Boost MPL：调用（成员）函数，只有存在 [英] Boost MPL: Call a (member) function only if it exists

问题描述

我有一个类A，它有一个模板参数T.有一些用例，类T提供了一个函数func1（），有一些用例，T不提供它。
A中的函数f（）应该调用func1（），iff它存在。我认为这应该是可能的提升mpl，但我不知道如何。
这里有一些伪代码：

I have a class A that has a template parameter T. There are use cases where the class T offers a function func1() and there are use cases where T doesn't offer it. A function f() in A should call func1(), iff it exists. I think this should be possible with boost mpl, but I don't know how. Here some pseudo code:

template<class T>
class A
{
    void f(T param)
    {
        if(T::func1 is an existing function)
            param.func1();
    }
};

更好的是一个else- case：

Even better would be an else-case:

template<class T>
class A
{
    void f(T param)
    {
        if(T::func1 is an existing function)
            param.func1();
        else
            cout << "func1 doesn't exist" << endl;
    }
};

推荐答案

Boost.MPL不处理，因为它严格用于TMP，您不能在TMP中调用成员。 Boost.Fusion和Boost.TypeTraits没有任何东西;我认为其中一个会显然是我错误的。

Boost.MPL doesn't deal with that as it's strictly for TMP and you can't call members in TMP. Boost.Fusion and Boost.TypeTraits don't have anything either; I thought one of them would but apparently I'm misremembering.

这里和这里是一些解决方案，如何编写一个trait来检测C ++ 03中的成员。一旦你有这样的特质（我会称为 has_func1_member ），你可以使用它SFINAE：

Here and here are some solutions on how to write a trait to detect a member in C++03. Once you have such a trait (I'll call it has_func1_member), you can use it for SFINAE:

template<typename T>
typename boost::enable_if<has_func1_member<T> >::type
maybe_call(T& t)
{ t.func1(); }

template<typename T>
typename boost::disable_if<has_func1_member<T> >::type
maybe_call(T&)
{
    // handle missing member case
}

// your example code would then do:
maybe_call(param);

请注意，使用C ++ 11，首先写trait更容易，有点神秘。

Note that with C++11 it's easier to write the trait in the first place, although it's still somewhat arcane.

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