如何编写可变参数模板的递归函数？ [英] How to write a variadic template recursive function?

问题描述

我正在尝试编写可变参数模板 constexpr 函数，该函数计算给定模板参数的总和。这是我的代码：

I'm trying to write a variadic template constexpr function which calculates sum of the template parameters given. Here's my code:

template<int First, int... Rest>
constexpr int f()
{
    return First + f<Rest...>();
}

template<int First>
constexpr int f()
{
    return First;
}

int main()
{
    f<1, 2, 3>();
    return 0;
}

不幸的是，它不会编译并报告错误消息错误C2668： f：尝试解析 f< 3，>（）调用时对重载函数的模糊调用。

Unfortunately, it does not compile reporting an error message error C2668: 'f': ambiguous call to overloaded function while trying to resolve f<3,>() call.

我还尝试将递归基本情况更改为接受0个模板参数，而不是1个：

I also tried to change my recursion base case to accept 0 template arguments instead of 1:

template<>
constexpr int f()
{
    return 0;
}

但此代码也无法编译（消息错误C2912：显式专业化'int f（void）'不是函数模板的专业化）。

But this code also does not compile (message error C2912: explicit specialization 'int f(void)' is not a specialization of a function template).

我可以提取第一个和第二个模板使其编译和工作的参数，例如：

I could extract first and second template arguments to make this compile and work, like this:

template<int First, int Second, int... Rest>
constexpr int f()
{
    return First + f<Second, Rest...>();
}

但这似乎不是最佳选择。因此，问题是：如何以一种简洁的方式编写此计算？

But this does not seem to be the best option. So, the question is: how to write this calculation in an elegant way?

UP ：我也尝试将其作为单个函数编写：

UP: I also tried to write this as a single function:

template<int First, int... Rest>
constexpr int f()
{
    return sizeof...(Rest) == 0 ? First : (First + f<Rest...>());
}

这也不起作用：错误C2672： 'f'：找不到匹配的重载函数。

推荐答案

您的基本情况错误。您需要为空列表添加一个大小写，但是正如编译器建议的那样，您的第二次尝试不是有效的模板专业化。为零参数定义有效实例化的一种方法是创建一个接受空列表的重载

Your base case was wrong. You need a case for the empty list, but as the compiler suggests, your second try was not a valid template specialization. One way to define a valid instantiation for zero arguments is to create an overload that accepts an empty list

template<class none = void>
constexpr int f()
{
    return 0;
}
template<int First, int... Rest>
constexpr int f()
{
    return First + f<Rest...>();
}
int main()
{
    f<1, 2, 3>();
    return 0;
}

---

编辑：出于完整性考虑还有我的第一个答案，即@ alexeykuzmin0通过添加条件来修复：

---

for completeness sake also my first answer, that @alexeykuzmin0 fixed by adding the conditional:

template<int First=0, int... Rest>
constexpr int f()
{
    return sizeof...(Rest)==0 ? First : First + f<Rest...>();
}

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