C语言 atoi 函数的实现

首先说下atoi(const char *)这个函数的主要功能：

该函数是用于将含有数字字符的字符串转为整数（可正可负）

/********************
this func is completation for the atoi(const char *);
@parameter :
const char *
@return
output: num
by alfredseng 2016-9-08
***********************/

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

#define RET_VALID 0
#define RET_INVALID 1
#define true 1
#define false 0

int ret_status = RET_INVALID;

int _atoi(char *argv)
{
int flag = false;//false means no any digit
int minus = false;
int sub_flag = false;// default operation with plus
char *ptr = (char *)argv;
int num = 0;
int i = 0;

if(ptr)
{
if (*ptr != '\0')
{
while(' ' == ptr[i])
i++;//跳过前面的空格
if(ptr[i] == '+')
{
i++;
}
else if(ptr[i] == '-')
{
i++;
sub_flag = true;
}
}
int flag=sub_flag?-1:1;
while(i<strlen(ptr))
{
if(isdigit(ptr[i]))
{
num=flag*(ptr[i]-'0')+num*10;
ret_status = RET_VALID;

if((flag&&num > 0x7FFFFFFF) ||(!flag&&( num<0x80000000)))//检查溢出
{
num = 0;
break;
}
}
else
{
num = 0;
break;
}
i++;
}
}
if(ret_status)
{
printf("Sorry,invalid input data detected!\n");
return 0;
}
printf("Valid data detected,and the result:%d\n",num);
return num;
}
int main(int argc,char *argv[])
{
int a=_atoi(argv[1]);
return 0;
}