为什么传递临时对象时不调用复制构造函数

Question

class MyClass
{
  public:    
  int a;
  MyClass(int r): a(r) {}

  MyClass(const MyClass& ref)
  {
      cout << "Copy Constructor\n";
      a= ref.a;
  }
};
int main()
{
    MyClass obj(5);
    MyClass obj1(MyClass(5));  //Case 1
    MyClass obj2(obj);        //Case 2
    return 0;
}

为什么copy constructor是在Case 2中调用的，而不是在Case 1中。在情况 1 中，临时对象作为参数传递。

Answer 1

在 MyClass obj1(MyClass(5)); 中，编译器省略了临时对象 MyClass(5)，因为它是允许的。

特别是，C++ 标准 2014 §12.8 第 31 段定义了可以执行 复制省略 的情况：

When certain criteria are met, an implementation is allowed to omit the copy/move construction of a class object, even if the constructor selected for the copy/move operation and/or the destructor for the object have side effects... This elision of copy/move operations, called copy elision, is permitted in the following circumstances (which may be combined to eliminate multiple copies):

when a temporary class object that has not been bound to a reference (12.2) would be copied/moved to a class object with the same cv-unqualified type, the copy/move operation can be omitted by constructing the temporary object directly into the target of the omitted copy/move.

为什么传递临时对象时不调用复制构造函数

Why copy constructor is not called when pass temporary object

c++

copy-constructor

c++11