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# development
f

fast-nail-55400

03/04/2022, 7:48 PM
@ancient-vegetable-10556: Is there an easy way to get the output filename(s) for a
ClasspathEntry
without actually building the underlying code?
w

witty-crayon-22786

03/04/2022, 7:58 PM
@ancient-vegetable-10556: the fact that exports are merged in will make it a bit trickier.
what’s the usecase?
f

fast-nail-55400

03/04/2022, 8:07 PM
BSP has to report the classpath for a build target via a metadata query before a compile is requested
(since I opted for now to output compiled thin jars for a target via the “classpath” attribute and not the “class directory” (unpacked classfiles) attribute in that metadata
I’ve copied the relevant line from the scalac rules into the BSP scala metadata rule. So no need to solve anything generally right now.
subject to change as I gain more experience with how to properly map into IntelliJ’s data model
w

witty-crayon-22786

03/04/2022, 8:10 PM
hm. and i guess scala doesn’t have exports. but yea, will affect java
Ah, you know what though: for
exports
, you can see how the Bazel BSP server does it, since they have explicit exports
👍 1