Cannot convert this pointer

Question

I have a class Foo which looks like this

class Foo
{
public:
    void set_bar(int val)
    {
        _bar = val;
    }

protected:
    int _bar;
};

which is inherited by class Baz which is declared as follows:

class Baz
    : public Foo
{
public:
    void set_baz(int val)
    {
        _baz = val;
    }

protected:
    int _baz;
};

and then i have a Factory class which contains a member variable of type Baz.

class Factory
{
protected:
    Baz _baz;
public:
    const Baz& get_baz() const 
    {
        return _baz;
    }
};

If I used Factory like this:

Factory f;
f.get_baz().set_bar(1);

It throws this error C2662: 'Foo::set_bar' : cannot convert 'this' pointer from 'const Baz' to 'Foo &'

score 1 · Answer 1 · answered Feb 05 '11 at 14:39

1

The object returned by get_baz is constant, you cannot modify its contents.

answered Feb 05 '11 at 14:39

adamax

3,835
2
19
21

Ouch, I didn't see that.. Thanks. – SimpleButPerfect Feb 05 '11 at 14:40

score 1 · Accepted Answer · edited May 23 '17 at 09:58

1

You need to provide a non-const version of Factory::get_baz that returns a non-const reference to Baz:

class Factory
{
protected:
    Baz _baz;
public:
    const Baz& get_baz() const 
    {
        return _baz;
    }

    Baz& get_baz()
    {
        return _baz;
    }
};

When using a const Factory, the const version of get_baz will get invoked, otherwise the non-const version of get_baz will get invoked.

You should also refrain from using leading underscores in your names. See What are the rules about using an underscore in a C++ identifier?.

edited May 23 '17 at 09:58

Community

1
1

answered Feb 05 '11 at 14:43

Emile Cormier

28,391
15
94
122

1

I don't prefix my code with underscores in real code, thanks for the read though. – SimpleButPerfect Feb 05 '11 at 14:47

score 0 · Answer 3 · answered Feb 05 '11 at 14:42

0

After calling get_baz you are getting constant object and constant object can call only constant function.

answered Feb 05 '11 at 14:42

Arunmu

6,837
1
24
46

score 0 · Answer 4 · answered Feb 06 '11 at 11:41

0

You can only change the value if you return a non-const reference to your object.

If you return a non-const reference, why not just make the member public? You'll get the same effect...

answered Feb 06 '11 at 11:41

Bo Persson

90,663
31
146
203

Cannot convert this pointer

4 Answers4