ubytes to ulong problem

[John Colvin]

On Sunday, 22 December 2013 at 03:57:38 UTC, Ali Çehreli wrote:
> 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.

It's just an unaligned load. If your target cpu architecture 
can't do unaligned loads then you're either using something very 
small or very old.