从C ++中的错误模板实例继承

Question

Consider the following code:

struct Foo
{
    Foo operator+(const Foo &rhs) const;
    // notice lack of: Foo operator*(const Foo &rhs) const;
};

template <class T>
struct Bar
{
    T x, y;
    T add() const { return x + y; }
    T mul() const { return x * y; }
};

I have two questions:

1. Can I inherit from Bar<Foo> and override mul() to something meaningful?

2. Can I inherit from Bar<Foo> without overriding mul() if I never use mul() anywhere?

Answer 1

1. Bar<Foo>::mul() isn't a virtual function, so it cannot be overridden.

2. Yes, if you don't use a template member function then it does not get instantiated and you don't get any errors that would result from instantiating it.

You can hide Bar<Foo>::mul() by providing a function of the same signature in a subclass, and because of 2 , Bar<Foo>::mul() won't be instantiated. However this is probably not a good practice. Readers are likely to get confused about the hiding vs. overriding, and there's not much benefit to doing this over simply using a different function name and never using mul() , or providing an explicit specialization of Bar for Foo.

Answer 2

1. sure
2. sure

Templates are really a kind of smart preprocessor, they're not compiled. If you don't use something, you can write complete (syntactically correct) rubbish, ie you may inherit from

template <class T>
struct Bar
{
    T x, y;
    T add() const { return x + y; }
    T mul() const { return x.who cares what-s in here; }
};

PS since your + operator is used in a const function, it should be declared as const too.

EDIT : OK, not all compilers support this, here's one that compiles with gcc:

template <class T>
struct Bar
{
    T x, y;
    T add() const { return x + y; }
    T mul() const { T::was_brillig & T::he::slith(y.toves).WTF?!0:-0; }
};