Disposable doesn't work?

[Kakhkhor Abdijalilov]

Disposable template is supposed to eliminate unnecessary copy
construction when returning temporal objects from functions. On my
platform it actually prevents RVO (elide optimization) and triggers
copy construction, whereas without Disposable RVO eliminates
unnecessary copy construction.

Explicit assignment triggers copying with or without Disposable.

The code is below.

Regards,
Kakhkhor Abdijalilov.

//-------------------------------------------
/*
    Compile with optimization ON.
*/
#include <iostream>
#include <ql/utilities/disposable.hpp>

using namespace std;
using namespace QuantLib;

struct A {
    A() {}
    A(const A&) { cout << "Triggered copy ctor A(const A&)\n"; }

    A& operator=(const A&) {
        cout << "Triggered assignment A::=\n";
        return *this;
    }

    void swap(A&) {}
};

Disposable<A> f1() {
    A a;
    return a;
}

A f2() {
    A a;
    return a;
}

int main() {

    cout << "\nTesting RVO with Disposable...\n";
    A x1 = f1();
    cout << "\nTesting RVO without Disposable...\n";
    A x2 = f2();

    cout << "\nTesting explicit assignment with Disposable...\n";
    A y1;
    y1 = f1();
    cout << "\nTesting explicit assignment without Disposable...\n";
    A y2;
    y2 = f2();

    // Possible solution/guideline to assignment problem?
    cout << "\nUsing swap explicitly without Disposable...\n";
    A z;
    z.swap(f2());

    return 0;
}

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[Luigi Ballabio]

On Fri, 2010-12-03 at 11:45 -0500, Kakhkhor Abdijalilov wrote:
> Disposable template is supposed to eliminate unnecessary copy
> construction when returning temporal objects from functions. On my
> platform it actually prevents RVO (elide optimization) and triggers
> copy construction, whereas without Disposable RVO eliminates
> unnecessary copy construction.
>
> Explicit assignment triggers copying with or without Disposable.

Of course it triggers assignment.  Your class must support Disposable
for it to work fully (see, e.g., the Array class.)  In your case, it
should be something like

struct A {
    A() {}
    A(const A&) { cout << "Triggered copy ctor A(const A&)\n"; }
    A(const Disposable<A>& a) { swap(const_cast<Disposable<A>&>(a)); }

    A& operator=(const A&) {
        cout << "Triggered assignment A::=\n";
        return *this;
    }
    A& operator=(const Disposable<A>& a) {
        swap(const_cast<Disposable<A>&>(a));
        return *this;
    }

    void swap(A&) {}
};

Hopefully, in a few years all compilers will support C++0X and we'll be
able to use rvalue references instead... (your class will still have to
support them. though.)

Luigi


--

For every problem there is one solution which is simple, neat, and
wrong.
-- H. L. Mencken



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[Kakhkhor Abdijalilov]

So, this how it works. Thank you for clarification.

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