A question about 'Return Value'

meili100 · #1 03-Jun-2007, 22:23

I have the following code just for fun

CPP / C++ / C Code:

class Base{
public:
	Base(){cout<<"Constructor"<<endl;}
	Base(const Base &obj){cout<<"Copy Constructor"<<endl;}
	Base & operator=(const Base& obj){
		cout<<"Operator="<<endl;
		return *this;
	}
};

Base foo(){
	Base x;
	return x;
}

void main(){
	Base tmp = foo();
}

=================
Why the outputs are only

Constructor
Copy Constructor

But I think the outputs should be:
Constructor
Copy Constructor // from return x
Copy Constructor // from Base tmp = foo();

Peter_APIIT · #2 04-Jun-2007, 03:41

I not familiar with this pointer, copy constructor and assignment.

I guess the x value and the function should inside the class.

I also want to know what is this pointer. The function of this pointer is returned the object which belongs to the function. (Just Guess)

Thanks for your help.

Your help is greatly appreciated by me and others.

davekw7x · #3 04-Jun-2007, 08:30

Quote:

Originally Posted by meili100

Why the outputs are only

Constructor
Copy Constructor

To get some insight, I respectfully suggest that you can investigate things that puzzle you rather than trying to get someone else to explain it.

For example try the following:

CPP / C++ / C Code:

#include <iostream>

using namespace std;

class Base {
  public:
    Base() {
        cout << "Constructor" << endl;
    }
    Base(const Base & obj) {
        cout << "Copy Constructor" << endl;
    }
    Base & operator=(const Base & obj) 
    {
        cout << "Operator=" << endl;
        return *this;
    }
    ~Base() {
        cout << "Destructor" << endl;
    }
};

Base foo()
{
    Base x;
    cout << "foo() is returning" << endl;
    return x;
}

int main()
{
    cout << "Before tmp=foo() declaration" << endl << endl;
    Base tmp = foo();
    cout << endl << "After tmp=foo() declaration" << endl;

    return 0;
}

Output with program compiled by Borland and Microsoft compilers on my Windows XP system:

Code:

Before tmp=foo() declaratio

Constructor
foo() is returning
Copy Constructor
Destructor

After tmp=foo() declaration
Destructor

The initialization statement for tmp apparently calls foo(), which calls the constructor to create a temporary object. Then it calls the copy instructor with the value returned by foo() and destructs the temporary object.

An interesting thing (to me at least) is what happens when the program is compiled with GNU g++. The output is

Code:

Before tmp=foo() declaration

Constructor
foo() is returning

After tmp=foo() declaration
Destructor

g++ is apparently smart enough to realize that it's silly to call a constructor for a temporary object and then use its value to call another constructor and then destruct the temporary object. So, its code just does everything in one step! See footnote.

Want more? Then try the following:

CPP / C++ / C Code:

#include <iostream>

using namespace std;

class Base {
  public:
    Base() {
        cout << "Constructor" << endl;
    }
    Base(const Base & obj) {
        cout << "Copy Constructor" << endl;
    }
    Base & operator=(const Base & obj) 
    {
        cout << "Operator=" << endl;
        return *this;
    }
    ~Base() {
        cout << "Destructor" << endl;
    }
};

Base foo()
{
    Base x;
    cout << "foo() is returning" << endl;
    return x;
}

int main()
{
    cout << "Before tmp2(foo()) declaration" << endl << endl;
    Base tmp2(foo());
    cout << endl << "After tmp2(foo()) declaration" << endl;

    return 0;
}

Now, believe it or not, I get three different results with the three compilers

Borland:

Code:

Before tmp2(foo()) declaration

Constructor
foo() is returning
Copy Constructor
Destructor
Copy Constructor
Destructor

After tmp2(foo()) declaration
Destructor

Microsoft:

Code:

Before tmp2(foo()) declaration

Constructor
foo() is returning
Copy Constructor
Destructor

After tmp2(foo()) declaration
Destructor

GNU:

Code:

Before tmp2(foo()) declaration

Constructor
foo() is returning

After tmp2(foo()) declaration
Destructor

It appears to me that Microsoft and GNU treat the two examples the same, but Borland actually does more work for the second one than for the first. (If I had been guessing, I would have guessed that, if there were any differences between the first example and the second, the second would have been more efficient. Don't ask me why I would have guessed this, and, in particular, don't bet on anything that I or anyone else on this beautiful planet would guess.)

Bottom line: You can investigate and see what your compiler is actually doing with your code. But, and this is important, don't draw general conclusions based on the behavior of a single version of a single compiler. The C++ standard doesn't say what kinds of code must be generated from the source given in the two examples; it only says what the result must be.

So, how the heck can you say "I expected..." when each compiler writer is free to do whatever the heck he/she wants to as long as the final result is the required one? Whatever may seem obvious to you (or me) may or may not be what the compiler actually does.

Regards,

Dave

Footnote: The process of eliminating unnecessary storing and retrieval of objects can be eliminated by even the simplest optimization steps in what we used to call a "peephole" optimizer. Just keep track of whether an object is loaded again after storing it without any destructive steps in between times. Then, simply, don't store it.

I learned this from studying the "Small C" compiler published about a thousand years ago (that's how long ago it seems to me). This was a very primitive compiler for 8088/8086 machines when the world was young and CP/M was the Big Thing and Microsoft compilers cost USD 700 and Borland hadn't hit the scene with USD 50 Turbo Pascal (for CP/M and DOS) and, later, Turbo C (alas, no CP/M---that's when I built my first 8088 system for DOS and abandoned CP/M forever).

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A question about 'Return Value'

Re: A question about 'Return Value'

Re: A question about 'Return Value'