How to avoid decay with template parameter deduction

Question

Simplified:

// CHAR_TYPE == char, wchar_t, ...
template 
void Foo(CHAR_TYPE const (&value)[CHAR_COUNT]) no         


        

Answer 1

One idea that works is to remove the pointer and simply have T instead, with a std::enable_if_t<std::is_pointer<T>::value> guard. Simplified example below:

#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

Taking the argument by (const) reference blocks array-to-pointer decay during template argument deduction. See [temp.deduct.call]/2. So:

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