercise 1-6
Verify that the expression getchar()!= EOF is 0 or 1./* This program prompts for input, and then captures a character * from the keyboard.If EOF is signalled(typically through a * control-D or control-Z character, though not necessarily), * the program prints 0.Otherwise, it prints 1.*
* If your input stream is buffered(and it probably is), then * you will need to press the ENTER key before the program will * respond.*/
#include
int main(void){
printf(”Press a key.ENTER would be nice :-)nn“);
printf(”The expression getchar()!= EOF evaluates to %dn“, getchar()!= EOF);
return 0;}
Answer to Exercise 1-7
Write a program to print the value of EOF.#include
int main(void){
printf(”The value of EOF is %dnn“, EOF);
return 0;}
Exercise 1-8
Write a program to count blanks, tabs, and newlines.#include
int blanks, tabs, newlines;
int c;
int done = 0;
int lastchar = 0;
blanks = 0;
tabs = 0;
newlines = 0;
while(done == 0)
{
c = getchar();
if(c == ' ')
++blanks;
if(c == 't')
++tabs;
if(c == 'n')
++newlines;
if(c == EOF)
{
if(lastchar!= 'n')
{
++newlines;/* this is a bit of a semantic stretch, but it copes * with implementations where a text file might not * end with a newline.Thanks to Jim Stad for pointing * this out.*/
}
done = 1;
}
lastchar = c;
}
printf(”Blanks: %dnTabs: %dnLines: %dn“, blanks, tabs, newlines);
return 0;}
Exercise 1-9 Write a program to copy its input to its output, replacing each string of one or more blanks by a single blank.#include
int main(void){
int c;
int inspace;
inspace = 0;
while((c = getchar())!= EOF)
{
if(c == ' ')
{
if(inspace == 0)
{
inspace = 1;
putchar(c);
}
}
/* We haven't met 'else' yet, so we have to be a little clumsy */
if(c!= ' ')
{
inspace = 0;
putchar(c);
}
}
return 0;}
Chris Sidi writes: ”instead of having an “inspace” boolean, you can keep track of the previous character and see if both the current character and previous character are spaces:“
#include
/* count lines in input */ int main(){
int c, pc;/* c = character, pc = previous character */
/* set pc to a value that wouldn't match any character, in case this program is ever modified to get rid of multiples of other characters */
pc = EOF;
while((c = getchar())!= EOF){
if(c == ' ')
if(pc!= ' ')/* or if(pc!= c)*/
putchar(c);
/* We haven't met 'else' yet, so we have to be a little clumsy */
if(c!= ' ')
putchar(c);
pc = c;
}
return 0;}
Stig writes: ”I am hiding behind the fact that break is mentioned in the introduction“!
#include
int main(void){
int c;
while((c = getchar())!= EOF){
if(c == ' '){
putchar(c);
}
while((c = getchar())== ' ' && c!= EOF)
;}
if(c == EOF)
break;/* the break keyword is mentioned
* in the introduction...* */
putchar(c);
}
return 0;
Exercise 1-10
Write a program to copy its input to its output, replacing each tab by t , each backspace by b , and each backslash by.This makes tabs and backspaces visible in an unambiguous way.Category 0 Gregory Pietsch pointed out that my solution was actually Category 1.He was quite right.Better still, he was kind enough to submit a Category 0 solution himself.Here it is:
/* Gregory Pietsch
/*
* Here's my attempt at a Category 0 version of 1-10.*
* Gregory Pietsch */
#include
int main(){
int c, d;
while((c=getchar())!= EOF){
d = 0;
if(c == ''){
putchar('');putchar('');
d = 1;
}
if(c == 't'){
putchar('');
putchar('t');
d = 1;
}
if(c == 'b'){
putchar('');
putchar('b');
d = 1;
}
if(d == 0)
putchar(c);
}
return 0;}
Category 1
This solution, which I wrote myself, is the sadly discredited Cat 0 answer which has found a new lease of life in Category 1.#include
#define ESC_CHAR ''
int main(void){
int c;
while((c = getchar())!= EOF)
{
switch(c)
{ case 'b':
/* The OS on which I tested this(NT)intercepts b characters.*/
putchar(ESC_CHAR);
putchar('b');
break;
case 't':
putchar(ESC_CHAR);
putchar('t');
break;
case ESC_CHAR:
putchar(ESC_CHAR);
putchar(ESC_CHAR);
break;
default:
putchar(c);
break;
}
}
return 0;}
Exercise 1-11
How would you test the word count program? What kinds of input are most likely to uncover bugs if there are any?
It sounds like they are really trying to get the programmers to learn how to do a unit test.I would submit the following:
0.input file contains zero words 1.input file contains 1 enormous word without any newlines 2.input file contains all white space without newlines 3.input file contains 66000 newlines 4.input file contains word/{huge sequence of whitespace of different kinds}/word 5.input file contains 66000 single letter words, 66 to the line 6.input file contains 66000 words without any newlines 7.input file is /usr/dict contents(or equivalent)8.input file is full collection of moby words 9.input file is binary(e.g.its own executable)10.input file is /dev/nul(or equivalent)
66000 is chosen to check for integral overflow on small integer machines.Dann suggests a followup exercise 1-11a: write a program to generate inputs(0,1,2,3,4,5,6)
I guess it was inevitable that I'd receive a solution for this followup exercise!Here is Gregory Pietsch's program to generate Dann's suggested inputs:
#include
int main(void){
FILE *f;
unsigned long i;
static char *ws = ” ftv“;
static char *al = ”abcdefghijklmnopqrstuvwxyz“;
static char *i5 = ”a b c d e f g h i j k l m “ ”n o p q r s t u v w x y z “ ”a b c d e f g h i j k l m “ ”n o p q r s t u v w x y z “ ”a b c d e f g h i j k l m “ ”nn“;
/* Generate the following: */
/* 0.input file contains zero words */
f = fopen(”test0“, ”w“);
assert(f!= NULL);
fclose(f);
/* 1.input file contains 1 enormous word without any newlines */
f = fopen(”test1“, ”w“);
assert(f!= NULL);
for(i = 0;i <((66000ul / 26)+ 1);i++)
fputs(al, f);
fclose(f);
/* 2.input file contains all white space without newlines */
f = fopen(”test2“, ”w“);
assert(f!= NULL);
for(i = 0;i <((66000ul / 4)+ 1);i++)
fputs(ws, f);
fclose(f);
/* 3.input file contains 66000 newlines */
f = fopen(”test3“, ”w“);
assert(f!= NULL);
for(i = 0;i < 66000;i++)
fputc('n', f);
fclose(f);
/* 4.input file contains word/
* {huge sequence of whitespace of different kinds} * /word */
f = fopen(”test4“, ”w“);
assert(f!= NULL);
fputs(”word“, f);
for(i = 0;i <((66000ul / 26)+ 1);i++)
fputs(ws, f);
fputs(”word“, f);
fclose(f);
/* 5.input file contains 66000 single letter words, * 66 to the line */
f = fopen(”test5“, ”w“);
assert(f!= NULL);
for(i = 0;i < 1000;i++)
fputs(i5, f);
fclose(f);
/* 6.input file contains 66000 words without any newlines */
f = fopen(”test6“, ”w“);
assert(f!= NULL);
for(i = 0;i < 66000;i++)
fputs(”word “, f);
fclose(f);
return 0;}
Exercise 1-12
Write a program that prints its input one word per line.#include
int c;
int inspace;
inspace = 0;
while((c = getchar())!= EOF)
{
if(c == ' ' || c == 't' || c == 'n')
{
if(inspace == 0)
{
inspace = 1;
putchar('n');
}
/* else, don't print anything */
}
else
{
inspace = 0;
putchar(c);
}
}
return 0;}
Exercise 1-13
Write a program to print a histogram of the lengths of words in its input.It is easy to draw the histogram with the bars horizontal;a vertical orientation is more challenging./* This program was the subject of a thread in comp.lang.c, because of the way it handled EOF.* The complaint was that, in the event of a text file's last line not ending with a newline，* this program would not count the last word.I objected somewhat to this complaint, on the
* grounds that ”if it hasn't got a newline at the end of each line, it isn't a text file“.*
* These grounds turned out to be incorrect.Whether such a file is a text file turns out to
* be implementation-defined.I'd had a go at checking my facts, and had
* checked the wrong facts!(sigh)*
* It cost me an extra variable.It turned out that the least disturbing way to modify the
* program(I always look for the least disturbing way)was to replace the traditional
* while((c = getchar())!= EOF)with an EOF test actually inside the loop body.This meant
* adding an extra variable, but is undoubtedly worth the cost, because it means the program
* can now handle other people's text files as well as my own.As Ben Pfaff said at the * time, ”Be liberal in what you accept, strict in what you produce“.Sound advice.*
* The new version has, of course, been tested, and does now accept text files not ending in * newlines.*
* I have, of course, regenerated the sample output from this program.Actually, there's no
* ”of course“ about it1];
if(thisval > maxval)
{
maxval = thisval;
}
}
}
else
{
thisval = ++lengtharr[MAXWORDLEN];
if(thisval > maxval)
{
maxval = thisval;
}
}
} if(c == EOF)
{
done = 1;
}
}
else
{
if(inspace == 1 || firstletter == 1)
{
wordlen = 0;
firstletter = 0;
inspace = 0;
}
++wordlen;
}
}
for(thisval = maxval;thisval > 0;thisval--)
{
printf(”%4d | “, thisval);
for(thisidx = 0;thisidx <= MAXWORDLEN;thisidx++)
{
if(lengtharr[thisidx] >= thisval)
{
printf(”* “);
}
else
{
printf(” “);
}
}
printf(”n“);
}
printf(” +“);
for(thisidx = 0;thisidx <= MAXWORDLEN;thisidx++)
{
printf(”---“);
}
printf(”n “);
for(thisidx = 0;thisidx < MAXWORDLEN;thisidx++)
{
printf(”%2d “, thisidx + 1);
}
printf(”>%dn“, MAXWORDLEN);
return 0;}
Here's the output of the program when given its own source as input:
| * 112 | * 111 | * 110 | * 109 | * 108 | * 107 | * 106 | * 105 | * 104 | * 103 | * 102 | * 101 | * 100 | * 99 | * 98 | * 97 | * 96 | * 95 | * 94 | * * 93 | * * 92 | * * 91 | * * 90 | * * 89 | * * 88 | * * 87 | * * 86 | * * 85 | * * 84 | * * 83 | * * 82 | * * 81 | * * 80 | * * 79 | * * 78 | * * 77 | * * 76 | * * 75 | * * 74 | * * 73 | * * 72 | * * 71 | * * 70 | * * 69 | * * 68 | * * 67 | * * 66 | * * 65 | * * 64 | * * 63 | * * * 62 | * * * 61 | * * * 60 | * * * 59 | * * * 58 | * * * 57 | * * * 56 | * * * 55 | * * * 54 | * * * 53 | * * * 52 | * * * * 51 | * * * * 50 | * * * * 49 | * * * * 48 | * * * * 47 | * * * * 46 | * * * * 45 | * * * * 44 | * * * * 43 | * * * * * 42 | * * * * * 41 | * * * * * 40 | * * * * * 39 | * * * * * 38 | * * * * * 37 | * * * * * 36 | * * * * * 35 | * * * * * 34 | * * * * * 33 | * * * * * 32 | * * * * * 31 | * * * * * 30 | * * * * * * 29 | * * * * * * 28 | * * * * * * * 27 | * * * * * * * 26 | * * * * * * * 25 | * * * * * * * * 24 | * * * * * * * * 23 | * * * * * * * * 22 | * * * * * * * * * 21 | * * * * * * * * * 20 | * * * * * * * * * 19 | * * * * * * * * * 18 | * * * * * * * * * 17 | * * * * * * * * * 16 | * * * * * * * * * 15 | * * * * * * * * * 14 | * * * * * * * * * * 13 | * * * * * * * * * * 12 | * * * * * * * * * * 11 | * * * * * * * * * * 10 | * * * * * * * * * * 9 | * * * * * * * * * * * 8 | * * * * * * * * * * * 7 | * * * * * * * * * * * 6 | * * * * * * * * * * * 5 | * * * * * * * * * * * 4 | * * * * * * * * * * * 3 | * * * * * * * * * * * 2 | * * * * * * * * * * * 1 | * * * * * * * * * * * +--1 2 3 4 5 6 7 8 9 10 >10
Exercise 1-14 Write a program to print a histogram of the frequencies of different characters in its input.Naturally, I've gone for a vertical orientation to match exercise 13.I had some difficulty ensuring that the printing of the X-axis didn't involve cheating.I wanted to display each character if possible, but that would have meant using isprint(), which we haven't yet met.So I decided to display the value of the character instead.(The results below show the output on an ASCII system(100 *(thisidx / 100)))/ 10);
}
}
printf(”n “);
for(thisidx = 0;thisidx < NUM_CHARS;thisidx++)
{
if(freqarr[thisidx] > 0)
{
printf(”%d“, thisidx32.0);return c;}
int main(void){
float fahr, celsius;
int lower, upper, step;
lower = 0;
upper = 300;
step = 20;
printf(”F Cnn“);
fahr = lower;
while(fahr <= upper)
{
celsius = FtoC(fahr);
printf(”%3.0f %6.1fn“, fahr, celsius);
fahr = fahr + step;
}
return 0;}
Answer to Exercise 1-16, page 30
Revise the main routine of the longest-line program so it will correctly print the length of arbitrarily long input lines, and as much as possible of the text./* This is the first program exercise where the spec isn't entirely * clear.The spec says, 'Revise the main routine', but the true * length of an input line can only be determined by modifying * getline.So that's what we'll do.getline will now return the * actual length of the line rather than the number of characters * read into the array passed to it.*/
#include
int getline(char line[], int maxline);void copy(char to[], char from[]);
/* print longest input line */ int main(void){
int len;/* current line length */
int max;/* maximum length seen so far */
char line[MAXLINE];/* current input line */
char longest[MAXLINE];/* longest line saved here */
max = 0;
while((len = getline(line, MAXLINE))> 0)
{
printf(”%d: %s“, len, line);
if(len > max)
{
max = len;
copy(longest, line);
}
}
if(max > 0)
{
printf(”Longest is %d characters:n%s“, max, longest);
}
printf(”n“);
return 0;}
/* getline: read a line into s, return length */ int getline(char s[], int lim){
int c, i, j;
for(i = 0, j = 0;(c = getchar())!=EOF && c!= 'n';++i)
{
if(i < lim1)
{
s[j++] = c;
}
++i;
}
s[j] = '