文章标题

练习 3-4 在数的对二的补码表示中,我们编写的 itoa 函数不能处理最大的负数,即字长n 等于-2 -1 的情况。请解释其原因。修改该函数,使它在任何机器上运行时都能打印出正确的值。

#include <stdio.h>#include <string.h>void reverse(char []);void itoa(int , char []);int main(){    int n = -2147483648;    char s[100];    itoa(n, s);    printf("%s\n", s);    return 0;}/* reverse : reverse string s in place */void reverse(char s[]){    int c, i, j;    for(i = 0, j = strlen(s)-1; i < j; i++, j--){        c = s[i];        s[i] = s[j];        s[j] = c;    }}void itoa(int n, char s[]){    int i, sign = 1;    if(n < 0)        sign = n;    i = 0;    do {        s[i++] = abs(n%10) + '0';    } while((n /= 10) != 0);    if(sign < 0)        s[i++] = '-';    s[i] = '\0';    reverse(s);}

---

---

练习 4-2

对 atof 函数进行扩充,使它可以处理形如
123.45e-6
的科学表示法,其中,浮点数后面可能会紧跟一个 e 或 E 以及一个指数(可能有正负号) 。

#include <stdio.h>#include <ctype.h>#include <math.h>int pow1(int n){    int sum = 1;    int i = 0;    for(; i < n; i++)        sum *= 10;      return sum;}/* atof: convert string s to double */double atof(char s[]){    int i, sign;    double val, power;    int flag, index;    for(i = 0; isspace(s[i]); i++)        ;    sign = s[i] == '-' ? -1 : 1;    if(s[i] == '+' || s[i] == '-')        i++;    for(val = 0; isdigit(s[i]); i++)        val = 10*val + s[i] - '0';    if(s[i] == '.')        i++;    for(power = 1; isdigit(s[i]); i++){        val = 10*val + s[i] - '0';        power *= 10;    }    if(s[i] == 'e' || s[i] == 'E')        i++;    else         return sign*val/power;    flag = s[i] == '-' ? -1 : 1;    if(s[i] == '+' || s[i] == '-' )        i++;    for(index = 0; isdigit(s[i]); i++)        index = index*10 + s[i] - '0';    if(flag < 0)        return sign*val/power/pow1(index);    else        return sign*val/power*pow1(index);}int main(){    char s[] = "+123.456e-4";    double n = atof(s);    printf("%.4f\n", n);    return 0;}