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;
}