Why does std::panic::Location::caller not return an Option?

Question

The definition of Location::caller states:

pub fn caller() -> &'static Location<'static>`
This is a nightly-only experimental API. (track_caller #47809)

Returns the source location of the caller of this function. If that function's caller is annotated then its call location will be returned, and so on up the stack to the first call within a non-tracked function body.

That sounds to me like it should return &'static Option<Location<'static>>. Anything else seems to make walking the call stack much harder than it could be.

Is there a good reason for this decision?

You mention walking the call stack, but as far as I know that's not the usecase of this API. Given the usage of finding the specific caller of a given callsite of `Location::caller()`, what is the case where `None` would be the response value? — loganfsmyth, Mar 17 '20 at 03:20
Ah, you're right I was thinking the function was `caller(&self) -> Location` and that you'd get the root Location from the `PanicInfo::location`, so that you could walk the callstack via `while(location = location.caller()) { ... }`. But it doesn't give you that, just one point somewhere up the stack. — Michael Anderson, Mar 17 '20 at 05:30

score 2 · Accepted Answer · answered Mar 17 '20 at 17:39

It's impossible for a running function to not have a caller, so there's no time where None would be a possible return value.

You may be confused by the seemingly-poorly-worded clause:

If that function's caller is annotated [sic] then its call location will be returned

That probably should be worded as "annotated with #[track_caller]". What that means is that when the attribute is present, it's as if the function didn't exist and was textually inlined where it was called from.

Why does std::panic::Location::caller not return an Option?

1 Answers1