函数atoi,itoa 的实现

1.//整数转换成字符串itoa函数的实现

 

#include "stdafx.h"

#include <iostream>

using namespace std;

void itoaTest(int num,char str[] )

{

       int sign = num,i = 0,j = 0;

       char temp[11];

       if(sign<0)//判断是否是一个负数

       {

              num = -num;

       };

       do

       {

              temp[i] = num%10+'0';        

              num/=10;

              i++;

       }while(num>0);

       if(sign<0)

       {

              temp[i++] = '-';

       }

       temp[i] = '/0';

       i--;

       while(i>=0)

       {

              str[j] = temp[i];

              j++;

              i--;

       }

       str[j] = '/0';

}

 

 

2. //字符串转换成整数atoi函数的实现

 

int atoiTest(char s[])

{

       int i = 0,sum = 0,sign;    //输入的数前面可能还有空格或制表符应加判断

       while(' '==s[i]||'/t'==s[i])

       {

              i++;

       }

       sign = ('-'==s[i])?-1:1;

       if('-'==s[i]||'+'==s[i])

       {

              i++;

       }

       while(s[i]!='/0')

       {

              sum = s[i]-'0'+sum*10;

              i++;

       }    

       return sign*sum;

}