On Wed, May 16, 2012 at 11:03 AM, Regan Heath <regan@netmail.co.nz> wrote:
On Wed, 16 May 2012 15:24:33 +0100, ref2401 <refactor24@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


Unfortunately, this is undefined behavior because you're breaking alignment rules. On x86, this will just cause a slow load from memory. On ARM, this will either crash your program with a bus error on newer hardware or give you a gibberish value on ARMv6 and older.
Declaring a short, getting a pointer to it, and casting that pointer to a ubyte* to copy into it is fine, but casting a ubyte* to a short* will cause a 2-byte load from a 1-byte aligned address, which leads down the yellow brick road to pain.