Can I error if my bazel target generates files that aren't specified in its outs?

Question

Inside a .bzl file,
I specify a program to generate some code. It looks something like this:

def generate_code():
    native.genrule(
        name = "foo",
        outs = ["file.hpp"],
        tools = ["//path/to:tool"],
        cmd = $(location path/to:tool) $(@D)
    )

This works fine,
however the problem is that the tool might generate more files than are specified in outs.
I'm trying to find a way to either have bazel stop the build if more files are generated than specified, or to have the outs automatically be everything generated.

Answer 1

I'm not aware of a generalized way to have bazel error out if additional, unexpected files were generated.

One thing I could recommend is wrapping your tool in another tool which verifies the output files in the target directory and only returns success if no unexpected files were generated.

As for a generalized solution that intentionally includes all files in an output directory, consider using actions.declare_directory and creating a skylark rule instead of using native.genrule.