C++: How to convert wstring with md5 hash to byte* array?

Question

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

//....

byte digest[16];

How can I get my md5 hash in digest? My answer is:

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));
}

It's a `byte` array, not `byte*` array, and it will be too small for that string. — jrok, Feb 20 '12 at 17:36
@jrok : `digest` is sufficient size; 32 hex characters is 16 bytes. — ildjarn, Feb 20 '12 at 17:40
The answer here is a good one: http://stackoverflow.com/questions/194465/how-to-parse-a-string-to-an-int-in-c — John Zwinck, Feb 20 '12 at 17:45
@ildjarn What am I missing here? `hashStr.length()` is 32, right? — jrok, Feb 20 '12 at 17:45
@jrok : Yes, and each two characters represents one byte in hex; 32 / 2 == 16. — ildjarn, Feb 20 '12 at 17:46

score 1 · Answer 1 · answered Feb 20 '12 at 17:51

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);
}

score 0 · Answer 2 · answered Feb 20 '12 at 18:01

C-way (I didn't test it, but it should work (though I could've screwed up somewhere) and you will get the method anyway).

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)));
}

C++: How to convert wstring with md5 hash to byte* array?

2 Answers2