有没有办法在C ++编译器中禁用复制省略

问题描述

在c ++ 98中，预期以下程序将调用复制构造函数。

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;
}

如果在上述情况下显式声明了复制构造函数，则显而易见（编译器错误）。但是当未将其声明为显式时，我看不到副本构造函数的输出。我想那是因为复制省略。有什么方法可以禁用复制省略或标准是否规定了它？

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?

推荐答案

Pre C ++ 17

A a3 = 0;

将副本构造函数。从 C ++ 17 传递 -fno-elide-constructors 标志

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

c $ c>，可以保证复制省略。因此，您将复制构造函数被调用。

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

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