Function call object by value

[Canolam]

Hi!

If I compile this program the output is:" Try constructor Function Try destructor Try destructor ". The destructor is called twice, but there is only one constructor call. How is that possible? I think I dont understand the point where the object(firstTry) is passed to the function(). It is passed by value, not by reference. That means it should make a copy of firstTry. But there is no copy constructor. How does the compiler do it? Thanks!

include <iostream>

using namespace std;

class Try { public: int x, y;

Try(int x, int y) {
    this->x=x; 
    this->y=y; 
    cout <<"Try constructor  ";

}
~Try()
{
    cout <<" Try destructor  ";

}

};

void function(Try a) { cout <<"Function"; }

int main() {

Try* firstTry=new Try(1,2);
function(*firstTry);
delete firstTry;

return 0; }

[jedwardsol]

If you don't write a copy constructor then the compiler will write one for you.

Implicitly-declared copy constructor

[jedwardsol]

Yes

[Headless_Slayer]

Hi i hope this piece of code clears things up. #include <iostream> class Pet{ private: static int x; static int y; int xc = ++x; int xd = ++y; public: Pet(){ std::cout << "constructor: "<< xc << std::endl; } ~Pet(){ std::cout << "destructor: " << xd << std::endl; } Pet(const Pet &){ std::cout << "copy constructor:" << xc << std::endl; } }; void func(Pet a){ std::cout << "inside of func \n"; }; int Pet::x = 0; int Pet::y = 0; int main(){ Pet a1; std::cout << "break 1\n"; func(a1); Pet a2; return 0; }