Re: [boost] [boost::endian] Request for comments/interest

Subject: Re: [boost] [boost::endian] Request for comments/interest
From: Terry Golubiewski (tjgolubi_at_[hidden])
Date: 2010-06-03 10:46:30

Robert Stewart wrote:
>> I could make it more helpful; but then it would be called
>> swap_in_place and would be implemented just like Tomas'!
>
> Yours would still be in terms of the endian types.

What inteface would you suggest?

>> >> --- typed based ---
>> >>   for (int trial=0; trial != 1000; ++trial) {
>> >>     {
>> >>       ifstream input("array.dat", ios::binary);
>> >>       input.read(reinterpret_cast<char*>(&tmp_array),
>> >> sizeof(tmp_array));
>> >>       interface::copy(tmp_array.begin(), tmp_array.end(),
>> >> array2.begin());
>> >
>> > Does this actually swap anything?  Doesn't this just copy
>> > the data to unswapped objects that *would* swap on access?
>> > That's hardly a fair comparison.
>>
>> array2 is just a plain array<uint32_t, SIZE>.  tmp_array is an
>> array<endian<big, uint32_t>, SIZE>.  So the conversion/copy
>> happens in interface::copy.
>
> I don't think that's right.  The copy is done in interface::copy(), but 
> the conversion (swap) doesn't occur until you read from the endian 
> objects, right?

Not right. The conversion happens inside of interface::copy which is 
performing a swap-in-place in this example.  After the call to 
interface::copy, array2 is in native-endian format.
Both tests convert the disk file to native-endian format in memory, so no 
other comparison is really necessary after the conversion, other than the 
memcmp to verify that the conversion worked.

>  My point is that you need to walk the array summing the values, or 
> something, to read each value at least once.
>  Otherwise, your code merely copies the data as read from disk to the 
> endian objects but doesn't swap any bytes, and yet you compare that to 
> function-based code that does swap.

I think the memcmp does that.

terry