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

[mats petersson]

If the compiler can't say for sure that your operator new is identical (has
no side-effects beyond allocating memory), it would be correct for the
compiler to NOT optimise it out - as it can't determine that it has no
other effect that your program rely on in some way.

--
Mats

On 4 May 2015 at 11:37, 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?
>
> 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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