Does this "interpreted as a constant-expression" wording mean that the following will be treated as a trivial infinite loop?

(Note that std::is_constant_evaluated() returns true when evaluated in a constant-expression, and false when not manifestly constant-evaluated.)

#include  <type_traits>
int main() {
    while (std::is_constant_evaluated());
}

Yes, that is a "trivial infinite loop". That means, optimizations are reduced around that loop and each step of the loop execution will call yield. Except that the body of the loop will never actually be executed at runtime.