1
Fork 0

don't allow ZST in ScalarInt

There are several indications that we should not ZST as a ScalarInt:
- We had two ways to have ZST valtrees, either an empty `Branch` or a `Leaf` with a ZST in it.
  `ValTree::zst()` used the former, but the latter could possibly arise as well.
- Likewise, the interpreter had `Immediate::Uninit` and `Immediate::Scalar(Scalar::ZST)`.
- LLVM codegen already had to special-case ZST ScalarInt.

So instead add new ZST variants to those types that did not have other variants
which could be used for this purpose.
This commit is contained in:
Ralf Jung 2022-07-03 11:17:23 -04:00
parent c4693bc946
commit a422b42159
21 changed files with 78 additions and 61 deletions

View file

@ -49,11 +49,22 @@ impl<'a, 'tcx> Builder<'a, 'tcx> {
inferred_ty: ty,
})
});
let literal = ConstantKind::Val(ConstValue::Scalar(Scalar::Int(lit)), ty);
Constant { span, user_ty: user_ty, literal }
}
ExprKind::ZstLiteral { user_ty } => {
let user_ty = user_ty.map(|user_ty| {
this.canonical_user_type_annotations.push(CanonicalUserTypeAnnotation {
span,
user_ty,
inferred_ty: ty,
})
});
let literal = ConstantKind::Val(ConstValue::ZST, ty);
Constant { span, user_ty: user_ty, literal }
}
ExprKind::NamedConst { def_id, substs, user_ty } => {
let user_ty = user_ty.map(|user_ty| {
this.canonical_user_type_annotations.push(CanonicalUserTypeAnnotation {