itoa 实现

//int to ascii 整形转换成字符串 负数未实现
char *itoa(int num, char *str, int radix)
{
        //0的情况
        if (num==0)
        {
                str[0]="0";
                str[1]="\0";
                return str;
        }

        char string[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";

        char* ptr = str;
        int i;
        int j;


        while (num)
        {
                *ptr++ = string[num % radix];
                num /= radix;

                if (num < radix)
                {
                        *ptr++ = string[num];
                        *ptr = '\0';
                        break;
                }
        }
        //两边对调
        j = ptr - str - 1;

        for (i = 0; i < (ptr - str) / 2; i++)
        {
                int temp = str[i];
                str[i] = str[j];
                str[j--] = temp;
        }

        return str;
}