如何编写可变参数模板递归函数?

2021-12-16 00:00:00 递归 c++ c++11 variadic-functions variadic-templates

我正在尝试编写一个可变参数模板 constexpr 函数来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，它在尝试解析 f<3,>()error C2668: 'f': ambiguous call to重载函数> 打电话.

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 error C2912: explicit specialization 'int f(void)' is not a specialization of a function template).

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参数来编译和工作，就像这样:

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...>();
}

但这似乎不是最好的选择.那么，问题是:如何以优雅的方式编写此计算?

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

UP:我也试着把它写成一个单一的函数:

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:error C2672:'f':找不到匹配的重载函数.

推荐答案

您的基本情况是错误的.您需要一个空列表的案例，但正如编译器所建议的那样，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;
}

<小时>

为了完整起见，也是我的第一个答案，@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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