Reputation: 163
I have a code generator that produces three output files:
The genrule looks like this:
genrule(
name = 'code_gen',
tools = [ '//tools:code_gen.sh' ],
outs = [ 'client.cpp', 'server.cpp', 'data.h' ],
local = True,
cmd = '$(location //tools:code_gen.sh) $(@D)')
The 'client.cpp' and 'server.cpp' each have their own cc_library rule.
My question is how to depend on the genrule but only use a specific output file.
What I did is create a macro that defined the genrule with specific outs set to the file required, but this resulted in multiple execution of the genrule:
gen.bzl:
def code_generator(
name,
out):
native.genrule(
name = name,
tools = [ '//bazel:gen.sh' ],
outs = [ out ],
local = True,
cmd = '$(location //bazel:gen.sh) $(@D)')
BUILD
load(':gen.bzl', 'code_generator')
code_generator('client_cpp', 'client.cpp')
code_generator('server_cpp', 'server.cpp')
code_generator('data_h', 'data.h')
cc_library(
name = 'client',
srcs = [ ':client_cpp' ],
hdrs = [ ':data_h' ],
)
cc_library(
name = 'server',
srcs = [ ':server_cpp' ],
hdrs = [ ':data_h' ],
)
Is there a way to depend on a genrule making it run once and then use only selected outputs from it?
Upvotes: 1
Views: 1333
Reputation: 2370
You should be able to just use the filename (e.g. :server.cpp
) to depend on a specific output of a rule.
Upvotes: 1