c++ - Why is a VTABLE required when the derived class doesn't override the virtual function? -

c++ - Why is a VTABLE required when the derived class doesn't override the virtual function? -

class base of operations { public: void virtual fn(int i) { cout << "base" << endl; } }; class der : public base{ public: void fn(char i) { cout << "der" << endl; } }; int main() { base* p = new der; char = 5; p->fn(i); cout << sizeof(base); homecoming 0; }

here signature of function fn defined in base class different signature of function fn() defined in der class though function name same. therefore, function defined in der class hides base class function fn(). class der version of fn cannot called p->fn(i) call; fine.

my point why sizeof class base or der 4 if there no utilize of vtable? requirement of vtable here?

note highly implementation dependent & might vary each compiler.

the requirement presence of vtable base of operations class meant inheritance , extension, , class deriving might override method.

the 2 classes base of operations , derived might reside in different translation unit , compiler while compiling base of operations class won't know if method overidden or not. so, if finds keyword virtual generates vtable.

c++ virtual-functions vtable function-signature