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. 

 #include <stdio.h>
 #include <limits.h>
 int main()
 {
     printf("char max : %d  min : %d\n",CHAR_MAX, CHAR_MIN);
     printf("unchar max : %u  min : %u\n",UCHAR_MAX, );
     printf("int max : %d  min : %d\n",INT_MAX, INT_MIN);
     printf("unsigned int max : %lu  min : %d\n",UINT_MAX, );
     printf("long max : %ld  min : %ld\n",LONG_MAX, LONG_MIN);
     printf("unsigned long max : %lu  min : %d \n",ULONG_MAX ,);

     return ;
 }

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:

 int main(void)
 {
         /*
         for (i = 0; i < lim-1 && (c=getchar()) != '\n' && c != EOF; ++i)
                s[i] = c;
         */

         int i = ,
         lim = MAX_STRING_LENGTH,
         int c;
         char s[MAX_STRING_LENGTH];

         while (i < (lim - ))
         {
                c = getchar();

                if (c == EOF)
                        break;
                else if (c == '\n')
                        break;

                s[i++] = c;
         }

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

         return ;
 }

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 .

 #include <stdio.h>
 char A[];

 int main(int argc, char const *argv[])
 {
     scanf("%s", A);
     int v = _atoi(A);
     printf("%d\n", v);
     return ;
 }

 int _atoi(char s[]){
     int i, n, hex;
     n = ;
     for(i=;s[i]>='' && s[i]<='' || s[i]>='a' && s[i]<='z' || s[i]>='A' && s[i] <= 'Z'; ++i){
         if(s[i]>='' && s[i]<='')
             hex = s[i] - '';
         if(s[i]>='a' && s[i]<='z')
             hex = s[i] - 'a' + ;
         if(s[i]>='A' && s[i]<='Z')
             hex = s[i] - 'A' + ;
         n =  * n + hex;
     }
     return n;
 }

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 

 #include <stdio.h>
 #include <string.h>
 char a[];
 char b[];

 int lenth(char b[]);
 void strcat_1(char s[], char t[]);

 int main() {
     scanf("%s", a);
     scanf("%s", b);
     strcat_1(a ,b);
     printf("%s", a);
 }

 void strcat_1(char s[], char t[]){
     int i, j, k, l, tlen;
     tlen = lenth(t);
     for(i=;s[i] != '\0';i++){
         k=;
         if(s[i] != t[k])
             continue;
         l = i;
         while(s[l++] == t[k++]) {
             if(t[k]=='\0')
                 break;
         }
         if(t[k] == '\0'){
             while(s[l-tlen]!='\0'){
                  s[l-tlen] = s[l];
                 l++;
             }
             i+=tlen;
         }
     }
 }

 int lenth(char b[]){
     int i;
     for(i=;b[i]!='\0';++i);
     return i;
 }

Friend 's version

#include <stdio.h>

void squeeze(char s[], char t[]) {
    int i, j, k;

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

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.) 

 #include <stdio.h>
 #include <string.h>
 char s1[];
 char s2[];

 int any(char s1[], char s2[]){
     int i, j, len1, len2;
     len1 = strlen(s1);
     len2 = strlen(s2);
     for(i=;i<len1;++i){
         for(j=;j<len2;++j)
             if(s1[i] == s2[j])
                 return i+;

     }
     return -;
 }

 int main(int argc, char const *argv[])
 {
     int t;
     scanf("%s", s1);
     scanf("%s", s2);
     t = any(s1, s2);
     printf("%d\n", t);
     return ;
 }

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. 

 #include <stdio.h>
 unsigned getbits(unsigned x, int p, int n);
 unsigned setbits(int x, int p, int n, int y);
 int main(int argc, char const *argv[])
 {
     unsigned a, y, result;
     int b, c, count;
     scanf("%d %d %d %d", &a, &b, &c, &y);
     //result = getbits(a, b, c);
     result = setbits(a, b, c, y);
     printf("The result is : %d\n", result);
     return ;
 }

 unsigned setbits(unsigned x, int p, int n, int y){
     return  ( x & ~(~(~ << n) << (p+-n))) | ( (y & ~(~ << n)) << (p+-n) ) ;
 }

 unsigned getbits(unsigned x, int p, int n){
     return (x >> (p+-n)) & ~(~ << n);
 }

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. 

 #include <stdio.h>
 unsigned invert(unsigned x, int p, int n);
 int main(int argc, char const *argv[])
 {
     unsigned a, result;
     int b, c, count;
     scanf("%d %d %d", &a, &b, &c);
     result = invert(a, b, c);
     printf("The result is : %d\n", result);

     return ;
 }

 unsigned invert(unsigned x, int p, int n){
     return  (x & ~(~(~ << n) << (p+-n))) | (x >>(p+-n) ^ ~(~ << n)) << (p+-n) ;
 }

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：

 #include <stdio.h>

 void show(unsigned x, int step) {
     if(step < )
         show(x >> , step + );
     printf("%d", x&);
 }

 unsigned rightrot(unsigned x, int n) {
     return (x<<(-n))|(x>>n);
 }

 int main() {
     unsigned x = ;
     show(x, );
     printf("\n");
     show(rightrot(x, ), );
     printf("\n");
     return ;
 }

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 . 

 #include <stdio.h>

 int bitcount(unsigned x) {
     int b;

     for(b = ; x != ; x >>= )
         if(x & )
             b++;
     return b;
 }

 int bitcount2(unsigned x) {
     int b;

     for(b = ; x != ; x &= (x-))
         b++;
     return b;
 }

 int main() {
     printf("%d, %d\n", bitcount(), bitcount2());
     printf("%d, %d\n", bitcount(), bitcount2());
     return ;
 }

练习 2-10

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

 #include <stdio.h>
 char A[];
 int main(int argc, char const *argv[])
 {
     int i;
     scanf("%s", A);
     for(i=;A[i]!=;++i)
         A[i] = higher(A[i]);
     printf("%s\n", A);
     return ;
 }

 int lower(int c){
     return c >= 'A' && c <='Z' ? c + 'a' - 'A' : c;
 }

 int higher(int c){
     return c >= 'a' && c <='z' ? c + 'A' - 'a' : c;
 }