linux中atoi函数的实现 值得借鉴

int atoi(const char *str);

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

int my_atoi(const char *str) {
    int result = 0;
    int sign = 1;
    int i = 0;

    // Skip leading whitespace
    while (isspace(str[i])) {
        i++;
    }

    // Handle sign
    if (str[i] == '-') {
        sign = -1;
        i++;
    } else if (str[i] == '+') {
        i++;
    }

    // Convert digits and ignore other characters
    while (isdigit(str[i])) {
        result = result * 10 + (str[i] - '0');
        i++;
    }

    return result * sign;
}

int main() {
    const char *test_str = "   -123abc";
    int num = my_atoi(test_str);
    printf("Converted number: %d\n", num);  // Output: -123
    return 0;
}

#include <stdio.h>
#include <stdlib.h>

int main() {
    const char *test_str = "2147483648";  // 超出 int 范围
    char *endptr;
    long num = strtol(test_str, &endptr, 10);

    if (endptr == test_str || *endptr != '\0') {
        printf("Invalid input\n");
    } else if (num > INT_MAX || num < INT_MIN) {
        printf("Overflow or underflow\n");
    } else {
        printf("Converted number: %d\n", (int)num);
    }

    return 0;
}

#include <stdio.h>
#include <stdlib.h>

int main() {
    const char *test_str = "123abc";
    char *endptr;
    long num = strtol(test_str, &endptr, 10);

    if (endptr == test_str || *endptr != '\0') {
        printf("Invalid input: %s\n", endptr);
    } else {
        printf("Converted number: %ld\n", num);
    }

    return 0;
}