Home SFINAE works as return type, but not parameter type
Reply: 1

SFINAE works as return type, but not parameter type

Avi Ginsburg
1#
Avi Ginsburg Published in 2018-01-10 13:26:47Z

I was trying to use enable_if to avoid duplicating code. It works fine as long as placed in the return type, but not if it's in the parameters. Before this gets closed as a dup of this, the error I'm getting is not a redefinition, but a "no matching function for call." Here's my MCVE (not so "C", or so "M" for that matter) using VS2015 and g++ 7.2.0 (mingw):

#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>&)'

If I change the signature of Length to

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

it works fine. But the error it gives me seems to indicate that it was able to determine the correct signature (e.g. Length(Matrix<double, 3, 1>&)).

Why is the compiler unable to find a matching function if the enable_if is in the parameter list, but is able to if it's in the return type?

RustyX
2#
RustyX Reply to 2018-01-10 15:32:37Z

std::enable_if<..., T>::type is a nested name, and as such cannot be deduced:

See [temp.deduct.type]/5:

The non-deduced contexts are:

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

. . .

As a workaround, move the enable_if to a separate template argument:

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)
You need to login account before you can post.

About| Privacy statement| Terms of Service| Advertising| Contact us| Help| Sitemap|
Processed in 0.303953 second(s) , Gzip On .

© 2016 Powered by mzan.com design MATCHINFO