C ++ 11:解决方法使用此不完整的类型错误? [英] C++11: Workaround Use Of This Incomplete Type Error?

问题描述

#include <iostream>
#include <array>
using namespace std;

constexpr int N = 10;
constexpr int f(int x) { return x*2; }

typedef array<int, N> A;

template<int... i> struct F { constexpr A f() { return A{{ f(i)... }}; } };

template<class X, class Y> struct C;
template<int... i, int... j>
struct C<F<i...>, F<j...>> : F<i..., (sizeof...(i)+j)...> {};

template<int n> struct S : C<S<n/2>, S<n-n/2>> {}; // <--- HERE
template<> struct S<1> : F<0> {};

constexpr auto X = S<N>::f();

int main()
{
        cout << X[3] << endl;
}

我得到了:

test.cpp:15:24: error: invalid use of incomplete type ‘struct C<S<5>, S<5> >’

我怀疑这是因为S的定义将自身用作基类.(正确吗?)

I suspect this is because the definition of S is using itself as a base class. (Correct?)

解决此问题的最佳方法是什么?

What is the best way to fix this?

更新:

这是固定版本:

#include <iostream>
#include <array>
using namespace std;

constexpr int N = 10;
constexpr int f(int x) { return x*2; }

typedef array<int, N> A;

template<int... i> struct F { static constexpr A f() { return A{{ ::f(i)... }}; } };

template<class A, class B> struct C {};
template<int... i, int... j> struct C<F<i...>, F<j...>> : F<i..., (sizeof...(i)+j)...>
{
        using T = F<i..., (sizeof...(i)+j)...>;
};

template<int n> struct S : C<typename S<n/2>::T, typename S<n-n/2>::T> {};
template<> struct S<1> : F<0> { using T = F<0>; };

constexpr auto X = S<N>::f();

int main()
{
        cout << X[3] << endl;
}

推荐答案

定义 C 而不是仅仅声明它.

Define C instead of just declaring it.

template<class X, class Y> struct C {};

在使用它的地方，部分专业化不匹配，并且主模板被实例化，这只是一个声明.

In the place you use it the partial specialization does not match and the primary template is instantiated, which is just a declaration.

您可能想知道为什么不考虑专业化:专业化不考虑转换，而只考虑静态类型.这就是为什么它们与继承是如此背叛的原因.

You may wonder why that specialization is not considered: specializations don't consider conversions, but just the static type. That's why they are so treacherously incompatible with inheritance.

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