模拟实现atoi函数c语言

#include <stdio.h>

int atoi(char *s) {
    int i = 0, sign = 1, result = 0;
    char c;

    // skip white space
    while (s[i] == ' ') {
        i++;
    }

    // handle sign
    if (s[i] == '+' || s[i] == '-') {
        sign = (s[i++] == '+') ? 1 : -1;
    }

    // convert digits
    while (s[i] >= '0' && s[i] <= '9') {
        result = result * 10 + (s[i++] - '0');
    }

    return sign * result;
}

int main() {
    char s1[] = "123";
    char s2[] = "-456";
    char s3[] = "    789";
    char s4[] = "+0123";

    printf("%d\n", atoi(s1));   // expect 123
    printf("%d\n", atoi(s2));   // expect -456
    printf("%d\n", atoi(s3));   // expect 789
    printf("%d\n", atoi(s4));   // expect 123

    return 0;
}