C4 - 4.2 Functions Returning Non-integers

4.2 返回非整型值的函数

 

例子：

atoi和atof的实现（实现成了pzatoi和pzatof）

同时已经完成了Exercise 4-2 要求的对atof的扩充

 

 

#include<stdio.h>
#include<ctype.h>

int pzatoi(char *s, int *pn){
    int c, sign;

    while((c = *s++) == ' ')
        ;
    if(!isdigit(c) && c != '+' && c != '-')
        return -1;
    sign = (c == '-') ? -1 : 1;
    if(c == '+' || c == '-')
        c = *s++;
    for(*pn = 0; isdigit(c); c = *s++)
        *pn = 10 * *pn + (c - '0');
    *pn *= sign;
    return 0;
}

double power(int a, int b){
    int t = 1;
    if(b >= 0)
        for(; b > 0; b--)
            t *= a;
    else
        for(; b < 0; b++)
            t /= a;
    return t;
}

int pzatof(char *s, double *pd){
    int a, c, sign;
    double t;

    while((c = *s++) == ' ')
        ;
    if(!isdigit(c) && c != '+' && c != '-' && c != '.')
        return -1;
    sign = (c == '-') ? -1 : 1;
    if(c == '+' || c == '-')
        c = *s++;

    for(a = 0; isdigit(c); c = *s++)
        a = 10 * a + (c - '0');
    *pd = a;
    if(c == '.'){
        c = *s++;
        for(t = 0.1; isdigit(c); c = *s++){
            *pd += t * (c - '0');
            t *= 0.1;
        }
    }
    if(c == 'e'){
        c = *s++;
        pzatoi(s, &a);
        *pd *= power(10, a);
    }
    *pd *= sign;
    return 0;
}

int main(){
    char s[50];
    double a;
    scanf("%s", s);
    pzatof(s, &a);
    printf("%f\n", a);
}