C# Bitwise Operators

Question

Could someone please explain what the following code does.

private int ReadInt32(byte[] _il, ref int position)
{
    return (((il[position++] | (il[position++] << 8)) | (il[position++] << 0x10)) | (il[position++] << 0x18));
}

I'm not sure I understand how the bitwise operators in this method work, could some please break it down for me?

msdn.microsoft.com/en-us/library/6a71f45d%28v=vs.110%29.aspx ? — Soner Gönül
– Soner Gönül, Commented Apr 10, 2013 at 10:08
I'd change it to return (((il[position] | (il[position+1] << 8)) | (il[position+2] << 0x10)) (not answering the question, someone has already done that, it's just that the position++ stuff is just hard to read and unnecessary) — Matthew Watson
– Matthew Watson, Commented Apr 10, 2013 at 10:10
@MatthewWatson, I agree that it makes far more sense, but it seems that it is used in a loop over a buffer from an external function, as the position argument is ref int so the callee likely depends on index being updated. still, I would do it like you and just add pos += 4 at the end — Zdeslav Vojkovic
– Zdeslav Vojkovic, Commented Apr 10, 2013 at 10:18
@ZdeslavVojkovic Ah yes, quite right. Probably position += sizeof(int) would be even more expressive. — Matthew Watson
– Matthew Watson, Commented Apr 10, 2013 at 10:31
@MatthewWatson, right, I completely forgot that C# also has sizeof operator. — Zdeslav Vojkovic
– Zdeslav Vojkovic, Commented Apr 10, 2013 at 11:49

Zdeslav Vojkovic · Accepted Answer · 2013-04-10 10:14:02Z

8

The integer is given as a byte array. Then each byte is shifted left 0/8/16/24 places and these values are summed to get the integer value.

This is an Int32 in hexadecimal format:

0x10203040

It is represented as following byte array (little endian architecture, so bytes are in reverse order):

[0x40, 0x30, 0x20, 0x10]

In order to build the integer back from the array, each element is shifted i.e. following logic is performed:

a = 0x40        = 0x00000040
b = 0x30 << 8   = 0x00003000
c = 0x20 << 16  = 0x00200000
d = 0x10 << 24  = 0x10000000

then these values are OR'ed together:

int result = a | b | c | d;

this gives:

0x00000040 |
0x00003000 | 
0x00200000 |
0x10000000 |
------------------
0x10203040

edited Apr 10, 2013 at 10:14

answered Apr 10, 2013 at 10:08

Zdeslav Vojkovic

14.7k37 silver badges45 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

jjdev80 Over a year ago

Thank you very much, it makes much more sense now! :)

Phil · Accepted Answer · 2013-04-10 10:10:55Z

3

Think of it like this:

var i1 = il[position];
var i2 = il[position + 1] << 8;  (<< 8 is equivalent to * 256)
var i3 = il[position + 2] << 16; 
var i4 = il[position + 3] << 24;

position = position + 4;

return i1 | i2 | i3 | i4;

answered Apr 10, 2013 at 10:10

Phil

43.2k9 gold badges102 silver badges102 bronze badges

Collectives™ on Stack Overflow

C# Bitwise Operators

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related