Why initialization of constexpr variable from non constexpr variable succeeds in case of class

Question

I have learnt about constexpr variables in c++ and read that int i =0; constexpr int j = i; fails because i is not a constant expression which makes sense. But then when I did the same with a variable of a class, it worked.

struct C{};
int main()
{
    int i = 0; //i is not a constant expression as expected 
    //constexpr int j = i; //this fails AS EXPECTED

    C c; 
    constexpr C d = c; //WHY DOESN'T THIS FAIL?? 
}

As you can see, constexpr C d = c; compiles without any problem unlike constexpr int j = i; even though c is also not a constant expression.

I want to know the reason behind this.

Answer 1

I want to know the reason behind this.

Because the class C has an implicit constexpr copy constructor constexpr C::C(const C&) that can be used in constexpr context like this.

Basically the initialization constexpr C d = c; is a constexpr context and the constexpr ctors are designed exactly for these purposes.

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You can also see this by looking at the generated code at cppinsights:

struct C
{
  // inline constexpr C() noexcept = default;
  // inline constexpr C(const C &) noexcept = default;
};

As we can see in the above generated code, C has implicitly declared constexpr constructors that can be used in constexpr contexts.

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From constexpr:

A constexpr variable must satisfy the following requirements:

• ...
• the full-expression of its initialization, including all implicit conversions, constructors calls, etc, must be a constant expression