atoi implementation

2015-12-06

需要注意的地方

1、检查字符串是否为空
2、对非法输入，返回0，并设置全局变量
3、溢出
4、空字符串””
输入字符串只有”+”或”-“号

long long StrToIntCore(const char* str, bool minus);

enum Status {kValid = 0, kInvalid};
int g_nStatus = kValid;

int StrToInt(const char* str)
{
    g_nStatus = kInvalid;
    long long num = 0;

    if(str != NULL && *str != '\0') 
    {
        bool minus = false;
        if(*str == '+')
            str ++;
        else if(*str == '-') 
        {
            str ++;
            minus = true;
        }

        if(*str != '\0') 
        {
            num = StrToIntCore(str, minus);
        }
    }

    return (int)num;
}

long long StrToIntCore(const char* digit, bool minus)
{
    long long num = 0;

    while(*digit != '\0') 
    {
        if(*digit >= '0' && *digit <= '9') 
        {
            int flag = minus ? -1 : 1;
            num = num * 10 + flag * (*digit - '0');

            if((!minus && num > 0x7FFFFFFF) 
                || (minus && num < (signed int)0x80000000))
            {
                num = 0;
                break;
            }

            digit++;
        }
        else 
        {
            num = 0;
            break;
        }
    }

    if(*digit == '\0') 
    {
        g_nStatus = kValid;
    }

    return num;
}