c++ - C++：如何将带有 md5 哈希的 wstring 转换为 byte* 数组？

Question

std::wstring hashStr(L"4727b105cf792b2d8ad20424ed83658c");

//....

byte digest[16];

如何在摘要中获取我的 md5 哈希？我的回答是：

wchar_t * EndPtr;

for (int i = 0; i < 16; i++) {
std::wstring bt = hashStr.substr(i*2, 2);
digest[i] = static_cast<BYTE>(wcstoul(bt.c_str(), &EndPtr, 16));
}

Answer 1

You need to read two characters from hashStr, convert them from hex to a binary value, and put that value into the next spot in digest -- something on this order:

for (int i=0; i<16; i++) {
    std::wstring byte = hashStr.substr(i*2, 2);
    digest[i] = hextobin(byte);
}

Answer 2

C-way（我没有测试它，但它应该可以工作（尽管我可能在某个地方搞砸了），无论如何你都会得到这个方法）。

memset(digest, 0, sizeof(digest));

for (int i = 0; i < 32; i++)
{
    wchar_t numwc = hashStr[i];
    BYTE    numbt;

    if (numwc >= L'0' && numwc <= L'9')             //I assume that the string is right (i.e.: no SGJSGH chars and stuff) and is in uppercase (you can change that though)
    {
        numbt = (BYTE)(numwc - L'0');
    }
    else
    {
        numbt = 0xA + (BYTE)(numwc - L'A');
    }

    digest[i/2] += numbt*(2<<(4*((i+1)%2)));
}