Re: ubytes to ulong problem

[Ali Çehreli]

On 12/21/2013 05:44 PM, Charles Hixson wrote:

On 12/21/2013 03:52 PM, Ali Çehreli wrote:

On 12/21/2013 03:13 PM, John Colvin wrote:

> Ideally the compiler will optimise your version to be fast, but you may
> find you get better performance by doing the bit manipulations
eplicitly:

Assuming that the program needs to support only big endian and little
endian systems (i.e. excluding systems where no D compiler exists :)),
the following is less wordy and should be equally fast:

import std.bitmanip;
import std.system;

ulong ubytesToUlong(ubyte[] block, size_t n = 0)
in
{
    assert (n >= 0);
    assert (n + 8 <= block.length);
}
body
{
    ulong value = *cast(ulong*)(block.ptr + n);

    if (std.system.endian == Endian.littleEndian) {
        return *cast(ulong*)(value.nativeToBigEndian.ptr);

    } else {
        return value;
    }
}

Ali

Will that work even when the alignment is to odd bytes?  Because that's
the case I was really worried about.  The ubyte array is a packed
mixture of types, some of which are isolated bytes.

No, it is not guaranteed to work unless the alignment is right.

How about this, then: :)

import std.array;

ulong ubytesToUlong(ubyte[] block, size_t n = 0)
in
{
    assert (n >= 0);
    assert (n + 8 <= block.length);
}
body
{
    ulong value = block.front;
    block.popFront();

    foreach (ub; block) {
        value <<= 8;
        value |= ub;
    }

    return value;
}

Ali