Is there a built in function to expand a byte[] leaving zeros at the beginning?

Question

I've got this code

BitConverter.GetBytes(width).CopyTo(resultBytes, 0);

If the width is 12 it returns one byte not 4, is there a built in function to resize the array leaving 0's at the beginning to output [0, 0, 0, 12] instead of [12].

you should get 4 bytes then. have you tried it without the `CopyTo`? — thumbmunkeys, Jul 23 '15 at 18:34
If 12 was the 4th byte, that wouldn't equal `12`. It would equal `49152`. Are you trying to switch Endianness? — crush, Jul 23 '15 at 18:38
Well, yes, but I'm writing the result to a file and then my other program reads it wrong like that. Some weird big number(incorrect probably that) — trinalbadger587, Jul 23 '15 at 18:40
You probably should look into [Endianness](https://en.wikipedia.org/wiki/Endianness). It's likely written to the file in Little Endian. — crush, Jul 23 '15 at 18:41
how do you read and reconstruct the written value in the other program? — thumbmunkeys, Jul 23 '15 at 18:44
Have you tried using the [BinaryReader](https://msdn.microsoft.com/en-us/library/system.io.binaryreader(v=vs.110).aspx) class instead? — crush, Jul 23 '15 at 18:46
Is it written as `0C 00 00 00` or `00 00 00 0C` in your file? (assuming you are looking at it in hex) — crush, Jul 23 '15 at 18:51

score 2 · Answer 1 · answered Jul 23 '15 at 18:44

What is the type of width? Bit converter just converts the type to an array of appropriate size. If you say

long  x = 1 ;
int   y = 2 ;
short z = 3 ;

byte[] x_bytes = BitConverter.GetBytes(x) ;
byte[] y_bytes = BitConverter.GetBytes(y) ;
byte[] z_bytes = BitConverter.GetBytes(z) ;

You'll get back 8-, 4- and 2-byte arrays, respectively. You can cast to the desired type:

byte[] bytes = BitConverter.GetBytes( (int) x ) ;

And if you say something like

byte[] bytes = BitConverter.GetBytes(1) ;

You'll get back an array of 4 bytes: the type of an unsuffixed integer literal is the smallest type in which will fit, in order of preference: int, uint, long, ulong. If the literal is suffixed, it will be the type specified by the suffix (e.g., 1L will give you an 8-byte long).

If you are converting an expression, such as:

byte[] bytes = BitConverter.GetBytes( ((3*x + 2&y + z) << 3 ) & 5L ) ;

What gets converted is, of course, the type produced by evaluating the expression.

thumbmunkeys · Answer 2 · 2015-07-23T18:47:45.810

1

you need to cast width to int in order to get 4 bytes, because the result of GetBytes() is dependent on the type passed in:

    BitConverter.GetBytes((int)width).CopyTo(resultBytes, 0);

edited Jul 23 '15 at 18:47

answered Jul 23 '15 at 18:33

thumbmunkeys

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score 0 · Answer 3 · answered Jul 23 '15 at 18:41

0

Maybe it's the simplest solution ever, but what about Array.Reverse:

BitConverter.GetBytes(4).CopyTo(resultBytes, 0); // [4, 0, 0, 0]
Array.Reverse(resultBytes); // [0, 0, 0, 4]

answered Jul 23 '15 at 18:41

Matías Fidemraizer

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What about if it was a big number though? It works in this case(your right). – trinalbadger587 Jul 23 '15 at 18:42
@user3208848 and why it shouldn't work with big numbers? 100k zeroes and the number. – Matías Fidemraizer Jul 23 '15 at 18:42

Is there a built in function to expand a byte[] leaving zeros at the beginning?

3 Answers3