atoi 函数的实现

atoi 函数的实现

atoi()函数实现的功能是：将字符串转换为相应的整形；

atoi代码块的实现

#define _CRT_SECURE_NO_WARNINGS 1
#include<iostream>
using namespace std;
//atoi()的实现(将字符串转为整形)
int my_atoi(char * str)
{
    int n=0;//保存整形  
    char firstChar = *str;//记录str的第一个字符
    if (firstChar == '+' || firstChar == '-')//+123 -123
    {
        ++str;
    }
    while(*str == '0')//00123  -00123
    {
        ++str;
    }

    while (*str)
    {
        char temp = *str;
        if (*str < '0' || *str> '9')
        {
            cout << "字符串不匹配" ;
            return n;//该字符串含有非数字字符，不能转为整数
        }
        else
        {
            temp -= '0';
            n = n * 10 + temp;//将字符串转为整形
        }
        str++;
    }
    if (firstChar == '-')
    {
        n = -n;
    }
    return n;
}

//测试代码块
int main()
{
    char* str1 = "+123";
    cout << "原字符串：" << str1 << "  ";
    cout << "转换后整形：" << my_atoi(str1) <<endl;

    char* str2 = "-123";
    cout << "原字符串：" << str2 << "  ";
    cout << "转换后整形：" << my_atoi(str2) << endl;

    char* str3 = "-000123";
    cout << "原字符串：" << str3 << "  ";
    cout << "转换后整形：" << my_atoi(str3) << endl;

    char* str4 = "+000123";
    cout << "原字符串：" << str4 << "  ";
    cout << "转换后整形：" << my_atoi(str4) << endl;

    char* str5 = "1002300";
    cout << "原字符串：" << str5 << "  ";
    cout << "转换后整形：" << my_atoi(str5) << endl;

    char* str6 = "12a3";
    cout << "原字符串：" << str6 << "  ";
    cout << "转换后整形：" << my_atoi(str6) << endl;
    system("pause");
    return 0;
}

atoi 结果显示

itoa函数的实现