c++ - SFINAE 用作返回类型，但不是参数类型

Question

我正在尝试使用 enable_if以避免重复代码。只要放在返回类型中它就可以正常工作，但如果放在参数中则不行。在这作为 this 的拷贝关闭之前，我得到的错误不是重新定义，而是“没有匹配的调用函数”。这是我使用 VS2015 和 g++ 7.2.0 (mingw) 的 MCVE(不是“C”，也不是“M”):

#include <cmath>
#include <array>
#include <algorithm>


template <typename T, size_t M, size_t N>
class Matrix
{
public:
    static const size_t ROWS = M;
    static const size_t COLS = N;
    typedef T SCALAR;

    SCALAR operator[](const size_t index) const
    {
        static_assert((COLS == 1 || ROWS == 1), "operator[] is only for vectors (single row or column).");
        return m_elements.at(index);
    }

    SCALAR& operator[](const size_t index)
    {
        static_assert((COLS == 1 || ROWS == 1), "operator[] is only for vectors (single row or column).");
        return m_elements.at(index);
    }

    std::array<T, M * N> m_elements;
};

template <typename T, size_t N, size_t M>
static inline T Length(
    const Matrix<typename std::enable_if<(M == 1 || N == 1), T>::type, N, M> & input)
{
    T value = 0;
    for (size_t i = 0; i < std::max(N, M); ++i)
    {
        value += (input[i] * input[i]);
    }
    return std::sqrt(value);
}

template <typename T, size_t M, size_t N>
static inline 
Matrix<typename std::enable_if<(M == 3 && N == 1) || (M == 1 && N == 3), T>::type , M, N>
CrossProduct(const Matrix<T, M, N> & a, const Matrix<T, M, N> & b)
{
    Matrix<T, M, N> result;
    result[0] = a[1] * b[2] - a[2] * b[1];
    result[1] = a[2] * b[0] - a[0] * b[2];
    result[2] = a[0] * b[1] - a[1] * b[0];
    return result;
}


Matrix<double, 1, 1> m11;
Matrix<double, 3, 1> m31;
Matrix<double, 1, 3> m13;
Matrix<double, 3, 3> m33;


auto l0 = Length(m11);  // Should work, but doesn't: no matching function for call to 'Length(Matrix<double, 1, 1>&)'
auto l1 = Length(m31);  // Should work, but doesn't: no matching function for call to 'Length(Matrix<double, 3, 1>&)'
auto l2 = Length(m13);  // Should work, but doesn't: no matching function for call to 'Length(Matrix<double, 1, 3>&)'
//auto l3 = Length(m33);    // Shouldn't work, and doesn't: no matching function for call to 'Length(Matrix<double, 3, 3>&)'

auto v1 = CrossProduct(m13, m13); //Works, as expected
//auto v2 = CrossProduct(m11, m11); // As expected: enable_if.cpp:71:32: error: no matching function for
                                    // call to 'CrossProduct(Matrix<double, 1, 1>&, Matrix<double, 1, 1>&)'

如果我更改 Length 的签名到

static inline typename std::enable_if<(M == 1 || N == 1), T>::type \
Length(const math::Matrix<T, N, M> & input)

它工作正常。但它给我的错误似乎表明它能够确定正确的签名(例如 Length(Matrix<double, 3, 1>&) )。

如果 enable_if 编译器为什么找不到匹配的函数在参数列表中，但如果它在返回类型中则能够？

Answer 1

std::enable_if<..., T>::type是一个嵌套名称，因此无法推导出:

参见 [temp.deduct.type]/5 :

The non-deduced contexts are:

— The nested-name-specifier of a type that was specified using a qualified-id.

. . .

作为解决方法，移动 enable_if到一个单独的模板参数:

template <typename T, size_t N, size_t M, typename std::enable_if<(M == 1 || N == 1), int>::type = 0>
static inline T Length(
    const Matrix<T, N, M> & input)