Explicit operator= call (T::operator=)

Question

I am reading qt sources and I've seen a code like this many times:

buttonOpt.QStyleOption::operator=(*opt);

So, I guess it is something like buttonOpt = *opt but why do they use this syntax instead of default and user-friendly? Is this faster or any other profit exists?

Answer 1

This is because they are explicitly calling the operator= from the base class of buttonOpt, which is QStyleOption.

buttonOpt.QStyleOption::operator=(*opt);

//similar behavior
class Base
{
public:
    virtual bool operator<(Base & other)
    {
        std::cout << "Base";
    }
};

class Derived : public Base
{
public:
    bool operator<(Base & other) override
    {
        std::cout << "Derived";
    }
};

int main()
{
    Derived a;
    Derived b;
    a < b; //prints "Derived"
    a.Base::operator <(b); //prints "Base"
}