問題描述

我正在嘗試編寫一個可變參數模板 constexpr 函數來計算給定模板參數的總和.這是我的代碼:

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，它在嘗試解析 f<3,>()error C2668: 'f': ambiguous call to重載函數> 打電話.

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

我還嘗試將遞歸基本情況更改為接受 0 個模板參數而不是 1 個:

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

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

但此代碼也無法編譯(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).

我可以提取第一個和第二個模板參數來編譯和工作，就像這樣:

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...>();
}

但這似乎不是最好的選擇.那么，問題是:如何以優雅的方式編寫此計算?

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

UP:我也試著把它寫成一個單一的函數:

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...>());
}

這也不起作用:error C2672:'f':找不到匹配的重載函數.

推薦答案

您的基本情況是錯誤的.您需要一個空列表的案例，但正如編譯器所建議的那樣，您的第二次嘗試不是有效的模板專業化.為零參數定義有效實例化的一種方法是創建一個接受空列表的重載

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;
}

<小時>

為了完整起見，也是我的第一個答案，@alexeykuzmin0 通過添加條件來修復:

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...>();
}

這篇關于如何編寫可變參數模板遞歸函數?的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！