c++ - 从一个值和一个大小创建递归函数参数列表

Question

我的目的是创建一个大小为 n 的函数参数列表，以便我可以将其传递给使用折叠表达式以递归方式将这些值相乘的助手。

我对如何将参数列表传递给助手有点困惑。 有没有办法在没有包表达式的情况下创建函数参数列表？ 也许通过创建数组或元组？

这是我到目前为止所想到的。

template<typename T, typename N>
T SmoothStart(const T& t, const N& n) {
    static_assert(std::is_integral_v<N>, "templatized SmoothStart requires type of N to be integral.");
    static_assert(n >= 0, "templatized SmoothStart requires value of N to be non-negative.");

    if constexpr (n == 0) {
        return 1;
    }
    if constexpr (n == 1) {
        return t;
    }
    return SmoothStart_helper((t, ...)); //<-- obviously this doesn't work but it would be awesome to have!
}

template<typename T, typename... Args>
T SmoothStart_helper(Args&&... args) {
    return (args * ...);
}

Answer 1

首先，n如果要使用折叠表达式，则必须在编译时知道。如果将其移至模板参数，则获得大小为 N 的参数包的最简单方法与 std::make_index_sequence :

// The helper has to be first so that the compiler can find SmoothStart_helper().
template<typename T, std::size_t... Is>
T SmoothStart_helper(const T& t, std::index_sequence<Is...>) {
    // You were taking t by value here; I think you might want to still
    // take it by reference

    // Use the comma operator to simply discard the current index and instead
    // yield t. The cast to void is to silence a compiler warning about
    // Is being unused
    return (((void) Is, t) * ...);
}

template<std::size_t N, typename T>
T SmoothStart(const T& t) {
    // std::size_t is unsigned, so no need to check for N >= 0.
    // We also don't need to special case when N == 1. The fold
    // expression handles that case and just returns t
    return SmoothStart_helper(t, std::make_index_sequence<N>{});
}

然后您可以像这样使用它:SmoothStart<N>(myThing); .

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