如何防止非专业模板实例化?

问题描述

我有一个模板化的class(称之为Foo)，它有几个专门化.如果有人尝试使用 Foo 的非专业版本，我希望编译失败.

I have a templated class (call it Foo) which has several specializations. I would like the compilation to fail if someone tries to use an unspecialized version of Foo.

这是我实际拥有的:

template <typename Type>
class Foo
{
  Foo() { cannot_instantiate_an_unspecialized_Foo(); }

  // This method is NEVER defined to prevent linking.
  // Its name was chosen to provide a clear explanation why the compilation failed.
  void cannot_instantiate_an_unspecialized_Foo();
};

template <>
class Foo<int>
{    };

template <>
class Foo<double>
{    };

所以:

int main()
{
  Foo<int> foo;
}

同时工作:

int main()
{
  Foo<char> foo;
}

没有.

显然，编译器链只会在链接过程发生时发出警告.但是有没有办法让它在之前抱怨?

Obviously, the compiler chain only complains when the linking process takes place. But is there a way to make it complain before ?

我可以使用boost.

推荐答案

不要定义类:

template <typename Type>
class Foo;

template <>
class Foo<int> { };

int main(int argc, char *argv[]) 
{
    Foo<int> f; // Fine, Foo<int> exists
    Foo<char> fc; // Error, incomplete type
    return 0;
}

为什么会这样?仅仅是因为没有任何通用模板.声明，是，但未定义.

Why does this work? Simply because there isn't any generic template. Declared, yes, but not defined.

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