C ++ std :: enable_if在类模板中，用于成员函数 [英] C++ std::enable_if in a class template, for a member function

问题描述

问题很简单。我有一个头文件包含保护和实现文件.impl。头文件包括实现文件。我要执行以下操作：

The problem is simple. I have a header file with include guards and a implementation file .impl. Header file includes implementation file. I want to do the following:

头文件：

template<size_t N>
class A
{
  void func();
};

.impl文件：

template<size_t N>
typename std::enable_if<(N <= 5), void>::type A<N>::func() { ... }

template<size_t N>
typename std::enable_if<(N > 5), void>::type A<N>::func() { ... }

然而，我还是不能用 std :: enable_if ，它似乎找不到原型 func ，因为我通过更改返回类型更改函数签名。

However I am yet to be good with std::enable_if and it seems to fail to find the prototype for func because I change the function signature by changing the return type. How can I have different implementations while providing the user with one function for the interface.

这实际上是MCU寄存器修改器，它对两个寄存器进行操作，因为它们没有容量。我宁愿不使用任何脏偏移量基于N在函数内部，并依赖于平面结构。

This is essentially MCU register modifier that operates on two registers because one does not have the capacity. I'd rather not use any dirty offsets based on N inside the function and depend on plain structure. Also I'd rather not use helper functions which would complicate things if it is possible without them.

推荐答案

你可以委托：

#include <iostream>

template<std::size_t N>
class A
{
    private:
    template <std::size_t I>
    typename std::enable_if<(I <= 5)>::type
    f() { std::cout << "Small\n"; }

    template <std::size_t I>
    typename std::enable_if<(I > 5)>::type
    f() { std::cout << "Big\n"; }

    public:
    void func() { f<N>(); }
};

int main()
{
    A<1> small;
    small.func();
    A<10> big;
    big.func();
}

如果您信任编译器优化：

And if you trust your compiler optimizations:

template<std::size_t N>
class A
{
    private:
    void small() { std::cout << "Small\n"; }
    void big() { std::cout << "Big\n"; }

    public:
    void func() {
        if(N <= 5) small();
        else big();
    }
};

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