constexpr result from non-constexpr call

Question

Recently I was surprised that the following code compiles in clang, gcc and msvc too (at least with their current versions).

struct A {
    static const int value = 42;
};

constexpr int f(A a) { return a.value; }

void g() {
    A a;  // Intentionally non-constexpr.
    constexpr int kInt = f(a);
}

My understanding was that the call to f is not constexpr because the argument i isn't, but it seems I am wrong. Is this a proper standard-supported code or some kind of compiler extension?

Answer 1

As mentioned in the comments, the rules for constant expressions do not generally require that every variable mentioned in the expression and whose lifetime began outside the expression evaluation is constexpr.

There is a (long) list of requirements that when not satisfied prevent an expression from being a constant expression. As long as none of them is violated, the expression is a constant expression.

The requirement that a used variable/object be constexpr is formally known as the object being usable in constant expressions (although the exact definition contains more detailed requirements and exceptions, see also linked cppreference page).

Looking at the list you can see that this property is required only in certain situations, namely only for variables/objects whose lifetime began outside the expression and if either a virtual function call is performed on it, a lvalue-to-rvalue conversion is performed on it or it is a reference variable named in the expression.

Neither of these cases apply here. There are no virtual functions involved and a is not a reference variable. Typically the lvalue-to-rvalue conversion causes the requirement to become important. An lvalue-to-rvalue conversions happens whenever you try to use the value stored in the object or one of its subobjects. However A is an empty class without any state and therefore there is no value to read. When passing a to the function, the implicit copy constructor is called to construct the parameter of f, but because the class is empty, it doesn't actually do anything. It doesn't access any state of a.

Note that, as mentioned above, the rules are stricter if you use references, e.g.

A a;
A& ar = a;
constexpr int kInt = f(ar);

will fail, because ar names a reference variable which is not usable in constant expressions. This will hopefully be fixed soon to be more consistent. (see https://github.com/cplusplus/papers/issues/973)