避免在默认模板中使用尖括号 [英] avoid angle brackets in default template

问题描述

如果我有一个模板类与默认模板类型，但我必须写模板尖括号。 是否有可能避免此问题？

if i have a template class with a default template type, however i have to write the template angle brackets. is it possible to avoid this?

例如：

template <typename T=int>
class tt {
public:
  T get() { return 5; }
};

...

tt<> t;  // how to avoid <>
std::cout << t.get() << std::endl;

到目前为止，我已经通过一个单独的命名空间并重新声明了类：

until now i've did this by a seperate namespace and redeclaring the class:

namespace detail_ {
template <typename T=int>
class tt {
public:
  T get() { return 5; }
};
}

class tt : public detail_::tt {}

...

tt t;  // now it works
std::cout << t.get() << std::endl;

问题是，如果我想使用其他类型的类，我必须去namepace detail_。 是有其他解决方案，我还没有看到。

the problem is, if i want to use the class with an other type i have to go ov er namepace detail_. is there an other solution, wich i didn't see yet.

推荐答案

。 ..如果我想使用类...

这是一个常见的混乱来源。类模板不是类，而是生成类的模板。尖括号是告诉编译器，要使用给定的模板参数生成类模板中的类，而不使用尖括号您将是一个模板。

This is a common source of confusion. A class template is not a class, but a template from which classes are generated. The angle brackets is what tells the compiler that you want to generate a class out of the class template with the given template arguments, without the angle brackets what you have is a template.

template <typename T = int>
struct TemplateClass {...};

template <template class T<typename> >
void f() {
   T<int> t; ...
}
template <typename T>
void g() {
   T t; ...
}

f<TemplateClass>();     // Accepts a template with a single type argument
g<TemplateClass<> >();  // Accepts a type, that can be generated out of the template

语言不允许共存的模板和在同一命名空间中具有相同名称的类型，因此答案是它不能完成。您可以创建类型别名，但必须为其指定其他名称。

The language does not allow the coexistence of a template and a type with the same name in the same namespace, so the answer is that it cannot be done. You can create a type alias but you will have to give it a different name.

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