Is there a way to disable copy elision in c++ compiler

问题

In c++98, the following program is expected to call the copy constructor.

#include <iostream>

using namespace std;
class A
{
  public:
    A() { cout << "default" ; }

    A(int i) { cout << "int" ; }


    A(const A& a) { cout << "copy"; }
};

int main ()
{
   A a1;
   A a2(0);
   A a3 = 0;

  return 0;
}

That is evident if you declare the copy constructor explicit in above case (the compiler errors out). But I don't I see the output of copy constructor when it is not declared as explicit. I guess that is because of copy elision. Is there any way to disable copy elision or does the standard mandates it?

回答1:

Pre C++ 17

A a3 = 0;

will call copy constructor unless copy is elided. Pass -fno-elide-constructors flag

from C++17, copy elision is guaranteed. So you will not see copy constructor getting called.

回答2:

You have wrong understanding what the copy elision is. Please refer to this question for more info.

In this particular case, if you define the constructor explicit, it will cause an error because A a3 = 0; on this line the compiler created a object using 0.

来源：https://stackoverflow.com/questions/50834215/is-there-a-way-to-disable-copy-elision-in-c-compiler