Dispatching C++ generated files into srcs and hdrs - bazel

Question

In the Bazel official documentation there is an example explaining how to create a Java library built from regular java files and files generated by a :gen_java_srcs rule. I rewrite this code here for ease of reading:
java_library(
name = "mylib",
srcs = glob(["*.java"]) + [":gen_java_srcs"],
deps = "...",
)
genrule(
name = "gen_java_srcs",
outs = [
"Foo.java",
"Bar.java",
],
...
)
Now in a C++ perspective, I am in a scenario where the genrule generates two kind of files: .hpp and .cpp:
genrule(
name = "gen_cpp_srcs",
outs = [
"myFile_1.hpp","myFile_2.hpp",...,"myFile_N.hpp",
"myFile.cpp","myFile_2.cpp",...,"myFile_N.cpp",
],
...
)
where N is some tens.
My problem/question is: how to write the cc_library rule, with an automatic dispatching of the hpp and cpp files into hdrs and srcs field?
I want something like:
cc_library(
name = "mylib",
srcs = glob(["*.cpp"]) + (howto: .cpp files of [":gen_cpp_srcs"]),
hdrs = glob(["*.hpp"]) + (howto: .hpp files of [":gen_cpp_srcs"]),
...
)
Some magic like:
output_filter(":gen_cpp_srcs","*.cpp")
would be perfect, but I do not know enough of Bazel to make it real.

Answer 1

Globs only get expanded when they're passed into rules, so you'll need to write a simple rule. I would package it like this (in a file named filter.bzl):
# The actual rule which does the filtering.
def _do_filter_impl(ctx):
return struct(
files = set([f for f in ctx.files.srcs if f.path.endswith(ctx.attr.suffix)]),
)
_do_filter = rule(
implementation = _do_filter_impl,
attrs = {
"srcs": attr.label_list(
mandatory = True,
allow_files = True,
),
"suffix": attr.string(
mandatory = True,
),
},
)
# A convenient macro to wrap the custom rule and cc_library.
def filtered_cc_library(name, srcs, hdrs, **kwargs):
_do_filter(
name = "%s_hdrs" % name,
visibility = ["//visibility:private"],
srcs = hdrs,
suffix = ".hpp",
)
_do_filter(
name = "%s_srcs" % name,
visibility = ["//visibility:private"],
srcs = srcs,
suffix = ".cpp",
)
native.cc_library(
name = name,
srcs = [ ":%s_srcs" % name ],
hdrs = [ ":%s_hdrs" % name ],
**kwargs
)
This is what my demo BUILD file looks like (I changed the globs so they both include *.cpp and *.hpp files; using the label of a genrule will work the same way):
load("//:filter.bzl", "filtered_cc_library")
filtered_cc_library(
name = "mylib",
srcs = glob(["*.*pp"]),
hdrs = glob(["*.*pp"]),
)
This is easy to extend to more sophisticated filtering by changing _do_filter_impl. In particular, changing suffix to an attr.string_list so you can accept multiple C/C++ source/header extensions seems like a good idea.

Answer 2

Depending on the genrule by name (:gen_cpp_srcs) will give you all of the outputs of the genrule, as you have noted. Instead, you can depend on the individual outputs of the genrule (e.g. hdrs = [:myFile.hpp] and srcs = [:myFile.cpp]).
See also the answer to Bazel & automatically generated cpp / hpp files.

Answer 3

Looks like you know the total number of files that should be generated. Can you put those in their own variables and then reuse them in both targets. Something like this in your BUILD file:
output_cpp_files = [
"myFile_1.cpp",
"myFile_2.cpp",
"myFile_3.cpp"
]
output_hpp_files = [
"myFile_1.hpp",
"myFile_2.hpp",
"myFile_3.hpp"
]
genrule(
name = "gen_cpp_srcs",
outs = output_cpp_files + output_hpp_files,
cmd = """
touch $(OUTS)
"""
)
cc_library(
name = "mylib",
srcs = output_cpp_files,
hdrs = output_hpp_files
)