Template method over non-template method in derived class

Question

class A {
public:
    template<typename T> void func(size_t n, T values[]) { ... }
};

class B : public A {
public:
    void func(size_t n, uint32_t values[]) { ... }
};

Why does function B::func() not take precedence over the function template A::func() when calling this code?

uint32_t values[5];
A* obj = new B();
obj->func(5, values);

Because overload resolution rules select most equal type match and most specific one (in this case, non-template one). — 9dan, Jan 01 '11 at 20:03
@peoro: Yes, I think. I re-phrased it after I had missed out the "not" and posted a harsh comment explaining that there's a "not" missing... — sbi, Jan 01 '11 at 20:30

score 6 · Accepted Answer · answered Jan 01 '11 at 20:08

6

Two reasons-

In C++ a member function only overrides a base class member function if the base class function is marked virtual. Otherwise, C++ treats the two as independent functions that coincidentally have the same name. This contrasts with Java, where functions atomatically override base class functions.
In C++, template member functions can't be marked virtual. This is partly due to the implementation of virtual functions that's most commonly used - vtables - combined with the C++ template instantiations system. C++ treats all instantiations of the same template over different type arguments as separate functions, and generates these instantiations lazily. This is a problem for the vtable implementation, since a vtable requires static, compile-time determination of the number of different virtual functions in a class, along with an ordering on them. Since a template virtual function would mean an unknown number and ordering of virtual functions in the class, C++ disallows this.

answered Jan 01 '11 at 20:08

templatetypedef

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re #2 see also [my answer](http://stackoverflow.com/questions/2354210/can-a-member-function-template-be-virtual/2354671#2354671) to an FAQ entry. – sbi Jan 01 '11 at 20:15
So the only solution to my problem is to make non-template version of func in A (calling template function) and override it in B? – mnn Jan 01 '11 at 20:25
That should work. Out of curiosity, why do you have a template member function that produces data of arbitrary types? – templatetypedef Jan 01 '11 at 20:26
That's just one use case of that function. – mnn Jan 01 '11 at 20:31

score 1 · Answer 2 · answered Jan 01 '11 at 20:05

1

If you call func on a object/pointer/reference of type A, A::func gets called, since func is not virtual (and it cannot be: it's a template function).

Did you really test your code?

answered Jan 01 '11 at 20:05

peoro

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score 0 · Answer 3 · answered Jan 01 '11 at 20:04

0

A::func() isn't declared as virtual, so the compiler won't generate the virtual table and any other code required to invoke B::func() at runtime. It has nothing to do with templates.

answered Jan 01 '11 at 20:04

chrisaycock

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score 0 · Answer 4 · answered Jan 01 '11 at 20:08

Because B::func is NOT an overload of A::func and can never be, no matter what you do. Don't even try declaring A::func virtual because you can't. Whatever it is you're trying to do, you can't. Static and dynamic polymorphism just don't mix in C++.

If your pointer was to a B*, instead of an A*, you would expect it's version to be called.

Template method over non-template method in derived class

4 Answers4