如何故意造成模板实例化时的编译错误[英] How to intentionally cause a compile-time error on template instantiation

问题描述

Sometimes when coding with C++ templates, you want to prevent users from instantiating a specific specialization or set of specializations, because the result would be nonsensical. So you can define a (specific or partial) specialization whose definition, if instantiated, would cause a compiler error. The goal would be, if a user "misuses" the template, to cause a compiler error right next to a comment in your header file explaining what not to do, rather than leaving the compiler to come up with some confusing error message by its own devices, or maybe allowing the questionable code to compile.

Example:

template <typename T> struct MyClassTemplate {
  // ...
};

template <typename T> struct MyClassTemplate<T*> {
  // Do not use MyClassTemplate with a pointer type!
  typedef typename T::intentional_error err;
};

There are a number of ways to do this (depending on whether your specialization is a complete or partial specialization of a class or function). But the syntax used must (?) depend on a template parameter, or else the compiler will complain when it first parses the intentional-error definition. The example above has a hole in that somebody could stubbornly define an intentional_error nested type or member typedef (though I'd say they would then deserve whatever problems come up as a result). But if you use a trick too fancy, you're likely to get an indecipherable and/or misleading compiler error message, which mostly defeats the purpose.

Are there better straightforward ways to disallow template instantiations?

I'm aware that in C++0x, template Concepts and deleted function declarations will provide much better control over this sort of thing, but I'm looking for answers that are valid C++03.

推荐答案

You could just omit defining it.

template <typename T> struct MyClassTemplate<T*>;

You could also derive from a non-defined specialization

template <typename T> struct invalid;
template <typename T> struct MyClassTemplate<T*> : invalid<T> { };

Note that explicit specializations that declare classes or functions will never depend on template parameters. So, stuff like this that depend on template parameters can't work anyway. In that case, declaring a non-defined explicit specialization should be sufficient

template<> struct MyClassTemplate<int*>;

其他推荐答案

For me this sounds like a typical case for static_assert from C++0x or BOOST_STATIC_ASSERT. The static_assert functionality has the advantage that you can pass a custom error message so that the reason for the error is more clear.

Both ways are giving you the opportunity to prematurely end the compilation process under some custom defined compile time condition.

with static_assert:

template <typename T> struct MyClassTemplate<T*> {
    static_assert(always_false<T>::value, "Do not use MyClassTemplate with a pointer type!");
};

with BOOST_STATIC_ASSERT

template <typename T> struct MyClassTemplate<T*> {
    // Do not use MyClassTemplate with a pointer type!
    BOOST_STATIC_ASSERT(always_false<T>::value);
};

Always false would look something like this:

template< typename T >
struct always_false { 
    enum { value = false };  
};

HTH

Edit: Fixed the examples to make them actually work ;-) Thanks to GMan!

其他推荐答案

If you don't want to use a library, this construct is pretty reliable (it's roughly what Boost does internally):

template <typename T>
void must_be_specialized(T const&)
{
   enum dummy { d = (sizeof(struct must_be_specialized_for_this_type)
                     == sizeof(T)) };
}

You can put something analogous in a specialization to disallow instantiation of the template with that type. I wouldn't, personally, worry about must_be_specialized_for_this_type gaining a definition from somewhere, but you could use a forward declaration to squirrel it away in a private namespace if you really wanted.