[Solved] Returning const QSharedPointer

[Joel Bodenmann]

Let's consider the following code:

class Foo {
	int a;	
};

class Bar {
public:
	Bar() {
		_foo = new Foo;
	}
	
	const Foo* foo() const {
		return _foo;
	}

private:
	Foo* _foo;
}

How would I implement the same behavior using QSharedPointer? I understand how to use QSharedPointer but I don't understand how I would return a QSharedPointer that is a const pointer (the return type of the public method in the Bar class) so that the caller of Bar::foo() gets just "const access" to the Foo object.

[DanWatkins]

This post is deleted!

[DanWatkins]

Here is a complete example. The trick is to use QSharedPointer<const Foo>.

#include <QSharedPointer>
#include <QDebug>

struct Foo
{
    ~Foo()
    {
        qDebug() << "Foo destroyed";
    }

    int count() { return 7; }
    int sum(int a, int b) const { return a + b; }
};


class Bar
{
public:
    Bar() : mFoo(new Foo) {}
    QSharedPointer<Foo> foo() { return mFoo; }
    QSharedPointer<const Foo> constFoo() const { return mFoo; }

private:
    QSharedPointer<Foo> mFoo;
};


int main()
{
    //intentially store it as a local variable
    //this shows how the reference count is incremented
    QSharedPointer<const Foo> constFoo(nullptr);
    {
        {
            Bar bar;
            constFoo = bar.constFoo();

            //error, discards const qualifiers of const Foo
            //qDebug() << constFoo->count();
            qDebug() << constFoo->sum(5, 6);
            qDebug() << bar.foo()->count();
        }

        qDebug() << "bar has gone out of scope, constFoo holds the last reference";
    }
}

[Joel Bodenmann]

Awesome, thank you very much!