[DannyBee]

"The GP post is explicitly claiming that the values don't matter, and that per the standard a constexpr function should be guaranteed by the compiler to be incapable of producing UB (or else compilation should fail and it should not be allowed to be declared constexpr).

They are so guaranteed - as i cited to you, your example of foo, as used in foo(100), is considered not a constant. There is no such thing as a constexpr expression that can produce UB.

If it can produce UB, it is not constexpr.

That does not mean it will not be validly usable in non-constant contexts! But it is not constexpr.

[tsimionescu]

But foo(1) is a constant expression, and the compiler accepts it.

Not sure what you mean by a "constexpr expression" exactly. My point is that a function can legally be constexpr-qualified even if it can produce UB when called with certain arguments. I think it's very common to call such a function "a constexpr function". Of course, we agree that not every expression where a constexpr function is called is a constant expression (such as foo(100) in my example). However, expressions where that function is called with arguments where it does not produce UB are legal constant expressions (such as foo(1) in my example).

Perhaps I misunderstood your original comment and we are in fact in violent agreement. If so I apologize.