Using class member name in a constructor

Question

I know this as this will assign the constructor parameter to the class member:

class A
{
    public:
        some_type B;
        A(some_type B)
        {
            this->B = B;
        }
}

But what will this do:

class A
{
    public:
        some_type B;
        A(some_type B) : B(B)
        {
        }
}

Will this assign the parameter to itself or the parameter to the class member or do something else?
How are the names in the list after construct thing (I have no idea how its called) resolved?

Somebody get Dr. Dobbs on the horn, stat!

synthesizerpatel
– synthesizerpatel

2011-09-13 15:14:11 +00:00
Commented Sep 13, 2011 at 15:14 — synthesizerpatel
– synthesizerpatel, Commented Sep 13, 2011 at 15:14

Alok Save · Accepted Answer · 2011-09-13 18:09:07Z

9

That Thing is called a Member Initializer List in C++.

There is a difference between Initializing a member using initializer list(2nd Example) and assigning it an value inside the constructor body(1st Example).

When you initialize fields via initializer list the constructors will be called once. The object gets constructed with the passed parameters.

If you use the assignment then the fields will be first initialized with default constructors and then reassigned (via assignment operator) with actual values.

As you see there is an additional overhead of creation & assignment in the latter, which might be considerable for user defined classes.

How are the names in the list after construct thing (I have no idea how its called) resolved?

public:
    some_type B;
    A(some_type B) : B(B)
    {
    }

In the above snipet, there are two entities named B:

First is the one that constructor receives as an argument &
Second is the member of the class A

The variable received as argument in Constructor of A gets passed as an argument for constructing B(by calling its constructor) which is member of class A. There is no ambiguity in names here, the all to constructor is:

this->B(B);

this is class A pointer.
B() is constructor of type B.
B inside the parenthesis is an instance of type B.

edited Sep 13, 2011 at 18:09

answered Sep 13, 2011 at 15:13

Alok Save

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5 Comments

tdammers Over a year ago

To add to that: If the constructors and / or assignment operators have side effects, there is also a functional difference.

tinman Over a year ago

You didn't answer the question: "what happens when the parameter name is the same as the member name?" Was that intentional?

Alok Save Over a year ago

@tinman: I don't understand, Where is that Question? Why would i not answer something intentionally?

tinman Over a year ago

After the second code section in the question it contains "Will this assign the parameter to itself or the parameter to the class member or do something else? How are the names in the list after construct thing (I have no idea how its called) resolved?". Maybe the question was edited after you answered?

Alok Save Over a year ago

@tinman: I don't know if it was edited or I missed it, but I edited the answer anyways. Hth.

Baltasarq · Accepted Answer · 2011-09-13 15:24:28Z

Will this assign the parameter to itself or the parameter to the class member or do something else?

It will assign the parameter to the class member.

How are the names in the list after construct thing (I have no idea how its called) resolved?

This is the initializers list, and though the example can lead to confusion, the id at the left of the parenthesis is a member of the class, while the id (or literal) inside the parenthesis can only be one of the parameters of the constructor: thought that way, there is no ambiguity at all.

The key here is that this list must initialize class members with some value, so if you think of

...: B(B)

as conceptually equivalent to a constructor call:

this->B(B)

... there is no ambiguity.

Tristram Gräbener · Accepted Answer · 2011-09-13 15:31:20Z

0

In the first situation, your objet will be constructed. In the second, it will be constructed (with default constructor) and assigned.

You might not want assignation (maybe the operator is not defined, or no default constructur or the behaviour is specific, or maybe, in some cases because of performance issues).

Concerning the visibility, in both cases, if you just use B, it's the parameter you're manipulating.

answered Sep 13, 2011 at 15:31

Tristram Gräbener

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