C#中的PHP包（'N'，数字）和二进制连接

Question

How do I convert the following code to C#?

return pack('N', $number1) . pack('N', $number2);

I've managed to convert the rest of the function, but I have no idea how the pack('N', number) works, nor do I know what the . -operator does when applied to binary variables in PHP.

Answer 1

You use BitConverter to get the byte representation of the integer, but than you have to flip it because on most machines it is little-endian. Since I don't know whether you're packing these into a MemoryStream or byte[] (though you should), I'll just show exactly that.

int myInt = 1234;
byte[] num1 = BitConverter.GetBytes( myInt );
if ( BitConverter.IsLittleEndian ) {
    Array.Reverse( num1 );
}

And then you can transfer that to your buffer, which for C# might be a byte[] . Here's how you might do 2 integers:

int myInt1 = 1234;
int myInt2 = 5678;
byte[] temp1 = BitConverter.GetBytes( myInt1 );
byte[] temp2 = BitConverter.GetBytes( myInt2 );

if ( BitConverter.IsLittleEndian ) {
    Array.Reverse( temp1 );
    Array.Reverse( temp2 );
}

byte[] buffer = new byte[ temp1.Length + temp2.Length ];
Array.Copy( temp1, 0, buffer, 0, temp1.Length );
Array.Copy( temp2, 0, buffer, temp1.Length, temp2.Length );
return buffer;

Answer 2

pack('N', $number1) returns the integer $number1 as a 4-byte binary string in big endian byte order.

The "." operator concatenates strings.