constructor call return

Code:

---

class test
{
public:
test()

};

void main()
{

test t = test();  // what is returned here?

}

---

Since constructor do not have return type. Then what is returned in t?

When you write

Code:

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test t = test();

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two things are happening:

1) test() creates a temporary of type test.
2) The test t = part creates a second instance of type test. This instance t is created via the class's copy constructor (test t = test(); is basically the same as test t( test() );). The argument for t's copy constructor is the temporary created in step 1.


EDIT: Note that in your case, just test t; would be sufficient. This would create instance t via test's default constructor.

EDIT #2: In C++ and C, it should always be int main(), not void main().

Quote:

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is basically the same as test t( test() );

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I didn't understand this point.

In my opnion it should be

test t(t); // This syntax is incorrect. because you cannot pass same object in constructor; But I thought default copy constructor would require this.

Please clarify.

A copy constructor requires as its argument another instance of the same type. The expression test() generates a temporary that satisfies that requirement.

Trying playing around with this program a bit.

Code:

---

#include <iostream>

class test
{
public:
    test()
    {
        std::cout << "Creating test instance at 0x" << std::hex << this << std::dec << " via default c'tor.\n";
    }
   
    ~test()
    {
        std::cout << "Destroying test instance at 0x" << std::hex << this << std::dec << ".\n";
    }

    test( const test& t )
    {
        std::cout << "Creating test instance at 0x" << std::hex << this << std::dec << " via copy c'tor. Address of source instance is 0x"
            << std::hex << &t << std::dec << ".\n";
    }

    test& operator = ( const test& t )
    {
        std::cout << "Copy assigning to test instance at 0x" << std::hex << this << std::dec << ". Address of source instance is 0x"
            << std::hex << &t << std::dec << ".\n";

        return *this;
    }
};


int main()
{
    test t = test();
    t = test();
}

---

Compile without optimizations.


P.S. The program might actually not use the copy constructor at all. This is because your compiler is allowed to remove the copy if it finds that it is unnecessary. This is called copy elision: http://en.wikipedia.org/wiki/Copy_elision

Output:

Code:

---

Creating test instance at 0x002AF7FB via default c'tor.
Destroying test instance at 0x002AF7FB.
Press any key to continue . . .

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I don't know why copy constructor is not called?

Quote:

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Originally Posted by freiza

I don't know why copy constructor is not called?

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You may have missed antred's postscript:

Quote:

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Originally Posted by antred

P.S. The program might actually not use the copy constructor at all. This is because your compiler is allowed to remove the copy if it finds that it is unnecessary. This is called copy elision: Copy elision - Wikipedia, the free encyclopedia

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Quote:

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Originally Posted by freiza

test t(t); // This syntax is incorrect. because you cannot pass same object in constructor; But I thought default copy constructor would require this.

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C++ doesn't support reincarnation constructors.

Oh, and in case you missed it, that was a joke.