Why can't the compiler deduce a template parameter from a conditional class type argument?

Question

I am surprised that this code doesn't compile:

#include <type_traits>

template <int number, bool isTest1>
class TestClass1
{
};

template <int number, bool isTest1>
class TestClass2
{
};

template <int number, bool isTest1>
using MyTestClass = typename std::conditional<true, TestClass1<number, isTest1>, TestClass2<number, isTest1>>::type;

template <int number, bool isTest1>
auto createTest() -> MyTestClass<number, isTest1>
{
    return MyTestClass<number, isTest1>();
}

template <int number, bool isTest1>
auto function(int something, const MyTestClass<number, isTest1>& test) -> int
{
    return number;
}

int main()
{
    const MyTestClass<2, true> test = createTest<2, true>();
    function(123, test);

    // This works:
    // function<2, true>(123, test);

    return 0;
}

The error message is:

main.cpp:31:23: error: no matching function for call to 'function(int, MyTestClass<2, true>&)'
   31 |     function(123, test);
      |                       ^
main.cpp:23:6: note: candidate: 'template<int number, bool isTest1> int function(int, MyTestClass<number, isTest1>&)'
   23 | auto function(int something, const MyTestClass<number, isTest1>& test) -> int
      |      ^~~~~~~~
main.cpp:23:6: note:   template argument deduction/substitution failed:
main.cpp:31:23: note:   couldn't deduce template parameter 'number'
   31 |     function(123, test);
      |                       ^

See in Compiler Explorer: https://godbolt.org/z/71avfordM

I am surprised because everything seems to be there for template parameter deduction. I can spot no ambiguity or whatsoever. Does the compiler have a good reason to refuse my code?

Why can't the compiler deduce a template parameter from a conditional class type argument?

0 Answers0