================
@@ -14544,6 +14538,23 @@ EvaluateComparisonBinaryOperator(EvalInfo &Info, const
BinaryOperator *E,
(LHSValue.Base && isZeroSized(RHSValue)))
return DiagComparison(
diag::note_constexpr_pointer_comparison_zero_sized);
+ // A constexpr-unknown reference can be equal to any other lvalue, except
+ // for variables allocated during constant evaluation. (The "lifetime
+ // [...] includes the entire constant evaluation", so it has to be
+ // distinct from anything allocated during constant evaluation.)
+ //
+ // Theoretically we could handle other cases, but the standard doesn't
say
+ // what other cases we need to handle; it just says an "equality
+ // operator where the result is unspecified" isn't a constant expression.
----------------
efriedma-quic wrote:
When we know two pointers point into same object, pointer equality is trivial.
When we know two pointers point into different objects, we can follow the rules
you outline.
The problem is if we don't know whether the two pointers point into different
objects. Then weird things start happening, and you have to start digging into
other sections of the standard. Like, consider:
```
void f() {
void* x = malloc(4);
int* ip = new(x)ip;
int & ir = *ip;
float *fp = new(x)fp;
float &fr = *fp;
static_assert(&ir != &fr)
}
```
If you take the rules literally, the static_assert is valid, and succeeds.
"the reference is treated as binding to an unspecified object of the referenced
type whose lifetime and that of all subobjects includes the entire constant
evaluation"... and a given object can only have one type, so the two objects
must be distinct, so the pointers can't be equal.
It would help if the standard said that an equality comparison is not a core
constant expression if it contains "an equality comparison where one of the
operands points into a constexpr-unknown object".
https://github.com/llvm/llvm-project/pull/147663
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