如何通过模板参数推导避免衰减

Question

简化:

// CHAR_TYPE == char, wchar_t, ...
template <typename CHAR_TYPE, unsigned CHAR_COUNT>
void Foo(CHAR_TYPE const (&value)[CHAR_COUNT]) noexcept
{
    TRACE("const ref array");
    // perform a bit of logic and forward...
}

template <typename CHAR_TYPE>
void Foo(CHAR_TYPE const* value) noexcept
{
    TRACE("const ptr");
    // perform a bit of logic and forward...
}

// [ several other overloads ]

调用点:

char const* ptr = ...
wchar_t const* wptr = ...

Foo(ptr);     // <-- good: "const ptr"
Foo(wptr);    // <-- good: "const ptr"

constexpr char const buffer[] { "blah blah blah" };
constexpr wchar_t const wbuffer[] { L"blah blah blah" };

Foo(buffer);  // <-- ambiguous
Foo(wbuffer); // <-- ambiguous

当然,我可以删除const ref数组重载.但是我想以不同的方式处理这些类型.我试图有条件地启用正确的过载,但我无法确定必要的条件.

template <typename CHAR_TYPE, unsigned COUNT>
typename std::enable_if</* std::is_?? */, void>::type
    Foo(CHAR_TYPE const (&value)[COUNT]) noexcept
{
    TRACE("by ref array");
    // perform a bit of logic and forward...
}

template <typename CHAR_TYPE>
typename std::enable_if</* std::is_?? */, void>::type
    Foo(CHAR_TYPE const* value) noexcept
{
    TRACE("ptr");
    // perform a bit of logic and forward...
}

消除这些重载歧义的最佳方法是什么？
(我不想使用数组包装器)

Answer 1

一个有效的想法是删除指针,而只需要T一个std::enable_if_t<std::is_pointer<T>::value>后卫.简化示例如下:

#include <iostream>
#include <type_traits>

template<class T, size_t N>
void f(T const (&) [N])
{
    std::cout << __PRETTY_FUNCTION__ << std::endl;
}

template<class T, std::enable_if_t<std::is_pointer<T>::value>* = nullptr >
void f(T)
{
    std::cout << __PRETTY_FUNCTION__ << std::endl;
}

int main()
{
    const char* str = "test";
    char str2[]{"test2"};

    f(str);
    f(str2);
}

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Answer 2

采用(const)引用的参数在模板参数推导期间阻止数组到指针的衰减.见[temp.deduct.call]/2.所以:

template <typename CHAR_TYPE>
void Foo(CHAR_TYPE const* const & value) noexcept
{
    TRACE("const ptr");
    // perform a bit of logic and forward...
}