Why is the construction of std::optional more expensive than a std::pair?

Question

Consider these two approaches that can represent an \"optional int\":

using std_optional_int = std::optional;
using my_optional_int =         


        

Answer 1

In Calling conventions for different C++ compilers and operating systems by Agner Fog it says that a copy constructor or destructor prevents from returning a structure in registers. This explains why optional is not returned in registers.

There has to be something else preventing the compiler from doing store merging (merges contiguous stores of immediate values narrower than a word into fewer wider stores to reduce the number of instructions)... Update: gcc bug 82434 - -fstore-merging does not work reliably.