How do I concatenate 2 bytes?

Question

I have 2 bytes:

byte b1 = 0x5a;  
byte b2 = 0x25;

How do I get 0x5a25 ?

Solution 1

It can be done using bitwise operators '<<' and '|'

public int Combine(byte b1, byte b2)
{
    int combined = b1 << 8 | b2;
    return combined;
}

Usage example:

[Test]
public void Test()
{
    byte b1 = 0x5a;
    byte b2 = 0x25;
    var combine = Combine(b1, b2);
    Assert.That(combine, Is.EqualTo(0x5a25));
}

Solution 2

Using bit operators: (b1 << 8) | b2 or just as effective (b1 << 8) + b2

Solution 3

A more explicit solution (also one that might be easier to understand and extend to byte to int i.e.):

using System.Runtime.InteropServices;
[StructLayout(LayoutKind.Explicit)]
struct Byte2Short {
  [FieldOffset(0)]
  public byte lowerByte;
  [FieldOffset(1)]
  public byte higherByte;
  [FieldOffset(0)]
  public short Short;
}

Usage:

var result = (new Byte2Short(){lowerByte = b1, higherByte = b2}).Short;

This lets the compiler do all the bit-fiddling and since Byte2Short is a struct, not a class, the new does not even allocate a new heap object ;)

Solution 4

byte b1 = 0x5a;
byte b2 = 0x25;

Int16 x=0;

x= b1;
x= x << 8;
x +=b2;

View more solutions

Comments

[futureelite7]

(b1 << 8) + b2 will do the job faster.

[paxdiablo]

@futureelite7, that's unlikely to be the case with modern compilers - in any case, you should have used (b1<<8)|b2 :-)

[paxdiablo]

I'm pretty certain you didn't need to wrap that in a function :-) But you're right, so +1.

[danatel]

Beware of en.wikipedia.org/wiki/Endianness. It really matters what the bytes will be used for. Alhough most c# work is done on the Intel platform, the bytes may be send out as a part of network protocol where endiannes matters.

[AK_]

oh the OR is nice :-) but it is still short... +1 for the OR :)

[danielschemmel]

maybe - but if you want to express that something can be understood more than one way on a bit-wise level it is the tool of choice. Personally I prefer the shift+or method, but it becomes quite cumbersome when your target is i.e. a uint64, while the union still remains very readable in that case.

[jason]

@futureelite7: I note that b1*256+b2 has one fewer character than (b1<<8)+b2 so actually b1*256+b2 is faster. :-)

[Jeppe Stig Nielsen]

There's also BitConverter.ToUInt16(new[] { b2, b1, }, 0).

[scegg]

Func<byte, byte, int> combine; if (BitConverter.IsLittleEndian) { combine = new Func<byte, byte, int>((i, j) => j << 8 | i); } else { combine = new Func<byte, byte, int>((i, j) => i << 8 | j); } int result = combine(b1, b2);

[siggi_pop]

This is ridiculous. Both are practically the same. running a trillion statments using: c = (a << 8) + b stopping timer @ 00:00:00.7638591 running a trillion statments using: c = a << 8 | b stopping timer @ 00:00:00.7634362 running a trillion statments using: c = a * 256 + b stopping timer @ 00:00:00.8592438 a: 1 00000000000000000000000000000001 b: 15 00000000000000000000000000001111 c: 271 00000000000000000000000100001111