How to call constructor of a nested class

Question

First off, I'm not talking about questions like these (1, 2). I'm talking about a class that already has a constructor and is only nested by value, not by definition...if that makes any sense.

Currently, I have this methodology, which works, but uses new and is ugly IMO:

class A {
private:
    int value;

public:
    A(int _value);
};

class B {
private:
    A *a;
public:
    B(int _value);
};

The constructor of B:

B::B(int _value)
{
    a = new A(_value);
}

If I were to change A *a; to A a, how would I create a B such that A's constructor is called in the same way?

I've tried:

B::B(int _value)
{
    a = A(_value);    // no matching function for call to ‘A::A()’
    a = A::A(_value); // cannot call constructor ‘A::A’ directly
    a(_value);        // no match for call to ‘(A) (int&)’
    a.A(_value);      // invalid use of `A::A`
}

It doesn't seem possible without:

using a pointer with new
using some explicit void A::init(_value) (would need an additional constructor for A to take no arguments)
extending class A (would require changing specific private members to protected and would mesh their "namespaces" in an unwanted manner)
questionable use of memcpy() (A temp(_value); memcpy(&a, &temp, sizeof (A));)

Is this possible without resorting to the above methods?

Use a reference member and the constructor initializer list. — πάντα ῥεῖ, Sep 16 '16 at 00:49
Is there any advantage a reference has over a value in this case? It seems like both get created upon creation of a `B`, and both have the same usage syntax and semantics. — Braden Best, Sep 16 '16 at 00:57
@BradenBest If you don't have a reason to use a reference, don't. A plain `A a;` is most certainly what you want here. — Baum mit Augen, Sep 16 '16 at 00:58

Ed Heal · Accepted Answer · 2016-09-16T00:51:12.313

2

You use initialisation lists.

I.e. B constructor looks like

class B {
   ...

   A a;
};

B::B(int _value) : a(_value) {
  ...
}

edited Sep 16 '16 at 00:51

answered Sep 16 '16 at 00:46

Ed Heal

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`a` is a pointer actually! – πάντα ῥεῖ Sep 16 '16 at 00:48
1

@πάνταῥεῖ Which is exactly what OP is trying to get rid of. – Baum mit Augen Sep 16 '16 at 00:49
@πάνταῥεῖ - He says we wants to change it from a pointer – Ed Heal Sep 16 '16 at 00:49
@EdHeal Then complete your answer and show how to use a reference. – πάντα ῥεῖ Sep 16 '16 at 00:50
It works. I need to look up the `:` operator. Thanks! – Braden Best Sep 16 '16 at 00:50
1

@πάνταῥεῖ Why would he need a reference? – Baum mit Augen Sep 16 '16 at 00:50
@BaummitAugen To keep semantics? – πάντα ῥεῖ Sep 16 '16 at 00:51
@πάνταῥεῖ He does not want to keep the semantics, he wants good old value semantics of a plain "by value" member. – Baum mit Augen Sep 16 '16 at 00:52
1

@πάνταῥεῖ what are you getting at? I want `a` to be a value (`A`), not a pointer (`A*`), not a reference (`A&`). This answer answers my question and gives me exactly what I asked for. Case closed. – Braden Best Sep 16 '16 at 00:53
@BradenBest Okey. – πάντα ῥεῖ Sep 16 '16 at 00:54

How to call constructor of a nested class

1 Answers1