c ++：模板声明中is_member_function_pointer的语法 [英] c++ : syntax for is_member_function_pointer in a template declaration

问题描述

我有一个类似这样声明的模板：

I have a template with a declaration similar to this:

template <typename Arg0, typename... Args>
class blah {};

我有两个版本的模板，当Arg0是一个成员函数指针，否则使用另一个。我试图使用std :: enable_if和std :: is_member_function_pointer，但我找不到正确的语法。这是我对真实情况：

I have two versions of the template, and I want to use one when Arg0 is a member function pointer, otherwise use the other one. I'm trying to use std::enable_if and std::is_member_function_pointer but I cannot find the correct syntax. This is what I have for the true case:

template<typename = typename std::enable_if< std::is_member_function_pointer<Arg0> >::type, typename... Args>
class blah() {}

但这显然不是句法上正确的。

But this obviously isn't syntactically correct.

推荐答案

在类中使用布尔谓词时，通常有两种方法可供选择：

When using the Boolean predicates with classes there are generally two approaches I use to make the choice:

1. 如果我只需要在两种类型之间选择，我使用sonething像

1. If I just need to choose between two types, I use sonething like

typename std::conditional<
    std::is_member_function_pointer<F>::value,
        type_when_true, type_when_false>::type

如果事情需要改变，我从一个专门的布尔值覆盖两个实现选择的基础派生：

If things need to change more than that I derive from a base which is specialized on a Boolean covering the two implementation choices:

template <bool, typename...>
struct helper;

template <typename... A>
struct helper<true, A...> {
    // implementation 1
};
template <typename... A>
struct helper<false, A...> {
    // the other 1
};
template <typename F, typename... A>
struct actual
    : helper<std::is_member_function_pointer<F>::value, F, A...>
{
    // typedefs, using ctors, common code, etc.
};

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