In version 1, the return type is "a_t", so a class construction is
required there (the caller will then destruct the returned object).
Construction and destruction can have side effects, so the compiler
would not drop them. For the following code,
template<typename a_t, typename b_t>
inline a_t& append (a_t & a, b_t const& b) {
a.insert (a.end(), b.begin(), b.end());
return a;
}
it does not require a construction, and would be as fast as version 2.
On 9/26/07, Neal Becker <[EMAIL PROTECTED]> wrote:
> gcc version 4.1.2 20070502 (Red Hat 4.1.2-12)
> I noticed the following code
>
> === version 1:
> template<typename a_t, typename b_t>
> inline a_t append (a_t & a, b_t const& b) {
> a.insert (a.end(), b.begin(), b.end());
> return a;
> }
>
> === version 2:
> template<typename a_t, typename b_t>
> inline void append (a_t & a, b_t const& b) {
> a.insert (a.end(), b.begin(), b.end());
> }
>
> When instantiated for a_t, b_t std::list<T>. When called by code that _did
> not use the return value_, I had assumed that since the returned value is
> not used, the 2 versions would be equivalent. Instead, (compiling
> with -O3), version 2 runs very fast, but version 1 is extremely slow. Is
> it really necessary to construct the returned value even when it is seen
> that it is not used?
>
>
