convert ubyte[k..k + 1] to int

[Regan Heath]

On Wed, 16 May 2012 15:24:33 +0100, ref2401 <refactor24 at gmail.com> wrote:

> i have an array of ubytes. how can i convert two adjacent ubytes from  
> the array to an integer?
>
> pseudocode example:
> ubyte[5] array = createArray();
> int value = array[2..3];
>
> is there any 'memcpy' method or something else to do this?

You don't need to "copy" the data, just tell the compiler to "pretend"  
it's a short (in this case, for 2 bytes) then copy the value/assign to an  
int. e.g.

import std.stdio;

void main()
{
	ubyte[5] array = [ 0xFF, 0xFF, 0x01, 0x00, 0xFF ];
	int value = *cast(short*)array[2..3].ptr;
	writefln("Result = %s", value);
}

The line:
   int value = *cast(short*)array[2..3].ptr;

1. slices 2 bytes from the array.
2. obtains the ptr to them
3. casts the ptr to short*
4. copies the value pointed at by the short* ptr to an int

You may need to worry about little/big endian issues, see:
http://en.wikipedia.org/wiki/Endianness

The above code outputs "Result = 1" on my little-endian x86 desktop  
machine but would output "Result = 256" on a big-endian machine.

R

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