Why doesn't an if constexpr make this core constant expression error disappear?

Question

In reference to this question. The core constant expression that is used to initialize the constexpr variable y is ill-formed. So much is a given.<

Answer 1

The standard doesn't say much about the discarded statement of an if constexpr. There are essentially two statements in [stmt.if] about these:

1. In an enclosing template discarded statements are not instantiated.
2. Names referenced from a discarded statement are not required ODR to be defined.

Neither of these applies to your use: the compilers are correct to complain about the constexpr if initialisation. Note that you'll need to make the condition dependent on a template parameter when you want to take advantage of the instantiation to fail: if the value isn't dependent on a template parameter the failure happens when the template is defined. For example, this code still fails:

template 
void f() {
    constexpr int x = -1;
    if constexpr (x >= 0){
        constexpr int y = 1<

However, if you make x dependent on the type T it is OK, even when f is instantiated with int:

template 
void f() {
    constexpr T x = -1;
    if constexpr (x >= 0){
        constexpr int y = 1<();
}