[LLVMdev] Memory Allocation Optimized away with new by not with ::operator new

Benjamin Kramer benny.kra at gmail.com
Mon May 4 04:45:31 PDT 2015


On Mon, May 4, 2015 at 12:37 PM, François Fayard
<fayard.francois at icloud.com> wrote:
> Hi,
>
> I’ve made my own version of std::vector which is called il::Vector. Due to
> http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2013/n3664.html, LLVM can
> optimise away memory allocation. Therefore, the following code optimise away
> all memory allocation for w resulting in a single allocation during the
> whole program (for v).
>
> When using my own vector implementation, I realised that the allocation were
> not optimized away because I was using ::operator new. When I’ve switched
> back to new, the optimisation came back.
>
> Is it expected or a bug from LLVM?

Sadly this is a feature. The C++ standard has been unclear
historically about whether removing pairs of new/delete. The problem
is that the user may override them so this is an observable change,
but some compilers (LLVM) removed them anyways. As you said C++14
changed the wording so removing new/delete expression pairs is now
explicitly legal. Calls to ::operator new and delete aren't new/delete
expressions though, so it should behave as expected when overridden.

It looks like a glitch in the standard, but if I remember correctly it
was done intentionally.

- Ben

>
> François
>
> =====
>
> #include <iostream>
> #include <vector>
>
>
> std::vector<double> f_val(std::size_t i, std::size_t n) {
>     auto v = std::vector<double>(n);
>     for (std::size_t k = 0; k < v.size(); ++k) {
>         v[k] = static_cast<double>(i);
>     }
>     return v;
> }
>
> int main (int argc, char const *argv[])
> {
>     const auto n = std::size_t{10};
>     const auto nb_loops = std::size_t{300000000};
>
>     auto v = std::vector<double>( n, 0.0 );
>     for (std::size_t i = 0; i < nb_loops; ++i) {
>         auto w = f_val(i, n);
>         for (std::size_t k = 0; k < v.size(); ++k) {
>             v[k] += w[k];
>         }
>     }
>     std::cout << v[0] << " " << v[n - 1] << std::endl;
>
>     return 0;
> }
>
>
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