CODEWITHSUNDEEP
c++inheritanceconstructor
DotSlashSlash
DotSlashSlash

Reputation: 5203

Call parameterised constructor after base class after some calculation

I want to be able to call a parameterised constructor of the base class AFTER I've done some calculation with the input parameters from the derived class constructor.

To what I believe, when you call a constructor of a derived class, the default constructor of the base class is called...

(unless its parameterised constructor is called specifically as so:)

DerivedClass(int a,int b) : BaseClass(a,b) {}

Is there anyway of doing something like:

DerivedClass(int a,int b) { 
   a += 2; 
   b += 5; 
   BaseClass(a, b); // <- this line I am questioning :(
}

And I assume what is above doesn't work because the default constructor of the BaseClass is already been called..

Upvotes: 1

Views: 355

Answers (2)

vsoftco
vsoftco

Reputation: 56557

No, you cannot defer calling a base constructor in a derived constructor. The only way to call a constructor is via the constructor initialization list, which is executed before the body of your constructor. If otherwise, you'd see probably lots of nightmarish code, where the base part of an object is not yet constructed but the derived part is trying already to initialize itself...

In your case, you can just do

DerivedClass(int a,int b) : BaseClass(a + 2, b + 2) {}

PS: You have an error in your code, when you call the BaseClass you shouldn't use BaseClass(int a, int b), but only BaseClass(a, b), as it is a function call and not a declaration.

Upvotes: 4

Zan Lynx
Zan Lynx

Reputation: 54325

You cannot defer the base constructor but you can use functions.

For example:

DerivedClass(int a, int b) : BaseClass(adjust_a(a), adjust_b(b)) {}

Upvotes: 2

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