C++：如何将带有 md5 哈希值的 wstring 转换为 byte* 数组？

发布于 2025-01-07 13:29:26 字数 340 浏览 2 评论 0原文

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

//....

byte digest[16];

如何获取摘要中的 md5 哈希值？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));
}

原文

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));
}

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っ〆星空下的拥抱 2025-01-14 13:29:26

您需要从 hashStr 中读取两个字符，将它们从十六进制转换为二进制值，然后将该值放入 digest 中的下一个位置 - 顺序如下：

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

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);
}

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羅雙樹 2025-01-14 13:29:26

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该方法）。

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)));
}

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)));
}

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