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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T11:35:50+00:00 2026-05-13T11:35:50+00:00

I’m not a veteran in socket programming, so while analyzing code I found in

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I’m not a veteran in socket programming, so while analyzing code I found in a database API I came across this code

    public static void WriteInt(int i, NetworkStream bufOutputStream) 
    {
        byte[] buffer = new byte[IntSize];
        WriteInt(i, buffer, 0);
        bufOutputStream.Write(buffer, 0, buffer.Length);
    }

    public static void WriteInt(int i, byte[] byte_array, int pos)
    {

        byte_array[pos] =(byte)( 0xff & (i >> 24)); byte_array[pos+1] = (byte)(0xff & (i >> 16)); byte_array[pos+2] = (byte)(0xff & (i >> 8)); byte_array[pos+3] = (byte)(0xff & i);
    }

I understand the bit-shifts what I don’t understand is how the ‘buffer’ var keeps getting a value when no ref is in the args or no return is made. the bitshifts are somehow editing the actual value of buffer?

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  1. Editorial Team
    Editorial Team
    2026-05-13T11:35:51+00:00Added an answer on May 13, 2026 at 11:35 am

    Your confusion is a very common one. The essential point is realising that “reference types” and “passing by refrence” (ref keyboard) are totally independent. In this specific case, since byte[] is a reference type (as are all arrays), it means the object is not copied when you pass it around, hence you are always referring to the same object.

    I strongly recommend that you read Jon Skeet’s excellent article on Parameter passing in C#, and all should become clear…

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