atoi C++实现

下面是一个C语言中实现atoi函数的例子：

#include <stdio.h>

int myAtoi(char* str) {
    int i = 0;
    int sign = 1;
    long ret = 0;
    
    // 跳过空格
    while (str[i == ' ' && str[i != '\0') {
        i++;
    }
    
    // 处理符号
    if (str[i == '-') {
        sign = -1;
        i++;
    } else if (str[i == '+') {
        sign = 1;
        i++;
    }
    
    // 计算数字
    while (str[i >= '0' && str[i <= '9') {
        ret = ret * 10 + (str[i - '0');
        
        // 检查溢出
        if (ret * sign < INT_MIN) {
            return INT_MIN;
        } else if (ret * sign > INT_MAX) {
            return INT_MAX;
        }
        
        i++;
    }
    
    return (int)(ret * sign);
}

int main() {
    char str[] = "123";
    int result = myAtoi(str);
    printf("%d\n", result);
    return 0;
}

这个函数实现了将一个字符串转换为整数的功能。它会跳过字符串前面的空格，并处理正负号。然后，它会逐个处理字符串中的字符，将其转换为数字。如果遇到非数字字符，则停止转换。如果转换后的结果超出了INT_MIN和INT_MAX的范围，则返回相应的边界值。最后，返回转换后的整数。<span class="em">1</span><span class="em">2</span><span class="em">3</span>
#### 引用[.reference_title]
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- *3* [自己实现的atoi函数](https://download.csdn.net/download/welove20101/4177714)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v92^chatsearchT3_1"}}] [.reference_item style="max-width: 33.333333333333336%"]
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