Will function dispatch via vtable if calling non-virtual implementation?

Question

Say I have the following:

struct Base
{
    virtual void callback() { /* */ }
};

struct Derived : public Base
{
    void callback() { /* */ }
};

Base* obj = new Derived;
static_cast<Derived*>(obj)->callback();

Will the call to callback invoke via the vtable or a straight function call as it's not marked as virtual in the function signature?

James McNellis · Accepted Answer · 2011-02-07 18:10:17Z

4

Derived::callback is implicitly virtual because the Base::callback function with a the same return type and parameters is declared as virtual.

That said, whether or not the vtable is used depends on the compiler and what optimizations it performs. Since the compiler can know here that obj points to a Derived object, it knows what the final overrider for the callback virtual function is, so it could potentially make the call without looking up the function in the vtable. Whether or not it does so depends on the compiler and optimization settings, though.

answered Feb 7, 2011 at 18:10

James McNellis

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