missing_const_for_fn: consider constness of instance
#14759
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When determining when a function or method can be called from a `const` context, the determination must be made on the instance, not on the declaration. This makes a difference, for example, with `const_trait` traits whose implementations may or may not be `const`.
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May 14, 2025
| if let ty::FnDef(fn_def_id, fn_substs) = *fn_ty.kind() { | ||
| let fn_def_id = match Instance::try_resolve(cx.tcx, cx.typing_env(), fn_def_id, fn_substs) { | ||
| Ok(Some(fn_inst)) => fn_inst.def_id(), | ||
| Ok(None) => return Err((span, format!("cannot resolve instance for {func:?}").into())), |
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Not being able to resolve to an instance doesn't mean the function can't be const. Not sure what the syntax is but there should be some way to constrain the type to implement the trait as a const trait.
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I don't see how to do that.
I think this is still better to get false negatives rather than false positives, especially when none of the tested cases fails because of this change. There were also no lintcheck misses.
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I think the current form of const traits is recentish so I wouldn't expect many uses. Can you just add a comment about this doesn't handle where predicates?
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When determining when a function or method can be called from a
constcontext, the determination must be made on the instance, not on the declaration. This makes a difference, for example, withconst_traittraits whose implementations may or may not beconst.changelog: [
missing_const_for_fn]: when checking if a function or method can be called from aconstcontext, look at the concrete implementation rather than at the trait definitionFixes #14658
r? @Jarcho