Why is implicit conversion not applied to templated function parameter?

问题

I'm having an issue with some template stuff that I've narrowed down to the following example (C++17):

template <typename T> struct item {
  operator item<const T> () const { return item<const T>(); }
};

void conversionToConstRefWorks (const item<const int> &) { }

template <typename T> 
void butNotWhenTemplated (const item<const T> &) { }

int main () {

  item<int> i;
  item<const int> ci;

  // these all compile fine:
  conversionToConstRefWorks(ci);
  conversionToConstRefWorks(i);
  butNotWhenTemplated(ci);

  // but this one fails:
  butNotWhenTemplated(i); 

}

In that example:

item<T> has an implicit conversion operator to item<const T>, and
The conversion seems to work in conversionToConstRefWorks(), but
The conversion seems to be missed in butNotWhenTemplated(), where an item<const int> can be passed just fine but passing an item<int> fails to compile.

Compilation of that example fails (GCC 9.3) with:

g++ --std=c++17 -W -Wall -pedantic -Wno-unused-variable    const_interop.cpp   -o const_interop
const_interop.cpp: In function ‘int main()’:
const_interop.cpp:54:24: error: no matching function for call to ‘butNotWhenTemplated(item<int>&)’
   54 |   butNotWhenTemplated(i);
      |                        ^
const_interop.cpp:40:6: note: candidate: ‘template<class T> void butNotWhenTemplated(const item<const T>&)’
   40 | void butNotWhenTemplated (const item<const T> &) {
      |      ^~~~~~~~~~~~~~~~~~~
const_interop.cpp:40:6: note:   template argument deduction/substitution failed:
const_interop.cpp:54:24: note:   types ‘const T’ and ‘int’ have incompatible cv-qualifiers
   54 |   butNotWhenTemplated(i);
      |                        ^

The root error seems to be:

types ‘const T’ and ‘int’ have incompatible cv-qualifiers

I understand what that means in a literal sense, but I don't understand why it is happening. My expectation is that the item<int> :: operator item<const int> () const conversion operator would be applied when calling butNotWhenTemplated(i) just as it was applied when calling conversionToConstRefWorks(i), and that int would be selected for T.

My main question is: Why isn't this compiling?

My other question is: For reasons outside the scope of this post, butNotWhenTemplated has to be a template and has to specify <const T> to all item parameters, and I can't explicitly specify template parameters when calling it. Is there a way to make this work with those constraints?

Here it is on ideone (GCC 8.3).

回答1:

item<int> i;
template <typename T> void butNotWhenTemplated (const item<const T> &) { }
butNotWhenTemplated(i);

According to template argument substitution rules, no T could be found for item<const T> to match item<int>. This fails with an hard error way before any conversion (builtin or user-defined) could be considered.

Type deduction does not consider implicit conversions (other than type adjustments listed above): that's the job for overload resolution, which happens later. However, if deduction succeeds for all parameters that participate in template argument deduction, and all template arguments that aren't deduced are explicitly specified or defaulted, then the remaining function parameters are compared with the corresponding function arguments.

回答2:

Try this overload:

template <typename T>
void butNotWhenTemplated(const item<const T>&) { }

template <typename T>
void butNotWhenTemplated(const item<T>& x) {
    butNotWhenTemplated<const T>(x);
}

Addendum:

You're trying to pass by reference to const, but implicit conversion creates a copy of your object, even in the non-template case. You might want to rethink your design here.

来源：https://stackoverflow.com/questions/65401573/why-is-implicit-conversion-not-applied-to-templated-function-parameter

标签

c++

templates

gcc

c++17

implicit-conversion