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