Workaround the case when the `this` pointer is actually a `NonLoc`, by
returning `Unknown` instead.
The solution isn't ideal, as `this` should be really a `Loc`, but due to
how casts work, I feel this is our easiest and best option.
As this patch presents, I'm evaluating a cast to transform the `NonLoc`.
However, given that `evalCast()` can't be cast from `NonLoc` to a
pointer type thingy (`Loc`), we end up with `Unknown`.
It is because `EvalCastVisitor::VisitNonLocSymbolVal()` only evaluates
casts that happen from NonLoc to NonLocs.
When I tried to actually implement that case, I figured:
1) Create a `SymbolicRegion` from that `nonloc::SymbolVal`; but
`SymbolRegion` ctor expects a pointer type for the symbol.
2) Okay, just have a `SymbolCast`, getting us the pointer type; but
`SymbolRegion` expects `SymbolData` symbols, not generic `SymExpr`s, as
stated:
> // Because pointer arithmetic is represented by ElementRegion layers,
> // the base symbol here should not contain any arithmetic.
3) We can't use `ElementRegion`s to perform this cast because to have an
`ElementRegion`, you already have to have a `SubRegion` that you want to
cast, but the point is that we don't have that.
At this point, I gave up, and just left a FIXME instead, while still
returning `Unknown` on that path.
IMO this is still better than having a crash.
Fixes#69922
Previously `LValueToRValueBitCast`s were modeled in the same way how
a regular `BitCast` was. However, this should not produce an l-value.
Modeling bitcasts accurately is tricky, so it's probably better to
model this expression by binding a fresh conjured value.
The following code should not result in a diagnostic:
```lang=C++
__attribute__((always_inline))
static inline constexpr unsigned int_castf32_u32(float __A) {
return __builtin_bit_cast(unsigned int, __A); // no-warning
}
```
Previously, it reported
`Address of stack memory associated with local variable '__A' returned
to caller [core.StackAddressEscape]`.
Differential Revision: https://reviews.llvm.org/D105017
Reviewed by: NoQ, vsavchenko
