C中的const char *的unsigned int

I'm using a function which receives a const char*.

我正在使用一个接收const char *的函数。

long hex2long(const char* hexString)

I have that hexString in unsigned int format and need to convert it to const char* to use that function.

我有无符号int格式的hexString,需要将它转换为const char *才能使用该函数。

I have also tried to use strtol() but it's the same problem.

我也试过使用strtol(),但这是同样的问题。

Any idea?

This is the function:

这是功能:

static const long hextable[] = 
{
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,         // 10-19
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,         // 30-39
        -1, -1, -1, -1, -1, -1, -1, -1,  0,  1,
         2,  3,  4,  5,  6,  7,  8,  9, -1, -1,         // 50-59
        -1, -1, -1, -1, -1, 10, 11, 12, 13, 14,
        15, -1, -1, -1, -1, -1, -1, -1, -1, -1,         // 70-79
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
        -1, -1, -1, -1, -1, -1, -1, 10, 11, 12,         // 90-99
        13, 14, 15, -1, -1, -1, -1, -1, -1, -1,
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,         // 110-109
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,         // 130-139
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,         // 150-159
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,         // 170-179
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,         // 190-199
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,         // 210-219
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,         // 230-239
        -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
        -1, -1, -1, -1, -1, -1
};


long hex2long(const char* hexString)
{
        long ret = 0; 

        while (*hexString && ret >= 0) 
        {
                ret = (ret << 4) | hextable[*hexString++];
        }

        return ret; 
}

1 个解决方案

#1

Your question is confused. Your function parses the hexadecimal representation of a number and returns the value as a long. You should stop parsing when you encounter a non-hexadecimal digit, and you could ignore any leading white space characters. The library function strtol() does just that, with a value of 16 for the last argument (conversion base).

你的问题很困惑。您的函数解析数字的十六进制表示形式,并将值作为long返回。遇到非十六进制数字时应该停止解析,并且可以忽略任何前导空白字符。库函数strtol()就是这样做的,最后一个参数(转换基数)的值为16。

What else are you trying to achieve?

还有什么想要实现的?