Why does if constexpr require an else to work?

Question

I am trying to use if constexpr in the following way:

template<template <typename First, typename Second> class Trait,
    typename First, typename Second, typename... Rest>
constexpr bool binaryTraitAre_impl()
{
    if constexpr (sizeof... (Rest) == 0)
    {
        return Trait<First, Second>{}();    
    }
    return Trait<First, Second>{}() and binaryTraitAre_impl<Trait, Rest...>();
}

Example use case:

static_assert(binaryTraitAre_impl<std::is_convertible,
    int, int&,
    int*, void*>());

But this fails to compile

clang:

error: no matching function for call to 'binaryTraitAre_impl'
        return Trait<First, Second>{}() and binaryTraitAre_impl<Trait, Rest...>();
                                            ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

gcc:

prog.cc: In instantiation of 'constexpr bool binaryTraitAre_impl() [with Trait = std::is_convertible; First = int*; Second = void*; Rest = {}]':
prog.cc:9:80:   required from 'constexpr bool binaryTraitAre_impl() [with Trait = std::is_convertible; First = int; Second = int&; Rest = {int*, void*}]'
prog.cc:15:83:   required from here
prog.cc:9:80: error: no matching function for call to 'binaryTraitAre_impl<template<class _From, class _To> struct std::is_convertible>()'
    9 |         return Trait<First, Second>{}() and binaryTraitAre_impl<Trait, Rest...>();
      |                                             ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~
prog.cc:3:17: note: candidate: 'template<template<class First, class Second> class Trait, class First, class Second, class ... Rest> constexpr bool binaryTraitAre_impl()'
    3 |  constexpr bool binaryTraitAre_impl()
      |                 ^~~~~~~~~~~~~~~~~~~
prog.cc:3:17: note:   template argument deduction/substitution failed:
prog.cc:9:80: note:   couldn't deduce template parameter 'First'
    9 |         return Trait<First, Second>{}() and binaryTraitAre_impl<Trait, Rest...>();
      |                                             ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~

But I found the error goes away once I add else:

template<template <typename First, typename Second> class Trait,
    typename First, typename Second, typename... Rest>
constexpr bool binaryTraitAre_impl()
{
    if constexpr (sizeof... (Rest) == 0)
    {
        return Trait<First, Second>{}();
    }
    else
    {
        return Trait<First, Second>{}() and binaryTraitAre_impl<Trait, Rest...>();
    }
}

live demo

What happened? Why can the compiler not infer the else in this case?

Answer 1

This is the excerpt from cppreference on constexpr if:

Constexpr If The statement that begins with if constexpr is known as the constexpr if statement.

In a constexpr if statement, the value of condition must be a contextually converted constant expression of type bool. If the value is true, then statement-false is discarded (if present), otherwise, statement-true is discarded.

It is clear that only one of the two branches is discarded. In your case, the culprit code cannot be discarded because it's outside the else clause.