Copy Constructor in C++

I have this code

#include <iostream>
using namespace std;

class Test{
   public:
      int a;

      Test(int i=0):a(i){}
      ~Test(){
         cout << a << endl;
      }

      Test(const Test &){
         cout << "copy" << endl;
      }

      void operator=(const Test &){
         cout << "=" << endl;
      }

      Test operator+(Te开发者_运维知识库st& p){
         Test res(a+p.a);
         return res;
      }
};

int main (int argc, char const *argv[]){
   Test t1(10), t2(20);
   Test t3=t1+t2;
   return 0;
}

Output:

30
20
10

Why isn't the copy constructor called here?

This is a special case called Return Value Optimization in which the compiler is allowed to optimize away temporaries.

I assume you're wondering about the line Test t3=t1+t2;

The compiler is allowed to optimize the copy construction away. See http://www.gotw.ca/gotw/001.htm.

As the others said, its just optimizing the call to the copy constructor, here what happens if you disable those optimizations.

barricada ~$ g++ -o test test.cpp -O0 -fno-elide-constructors
barricada ~$ ./test
copy
30
copy
134515065
-1217015820
20
10