C程序设计语言（第二版）习题：第二章

这一章习题做着很舒服，毕竟很简单。所以很有感觉。

 

练习 2-1

Write a program to determine the ranges of char , short , int , and long variables, both signed and unsigned , by printing appropriate values from standard headers and by direct computation. Harder if you compute them: determine the ranges of the various floating-point types. 

 1 #include <stdio.h>

 2 #include <limits.h> 

 3 int main()

 4 {

 5     printf("char max : %d  min : %d\n",CHAR_MAX, CHAR_MIN);

 6     printf("unchar max : %u  min : %u\n",UCHAR_MAX, 0);

 7     printf("int max : %d  min : %d\n",INT_MAX, INT_MIN);

 8     printf("unsigned int max : %lu  min : %d\n",UINT_MAX, 0);

 9     printf("long max : %ld  min : %ld\n",LONG_MAX, LONG_MIN);

10     printf("unsigned long max : %lu  min : %d \n",ULONG_MAX ,0);

11 

12     return 0;

13 }

ps ：假如unsigned int 最大值不用 %lu 而用 %lld 会导致 后面变量发生无法预知的错误，  %ld 无法装下unsigned int 最大值

 

练习 2-2

Exercise 2-2 discusses a for loop from the text. Here it is:

 1 int main(void)

 2 {

 3         /*

 4         for (i = 0; i < lim-1 && (c=getchar()) != '\n' && c != EOF; ++i)

 5                s[i] = c;

 6         */

 7  

 8         int i = 0,

 9         lim = MAX_STRING_LENGTH,

10         int c;

11         char s[MAX_STRING_LENGTH];

12  

13         while (i < (lim - 1))

14         {

15                c = getchar();

16  

17                if (c == EOF)

18                        break;

19                else if (c == '\n')

20                        break;

21  

22                s[i++] = c;

23         }

24  

25         s[i] = '\0';   /* terminate the string */

26  

27         return 0;

28 }

 

练习 2-3

Write the function htoi(s) , which converts a string of hexadecimal digits (including an optional 0x or 0X) into its equivalent integer value. The allowable digits are 0 through 9,a through f, and A through F .

 1 #include <stdio.h>

 2 char A[100];

 3 

 4 int main(int argc, char const *argv[])

 5 {

 6     scanf("%s", A);

 7     int v = _atoi(A);

 8     printf("%d\n", v);

 9     return 0;

10 }

11 

12 int _atoi(char s[]){

13     int i, n, hex;

14     n = 0;

15     for(i=0;s[i]>='0' && s[i]<='9' || s[i]>='a' && s[i]<='z' || s[i]>='A' && s[i] <= 'Z'; ++i){

16         if(s[i]>='0' && s[i]<='9')

17             hex = s[i] - '0';

18         if(s[i]>='a' && s[i]<='z')

19             hex = s[i] - 'a' + 10;

20         if(s[i]>='A' && s[i]<='Z')

21             hex = s[i] - 'A' + 10;

22         n = 16 * n + hex;

23     }

24     return n;

25 }

 

练习 2-4

Write an alternate version of squeeze(s1,s2) that deletes each character in the string s1 that matches any character in the string s2 

 1 #include <stdio.h>

 2 #include <string.h>

 3 char a[100];

 4 char b[100];

 5 

 6 int lenth(char b[]);

 7 void strcat_1(char s[], char t[]);

 8 

 9 int main() {

10     scanf("%s", a);

11     scanf("%s", b);

12     strcat_1(a ,b);

13     printf("%s", a);

14 }

15 

16 void strcat_1(char s[], char t[]){

17     int i, j, k, l, tlen;

18     tlen = lenth(t);

19     for(i=0;s[i] != '\0';i++){

20         k=0;

21         if(s[i] != t[k]) 

22             continue;

23         l = i;

24         while(s[l++] == t[k++]) {

25             if(t[k]=='\0')

26                 break;

27         }

28         if(t[k] == '\0'){

29             while(s[l-tlen]!='\0'){

30                  s[l-tlen] = s[l];

31                 l++;

32             }

33             i+=tlen;

34         } 

35     }

36 }

37 

38 int lenth(char b[]){

39     int i;

40     for(i=0;b[i]!='\0';++i);

41     return i;

42 }

 

Friend 's version 

#include <stdio.h>



void squeeze(char s[], char t[]) {

    int i, j, k;



    k = 0;

    for(i = 0; s[i] != '\0'; ++i) {

        for(j = 0; t[j] != '\0'; ++j) {

            if(s[i] == t[j])

                break;

        }

        if(t[j] == '\0')

            s[k++] = s[i];

    }

    s[k] = '\0';

}



int main() {

    char s[] = "You are e a pig!!!";

    char t[] = "Not me";

    squeeze(s, t);

    printf("%s\n", s);

    return 0;

}

 

练习 2-5

Write the function any(s1,s2) , which returns the first location in the string s1 where any character from the string s2 occurs, or -1 if s1 contains no characters from s2 . (The standard library function strpbrk does the same job but returns a pointer to the location.) 

 1 #include <stdio.h>

 2 #include <string.h>

 3 char s1[100];

 4 char s2[100];

 5 

 6 int any(char s1[], char s2[]){

 7     int i, j, len1, len2;

 8     len1 = strlen(s1);

 9     len2 = strlen(s2);

10     for(i=0;i<len1;++i){

11         for(j=0;j<len2;++j)

12             if(s1[i] == s2[j])

13                 return i+1;

14 

15     }

16     return -1;

17 }

18 

19 

20 int main(int argc, char const *argv[])

21 {

22     int t;

23     scanf("%s", s1);

24     scanf("%s", s2);

25     t = any(s1, s2);

26     printf("%d\n", t);

27     return 0;

28 }

 

练习 2-6

Write a function setbits(x,p,n,y) that returns x with the n bits that begin at position p set to the rightmost n bits of y, leaving the other bits unchanged. 

 1 #include <stdio.h>

 2 unsigned getbits(unsigned x, int p, int n);

 3 unsigned setbits(int x, int p, int n, int y);

 4 int main(int argc, char const *argv[])

 5 {

 6     unsigned a, y, result;

 7     int b, c, count;

 8     scanf("%d %d %d %d", &a, &b, &c, &y);

 9     //result = getbits(a, b, c);

10     result = setbits(a, b, c, y);

11     printf("The result is : %d\n", result);

12     return 0;

13 }

14 

15 unsigned setbits(unsigned x, int p, int n, int y){

16     return  ( x & ~(~(~0 << n) << (p+1-n))) | ( (y & ~(~0 << n)) << (p+1-n) ) ;

17 }

18 

19 unsigned getbits(unsigned x, int p, int n){

20     return (x >> (p+1-n)) & ~(~0 << n);

21 }

 

练习 2-7

Write a function invert(x,p,n) that returns x with the n bits that begin at position p inverted (i.e., 1 changed into 0 and vice versa), leaving the others unchanged. 

 1 #include <stdio.h>

 2 unsigned invert(unsigned x, int p, int n);

 3 int main(int argc, char const *argv[])

 4 {

 5     unsigned a, result;

 6     int b, c, count;

 7     scanf("%d %d %d", &a, &b, &c);

 8     result = invert(a, b, c);

 9     printf("The result is : %d\n", result);

10 

11 

12     return 0;

13 }

14 

15 unsigned invert(unsigned x, int p, int n){

16     return  (x & ~(~(~0 << n) << (p+1-n))) | (x >>(p+1-n) ^ ~(~0 << n)) << (p+1-n) ; 

17 }

 

练习 2-8

Write a function rightrot(x,n) that returns the value of the integer x rotated to the right by n bit positions. 

递归法：

 1 #include <stdio.h>

 2 

 3 void show(unsigned x, int step) {

 4     if(step < 31)

 5         show(x >> 1, step + 1);

 6     printf("%d", x&1);

 7 }

 8 

 9 unsigned rightrot(unsigned x, int n) {

10     return (x<<(32-n))|(x>>n);

11 }

12 

13 int main() {

14     unsigned x = 129923235;

15     show(x, 0);

16     printf("\n");

17     show(rightrot(x, 11), 0);

18     printf("\n");

19     return 0;

20 }

 

 

练习 2-9

In a two's complement number system, x&=(x-1) deletes the rightmost 1-bit in x . Explain why. Use this observation to write a faster version of bitcount . 

 1 #include <stdio.h>

 2 

 3 int bitcount(unsigned x) {

 4     int b;

 5 

 6     for(b = 0; x != 0; x >>= 1)

 7         if(x & 01)

 8             b++;

 9     return b;

10 }

11 

12 int bitcount2(unsigned x) {

13     int b;

14 

15     for(b = 0; x != 0; x &= (x-1))

16         b++;

17     return b;

18 }

19 

20 int main() {

21     printf("%d, %d\n", bitcount(13333), bitcount2(13333));

22     printf("%d, %d\n", bitcount(11111111), bitcount2(11111111));

23     return 0;

24 }

 

练习 2-10

Rewrite the function lower, which converts upper case letters to lower case, with a conditional expression instead of if-else . 

 1 #include <stdio.h>

 2 char A[100];

 3 int main(int argc, char const *argv[])

 4 {

 5     int i;

 6     scanf("%s", A);

 7     for(i=0;A[i]!=0;++i)

 8         A[i] = higher(A[i]);

 9     printf("%s\n", A);

10     return 0;

11 }

12 

13 int lower(int c){

14     return c >= 'A' && c <='Z' ? c + 'a' - 'A' : c;

15 }

16 

17 int higher(int c){

18     return c >= 'a' && c <='z' ? c + 'A' - 'a' : c;

19 }