Base class constructor

Question

I have the following situation (the below is just for illustration):

class Base() 
{
public:
    Base(int);
};

class Derived : public Base 
{
public:
    Derived(int, int, bool);
};

I want to initialise the base class depending on the boolean flag in the derived constructor. What (if any) is the correct way of doing so ..

I could do (but is the evaluation done before Base is initialised?):

Derived::Derived(int _x, int _y, bool _z) : Base(_z?_x:_y) {}

or (but this probably doesn't work properly)

Derived::Derived(int _x, int _y, bool _z) 
{
    if(_z)
      ::Base(_x);
    else
      ::Base(_y);
}

If there is no correct way of doing this then I could possibly get away with adding additional constructors to Derived.

The first solution is the correct one (using the ternary operator). If you have a more complex initialization to compute the value of the Base parameter, you can use a (free or not) function call (Derived::Derived(int _x, int _y, bool _z) : Base( compute_value() ) {}) — utnapistim
– utnapistim, Commented Feb 13, 2013 at 10:44
class Base() is wrong, remove the parens. The 1st proposal you've made is fine, the evaluation will be made before the Base() constructor is called. — πάντα ῥεῖ
– πάντα ῥεῖ, Commented Feb 13, 2013 at 10:47
The second one, in each branch of the if, creates a temporary object of type Base and immediately destroys it. — Pete Becker
– Pete Becker, Commented Feb 13, 2013 at 13:33

billz · Accepted Answer · 2013-02-13 10:44:24Z

3

 Derived::Derived(int _x, int _y, bool _z) : Base(_z?_x:_y) {}

Is the correct way. You could only call base constructor with parameter in member initializers list

answered Feb 13, 2013 at 10:44

billz

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Comments

Konrad Rudolph · Accepted Answer · 2013-02-13 10:44:09Z

2

Your first alternative is the correct one: use the initialisation list.

The second alternative doesn’t work.

answered Feb 13, 2013 at 10:44

Konrad Rudolph

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Comments

Pete Becker · Accepted Answer · 2013-02-13 13:33:08Z

2

The first alternative is the only one of your alternatives that will work. The constructor of the derived class will be called after the constructor of the base-class, so your second option is invalid.

Now, of course, the question is whether this is the "right" thing to do. Since it may be that you should have two derived classes, rather than one. I'm not saying what you are doing is wrong as such, it's just something to bear in mind.

edited Feb 13, 2013 at 13:33

Pete Becker

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answered Feb 13, 2013 at 10:47

Mats Petersson

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