Why are some non-constant expressions allowed as operands of a constexpr comma operator?

Question

Consider a simple example:

int foo() {
    return 3;
}

template <int>
struct Bar {};

int a;

int main() {
    int b;
    //Bar<((void)foo(), 1)> bar1;  //case 1. compilation error as expected
    Bar<((void)a, 2)> bar2;        //case 2. no error (long shot but `a' has a linkage so maybe expected)
    Bar<((void)b, 3)> bar3;        //case 3. no error ? (`b' does not have linkage) 
    (void)bar2;
    (void)bar3;
}

I would say it is a bug but both the latest [clang] and [gcc] accept the code so maybe I'm missing some relevant standard rule that makes the code valid?

Answer 1

The lvalue-to-rvalue conversion is not applied to the first argument of the comma operator unless it is volatile. Therefore, (void)a, 2 and (void)b, 3 are constant expressions.

See [expr.comma]/1

... the left expression is a discarded-value expression ...

and [expr]/12

... The lvalue-to-rvalue conversion is applied [to a discarded-value expression] if and only if the expression is a glvalue of volatile-qualified type and it is one of the following: ...