Question

好的，所以我正在构建一个Base64编码器/解码器，它将十六进制转换为base64并返回，但我发现了一个奇怪的问题，我试图理解，这里的代码是：

string b64_encode(string str)
{
    string newStr = "";
    string ref = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";

    unsigned long long h = 0;

    for(int i=0; i<str.size(); i+=3)
    {
        //Get every 3 chars
        char a = str[i];
        char b = str[i+1];
        char c = str[i+2];

        //Now, convert each hex character (base 16) to it's equivalent decimal number
        //and merge them into one variable
        h = strtoull(&a, nullptr, 16) << 8; //shift left by 8 bits
        h |= strtoull(&b, nullptr, 16) << 4; //shift left by 4 bits
        h |= strtoull(&c, nullptr, 16); //no shift required only the first 2 characters need

        cout << h << endl; //for testing purposes only
    }

    return newStr;
}

当我在Mac OSX上运行此代码时，我得到以下结果，这是错误的：

但是，我在Windows 8中的Visual Studio 2013上编写了相同的代码，它为我提供了正确的值：

我使用的十六进制字符串：

string str = "49276d206b696c6c696e6720796f757220627261696e206c696b65206120706f69736f6e6f7573206d757368726f6f6d";

所以我的问题是，有没有办法在Mac OSX上显示正确的数字？我在网上查了一下，但它没什么帮助。

Answer 1

所以根据@mch，问题是字符不是0终止的，它调用了未定义的行为，多亏了他，这个问题是修正的：

string b64_encode(string str)
{
    string newStr = "";
    string ref = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";

    unsigned long long h = 0;

    for(int i=0; i<str.size(); i+=3)
    {
        //Get every 3 chars
        char a[2] = {str[i], 0};
        char b[2] = {str[i+1], 0};
        char c[2] = {str[i+2], 0};

        //Now, convert each hex character (base 16) to it's equivalent decimal number
        //and merge them into one variable
        h = strtoull(a, nullptr, 16) << 8; //shift left by 8 bits
        h |= strtoull(b, nullptr, 16) << 4; //shift left by 4 bits
        h |= strtoull(c, nullptr, 16); //no shift required only the first 2 characters need

        cout << h << endl;
    }

    return newStr;
}

在C ++中将字符串中的char转换为unsigned long long

1 个答案: