Re: [boost] [integer] Promotion of endian library code from vault

From: Neil Mayhew (neil_mayhew_at_[hidden])
Date: 2008-05-29 15:54:27

On 28/05/08 05:50 PM Neil Mayhew wrote:
> Perhaps the simplest and best solution is therefore:
>
>     class endian< ... >
>       : cover_operators< ... >
>     {
>       public:
> #if defined(CXX_0X) || !defined(ENDIANS_IN_UNIONS)
>         endian() {}
>         endian(T val) { ... }
> #endif

I just discovered that an operator=(T) is needed as well:

endian& operator=(T i) { detail::store_big_endian<T, n_bits/8>(bytes, i); }

Without this, and without the endian(T val) constructor, a lot of things 
just don't work - for example, the binary arithmetic operators.

This makes me think that there should have been an operator= all along. 
For a start, this is just good practice: anywhere there's a constructor 
initializing from a particular type, there should usually also be an 
assignment operator taking the same type.

Second, I think the generated code for binary operators must have been 
suboptimal, since it seems that the computed result (a native integer) 
of adding two endians was being used to construct a temporary endian 
which was then copy-assigned into the actual result. At least, that's my 
interpretation of the compilation errors I was getting before I put 
operator= in. To test this, take out the constructors and do:

big32_t a, b;
nt32_t i;
i = a + b;

operator+(endian, endian) is implemented using 
cover_operators::operator+=, which is defined as x = +x + y, hence uses 
assignment of the result of adding two native integers converted from 
the respective endians. If you're assigning the result to a native type 
anyway, then returning an endian is inefficient, even with operator= 
defined. I'm not sure how to fix this, without abandoning the use of 
boost::operators. In fact, I am beginning to wonder whether the 
traditional approach to binary operators, which returns the same type as 
its two arguments, and implements by calling +=, is not actually 
appropriate for endian.

--Neil