不使用bigint类进行按位运算? [英] bitwise operation without the use of a bigint class?

问题描述

你好.

我知道我曾经问过一个类似的问题，但是我仍然面临着类似的问题，并且我设法将我的问题浓缩了起来，希望可以更好地理解它?

我有以下功能/代码:

Hello.

I know I have asked a simular question before, but I am still facing problems with it, and I have managed to condense my problem so that, hopefully, it is a little better to understand?

I have the following function/code:

private void button1_Click(object sender, EventArgs e)
        {

            // Raw license, fixed bytes compiled from productID, product Version, License Data, expiery and 2-3 random bytes.
            byte[] rawLicense = {0x00, 0x00, 0xA3, 0xE1};


            int  i = rawLicense[3] + (rawLicense[2] << 8) + (rawLicense[1] << 16) + (rawLicense[0] << 24);

        }

它采用十六进制值00 00 A3 E1并使用移位来形成该十六进制值的int表示，即:41953


现在，我该如何编辑上面的函数以允许我使用更大长度的字节数组，这将产生一个如下所示的int:

45434-59685-45565-59685-56923-49565

我宁愿不使用bigint类，而且我已经看到了与上述类似的代码来执行此操作，从而生成了一个表示较大int的字符串，或者至少是一个类似于上述的数字字符串，当然还有一个方法将字符串反转为字节数组.

非常感谢，
亲切的问候，
Stephen

This takes the hex value 00 00 A3 E1 and uses bitshifting to form the int representation of that hex value, which is: 41953


Now, how can I edit the above function to allow me to work with a byte array of a bigger length, that would produce an int like the one below:

45434-59685-45565-59685-56923-49565

I would prefer not to use a bigint class, and I have seen code simular to the above that does this, producing a string that represents a larger int, or at least a string of digits like the above, and, of course, a method to reverse the string into a byte array.

Thank you very much,
Kind Regards,
Stephen

推荐答案

为什么不将上面按钮中的代码封装为方法:

Why don''t you encapsulate the code in the button above into a method:

int GetIntFromByteArr(byte[] rawLicense)
{
  int  i = rawLicense[3] + (rawLicense[2] << 8) + (rawLicense[1] << 16)     + (rawLicense[0] << 24);
  return j;
}

并多次调用?

And call it multiple times?

int el1 = GetIntFromByteArr(new byte[]{0x00, 0x00, 0xA3, 0xE1});
int el2 = GetIntFromByteArr(new byte[]{0x00, 0x10, 0x00, 0xB1});
// ....

然后，您只需串联这些整数即可.

您当然可以通过将数组传递给方法和表示要转换的开始/结束位置的索引的方式来对此进行优化.

希望这会有所帮助，
Alex

Then you would just concatenate those integers.

You could of course optimize this in such a way that you are passing the array to the method and the indexes that represent the start/end position to convert.

Hope this helps,
Alex

好的，让我们再试一次:)

尝试这段代码:

All right, let''s try this again :)

Try this piece of code:

class Program
{
    static void Main(string[] args)
    {
        byte[] rawLicense = { 0x00, 0x00, 0xA3, 0xE1, 0xA5, 0xB6 };

        int shift = 0;
        long result = 0;
        for (int i = rawLicense.Length - 1; i >=0 ; i--, shift += 8)
        {
            long value = (long)rawLicense[i] << shift;
            result += value;
        }

        Console.WriteLine(result);
    }
}

您仍然受制于长时间不会溢出的情况，但请注意这一点.

You are still limited to what a long can hold without overflowing though so be careful about that.

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