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Home/ Questions/Q 7061055
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T04:25:38+00:00 2026-05-28T04:25:38+00:00

I have a simple class class Foo { public: float m; Foo(); } Foo::Foo(){

  • 0

I have a simple class

class Foo {
    public:
        float m;
        Foo();
}

Foo::Foo(){
    this->m = 1.0f;
}

Then I’m extending it with

class Bar: public Foo {
    public:
        float m;
        Bar()
}

Bar::Bar(){
    this->m = 10.0f;
}

I then instantiate Bar() but Bar.m is still 1.0f. Is there a reason for this?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-28T04:25:39+00:00Added an answer on May 28, 2026 at 4:25 am

    In C++, you cannot override a field. Only methods can be overridden. Consequently, your declaration of the variable m in the class Bar is a new field that hides the base class Foo‘s version of m.

    If you want to access Foo‘s m from Bar, then you could use this syntax:

    Bar::Bar(){
        this->Foo::m = 10.0f;
    }
    

    Which explicitly tells the compiler to write to Foo‘s version of m. Alternatively, you can drop the this-> and just write

    Bar::Bar(){
        Foo::m = 10.0f;
    }
    

    Hope this helps!

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