How to simulate generating a source-file in a Bazel action? - bazel

Question

Suppose I am writing a custom Bazel rule for foo-compiler.
The user provides a list of source-files to the rule:
foo_library(
name = "hello",
srcs = [ "A.foo", "B.foo" ],
)
To build this without Bazel, the steps would be:
Create a config file config.json that lists the sources:
{
"srcs": [ "./A.foo", "./B.foo" ]
}
Place the config alongside the sources:
$ ls .
A.foo
B.foo
config.json
Call foo-compiler in that directory:
$ foo-compiler .
Now, in my Bazel rule implementation I can declare a file like this:
config_file = ctx.actions.declare_file("config.json")
ctx.actions.write(
output = config_file,
content = json_for_srcs(ctx.files.srcs),
)
The file is created and it has the right content.
However, Bazel does not place config.json alongside the srcs.
Is there a way to tell Bazel where to place the file?
Or perhaps I need to copy each source-file alongside the config?

Answer 1

You can do this with ctx.actions.symlink e.g.
srcs = []
# Declare a symlink for each src files in the same directory as the declared
# config file.Then write that symlink.
for f in ctx.files.srcs:
src = ctx.actions.declare_file(f.basename)
srcs.append(src)
ctx.actions.symlink(
output = src,
target_file = f,
)
config_file = ctx.actions.declare_file("config.json")
ctx.actions.write(
output = config_file,
content = json_for_srcs(ctx.files.srcs),
)
# Run compiler
ctx.actions.run(
inputs = srcs + [config_file],
outputs = # TODO: Up to you,
tools = [ctx.file.__compiler], #TODO: Update this to match your rule.
command = ctx.file.__compiler.path,
args = ["."],
#...
)
Note that when you return your provider that you should only return the result of your compilation not the srcs. Otherwise, you'll likely run into problems with duplicate outputs.