C函数atoi的实现

首先，我们先看看微软对于该功能的实现。

C/C++ code

long atol(const char *nptr)
{
        int c;              /* current char */
        long total;         /* current total */
        int sign;           /* if '-', then negative, otherwise positive */

        /* skip whitespace */
        while ( isspace((int)(unsigned char)*nptr) )
            ++nptr;

        c = (int)(unsigned char)*nptr++;
        sign = c;           /* save sign indication */
        if (c == '-' || c == '+')
            c = (int)(unsigned char)*nptr++;    /* skip sign */

        total = 0;

        while (isdigit(c)) {
            total = 10 * total + (c - '0');     /* accumulate digit */
            c = (int)(unsigned char)*nptr++;    /* get next char */
        }

        if (sign == '-')
            return -total;
        else
            return total;   /* return result, negated if necessary */
}

int atoi(const char *nptr)
{
        return (int)atol(nptr);
}

下面是对该功能的收藏:

#include
#include

int myatoi(char *str)
{
        int ret = 0;
        int sign = 1;
        if(*str == '-')
                sign = -1;
        else
            ret = ret *10 + (*str - '0');
        str++;
        while(*str!= '\0')
        {
            ret = ret *10 + (*str - '0');
            str++;
        }

        return sign*ret;
}

void main()
{
        char *mystring = "123";
        int i = myatoi(mystring);
        printf("i value:%d\n",i);
        getch();
}