derived pass to supper move constructor

Asked Apr 23 '20 at 15:06

Active Apr 23 '20 at 15:08

Viewed 24 times

move derived obj d to derived dd, but match base's move constructor, do not match derived's constructor

// base class 
class base {
    int x;

public:
    base() = default;

    base(base && b) {
        x = b.x;
    }
};

// derived struct
struct derived : public base {
    int y;

    derived(int d) {
        y = d;
    }

    derived() = default;
};

int main()
{
    derived d{1};
    derived dd = std::move(d);
    spdlog::info(dd.y); // => 1
    return 0;
}

I do not know why dd.y will be 1, not 0 ?

how d.y copied to dd.y struct ?

edited Apr 23 '20 at 15:08

asked Apr 23 '20 at 15:06

langyu

related/dupe: https://stackoverflow.com/questions/3734247/what-are-all-the-member-functions-created-by-compiler-for-a-class-does-that-hap – NathanOliver Apr 23 '20 at 15:08
UB with copy/move of uninitialized `x`. – Jarod42 Apr 23 '20 at 15:09
`the compiler will generate a copy constructor for you unless you declare your own.` -> thanks very much – langyu Apr 23 '20 at 15:10
And generated move constructor of derived will call move constructor of base. – Jarod42 Apr 23 '20 at 15:11

derived pass to supper move constructor

0 Answers0