Can't understand compiler's behaviour (creating object)

Question

I wrote the following code:

#include <iostream>

class A
{
public:
    A(){ std::cout << "created" << std::endl; }
    A(A& other) { std::cout << "copied" << std::endl; }
    A& get(){ std::cout <<"got" << std::endl; return *this; }
    ~A(){ std::cout << "destroyed" << std::endl; }
};

Now, lines

A a = A().get();

and

A a;    
a = A();

compile and work correctly, but

A a = A();

claims:

no matching function for call to ‘A::A(A)’
note: candidates are: A::A(A&)
note:                 A::A()

And to make things explicit,

A a = (A&)A();

claims:

error: invalid cast of an rvalue expression of type ‘A’ to type ‘A&’

I completely don't understand this behaviour.

P.S. I know, that if I make const reference in copy c_tor, everything will be OK.

Fred Foo · Accepted Answer · 2011-10-22 15:03:55Z

3

The copy constructor should get its argument as a const reference (or a simple A value). In your current setup, the program would have to make a mutable reference to a temporary, which is invalid.

answered Oct 22, 2011 at 15:03

Fred Foo

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2 Comments

Lol4t0 Over a year ago

You know, g++ does not think so sometimes. For exasmple, private member of type std::auto_ptr<std::string> will make g++ create default copy c_tor with non-const reference

Gorpik Over a year ago

@Lol4t0: It is OK to have a copy ctor taking a non-const reference; the problem arrives when you try to use it with something that cannot be converted into a non-const ref, such as in this example.

PierreBdR · Accepted Answer · 2011-10-22 15:06:41Z

3

This is simple: a rvalue (i.e. something that must be on the right hand side of an equal sign), can convert to a constant reference or be copied. Only a lvalue (i.e. something that can be placed on the left of an equal sign) can be converted into a non-constant reference. The reason being that you may be tempted to modify the content of the non-constant reference, which would be invalid.

A() is a rvalue, and therefore cannot be converted into a non-constant reference. This is your error.

answered Oct 22, 2011 at 15:06

PierreBdR

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2 Comments

Lol4t0 Over a year ago

Ok,but why c_tor return r-value? I can call members of that r-value, and I can even write A() = A(); :) More, why does my get() function work? And If you look output, you can see, that no object is removed twice

Lol4t0 Over a year ago

Oh,It seems also, that r-value can_be l-value

Sarfaraz Nawaz · Accepted Answer · 2011-10-22 15:06:16Z

2

A a = A();

This line tries to call the copy-constructor passing a temporary object, and the copy-constructor takes argument by non-const reference. But a temporary object cannot be bound to a non-const reference. That is why you get the compilation error.

However, a temporary object can be bound to const reference. So, the solution is to make the parameter const reference as:

A(const A& other) { std::cout << "copied" << std::endl; }

answered Oct 22, 2011 at 15:06

Sarfaraz Nawaz

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1 Comment

Lol4t0 Over a year ago

this explanation sounds most logical for me

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