How do I make a bazel `sh_binary` target depend on other binary targets? - bazel

I have set up bazel to build a number of CLI tools that perform various database maintenance tasks. Each one is a py_binary or cc_binary target that is called from the command line with the path to some data file: it processes that file and stores the results in a database.
Now, I need to create a dependent package that contains data files and shell scripts that call these CLI tools to perform application-specific database operations.
However, there doesn't seem to be a way to depend on the existing py_binary or cc_binary targets from a new package that only contains sh_binary targets and data files. Trying to do so results in an error like:
ERROR: /workspace/shbin/BUILD.bazel:5:12: in deps attribute of sh_binary rule //shbin:run: py_binary rule '//pybin:counter' is misplaced here (expected sh_library)
Is there a way to call/depend on an existing bazel binary target from a shell script using sh_binary?
I have implemented a full example here:
https://github.com/psigen/bazel-mixed-binaries
Notes:
I cannot use py_library and cc_library instead of py_binary and cc_binary. This is because (a) I need to call mixes of the two languages to process my data files and (b) these tools are from an upstream repository where they are already designed as CLI tools.
I also cannot put all the data files into the CLI tool packages -- there are multiple application-specific packages and they cannot be mixed.

You can either create a genrule to run these tools as part of the build, or create a sh_binary that depends on the tools via the data attribute and runs them them.
The genrule approach
This is the easier way and lets you run the tools as part of the build.
genrule(
name = "foo",
tools = [
"//tool_a:py",
"//tool_b:cc",
],
srcs = [
"//source:file1",
":file2",
],
outs = [
"output_file1",
"output_file2",
],
cmd = "$(location //tool_a:py) --input=$(location //source:file1) --output=$(location output_file1) && $(location //tool_b:cc) < $(location :file2) > $(location output_file2)",
)
The sh_binary approach
This is more complicated, but lets you run the sh_binary either as part of the build (if it is in a genrule.tools, similar to the previous approach) or after the build (from under bazel-bin).
In the sh_binary you have to data-depend on the tools:
sh_binary(
name = "foo",
srcs = ["my_shbin.sh"],
data = [
"//tool_a:py",
"//tool_b:cc",
],
)
Then, in the sh_binary you have to use the so-called "Bash runfiles library" built into Bazel to look up the runtime-path of the binaries. This library's documentation is in its source file.
The idea is:
the sh_binary has to depend on a specific target
you have to copy-paste some boilerplate code to the top of the sh_binary (reason is described here)
then you can use the rlocation function to look up the runtime-path of the binaries
For example your my_shbin.sh may look like this:
#!/bin/bash
# --- begin runfiles.bash initialization ---
...
# --- end runfiles.bash initialization ---
path=$(rlocation "__main__/tool_a/py")
if [[ ! -f "${path:-}" ]]; then
echo >&2 "ERROR: could not look up the Python tool path"
exit 1
fi
$path --input=$1 --output=$2
The __main__ in the rlocation path argument is the name of the workspace. Since your WORKSPACE file does not have a "workspace" rule in, which would define the workspace's name, Bazel will use the default workspace name, which is __main__.

An easier approach for me is to add the cc_binary as a dependency in the data section. In prefix/BUILD
cc_binary(name = "foo", ...)
sh_test(name = "foo_test", srcs = ["foo_test.sh"], data = [":foo"])
Inside foo_test.sh, the working directory is different, so you need to find the right prefix for the binary
#! /usr/bin/env bash
executable=prefix/foo
$executable ...

A clean way to do this is to use args and $(location):
Contents of BUILD:
py_binary(
name = "counter",
srcs = ["counter.py"],
main = "counter.py",
)
sh_binary(
name = "run",
srcs = ["run.sh"],
data = [":counter"],
args = ["$(location :counter)"],
)
Contents of counter.py (your tool):
print("This is the counter tool.")
Contents of run.sh (your bash script):
#!/bin/bash
set -eEuo pipefail
counter="$1"
shift
echo "This is the bash script, about to call the counter tool."
"$counter"
And here's a demo showing the bash script calling the Python tool:
$ bazel run //example:run 2>/dev/null
This is the bash script, about to call the counter tool.
This is the counter tool.
It's also worth mentioning this note (from the docs):
The arguments are not passed when you run the target outside of bazel (for example, by manually executing the binary in bazel-bin/).

Related

Bazel: genrule that outputs a directory

I'm just getting started working with Bazel. So, I apologize in advance that I haven't been able to figure this out.
I'm trying to run a command that outputs a bunch of files to a directory and make this directory available for subsequent targets. I have two different attempts:
Use genrule
Write my own rule
I was naively hoping to just do this with a genrule. But, it doesn't seem you can say "I don't know exactly what this command is going to output" and put a directory in outs. Now I'm trying to write a rule that can use ctx.actions.declare_directory but I haven't gotten it quite right. I can't seem to get the tools over from my workspace and into my rule.
My genrule attempt looks something like this:
genrule(
name = "doit",
srcs = [
"doitConfigA",
"doitConfigB",
],
cmd = 'HOME=. ./$(location path/to/doit) install',
# Neither of the below outs work - seems like bazel wants to know
# exactly this list of files. I don't know the files that
# will be output ahead of time.
# This one looks at the `out_dir` that I already have and
# expects the files to be the same which they might not be
outs = glob(["out_dir/**/*.*"]),
# this fails with:
# "declared output 'out_dir' was not
# created by genrule. This is probably because the genrule actually
# didn't create this output, or because the output was a directory
# and the genrule was run remotely (note that only the contents of
# declared file outputs are copied from genrules run remotely)"
outs = ['out_dir'],
tools = ['path/to/doit'],
)
My custom rule attempt looks something like this:
def _impl(ctx):
dir = ctx.actions.declare_directory("out_dir")
ctx.actions.run_shell(
outputs=[dir],
progress_message="Running doit install ...",
command="HOME=. ./path/to/doit install",
tools=[ctx.attr.tools],
)
doit = rule(
implementation=_impl,
attrs={
"tools": attr.label_list(allow_files=True),
},
outputs={"out": "out_dir"},
)
Then, to run my doit rule, my BUILD file looks like this:
doit(
name = 'doit',
tools = ['path/to/doit'],
)
In my genrule, the command runs but it doesn't like my trying to use a directory in outs, it seems. In my custom rule, I can't seem to tell Bazel that I want to use ./path/to/doit as a tool from my workspace, eg expected type 'File' for 'tools' element but got type 'list' instead ...
Seems like I must be missing something basic because surely this is a common situation to run a command and output a bunch of unknown stuff to a directory?
The output of a genrule must be a fixed list of files. As a work-around, you can create a zip from the output directory.
I used this approach to manipulate the output of yarn install where the usual method was not viable:
genrule(
name = "node_modules",
srcs = [
"package.json",
"yarn.lock",
],
cmd = " && ".join([
"yarn install --pure-lockfile",
"zip -r $# node_modules",
]),
outs = [
"node_modules.zip",
],
)
Then a rule that consumes the zip:
# Rule that generates a list of the folders in node_modules
genrule(
name = "node_modules_ls",
srcs = [
":node_modules",
],
cmd = " && ".join([
"unzip $(location :node_modules) -d . ",
"ls > $#",
]),
outs = [
"out.txt",
],
)
A while ago I created this example showing how to use directories with skylark action: How to build static library from the Generated source files using Bazel Build. Maybe it still works :)
Genrule won't work, this is too advanced use case.
https://github.com/aspect-build/bazel-lib/blob/main/docs/run_binary.md has a similar API to genrule, and it supports directory outputs.

Binary shell rule dependency

We're using Bazel at work for wrapping and organizing some build systems that we use for our projects. Since, each project has its own build system (like bitbake for example), we can't use Bazel for the actual building (e.g creating cc_binary and cc_library rules that contain all the source code). But we do use it for wrapping and for a uniformed API.
I want to make a shell script (A) dependent of another shell script (B), and I want to do so, such that B will run before A. Since we have some configurations inside the BUILD files, I can't simply run from command line B then A. I need Bazel to do so, and inject the configuration to both scripts. Something like :
In the BUILD file of one component:
sh_binary(
name = "bazel_build_multivisor",
srcs = ["bazel-build-multivisor.sh"],
data = ["wrkspceinfo"],
deps = ["//core-build/components/bazel-pull:bazel_pull_multivisor"])
In the BUILD file of a second component which is in charge of pulling from git:
sh_binary(
name = "bazel_pull_multivisor",
srcs = ["pull_repo_repository.sh"],
data = ["wrkspceinfo", "pull_repo_repository.sh"],
args = [ARGS, "NAME=multivisor_repo;MODULE_NAME=*;GIT_BRANCH=*;MANIFEST=*.xml;PULL_PARAMS=8,8,8"],)
Meaning, I want a sh_binary rule to depend on a different sh_binary rule.
Is this possible? Is there a better way doing so?
Thanks.
Consider using a genrule.
You can specify both sh_binary targets in the genrule.tools attribute, and then in the genrule.cmd attribute, specify the shell command that calls them in order.
For example,
genrule(
name = "pull_and_build",
srcs = "some_config_file.txt",
outs = ["built_project.tar.gz"]
tools = [":pull_sh", ":build_sh"],
cmd = """
$(location :pull_sh) $< | \
xargs -1 $(location :build_sh) | \
xargs -1 tar -czvf $#"""
)
Another possible way is to wrap the dependency sh_binary in a sh_library instead, and aggregate them at the top level sh_binary.

bazel: How do you "request" that a cc_binary build the stripped version of the binary?

The docs for Bazel's cc_binary rule say:
Implicit output targets
<name>.stripped (only built if explicitly requested): A stripped version of the binary. strip -g is run on the binary to remove debug symbols. Additional strip options can be provided on the command line using --stripopt=-foo. This output is only built if explicitly requested.
How do I "explicitly request" that this stripped binary get built? Is there something I need to put in my cc_binary declaration in my BUILD file? I can't figure it out from the docs (or the Bazel source).
Okay I think I figured out how.
If my BUILD file has this:
cc_binary(
name = "mytool",
srcs = ["mytool.c"]
)
... then from the command line I can build the stripped binary with this:
bazel build //:mytool.stripped
or, the more common scenario, if I have another BUILD rule that needs the stripped binary as one of its inputs, I can just refer to it by that same label, :mytool.stripped. Here is sort of a weird contrived example:
genrule(
name = "mygenrule",
outs = ["genrule.out"],
srcs = [":tool1.stripped"],
# run tool1.stripped, sends its output to genrule.out:
cmd = "$(SRCS) > $#"
)

Verify step in bazel

I'm looking for a good recipe to run "checks" or "verify" steps in Bazel, like go vet, gofmt, pylint, cppcheck. These steps don't create any output file. The only thing that matters is the return code (like a test).
Right now I'm using the following recipe:
sh_test(
name = "verify-pylint",
srcs = ["verify-pylint.sh"],
data = ["//:all-srcs"],
)
And verify-pylint.sh looks like this:
find . -name '*.py' | xargs pylint
This has two problems:
The verify logic is split between the shell script and the BUILD file. Ideally I would like to have both in the same place (in the BUILD file)
Anytime one of the source file changes (in //:all-srcs), bazel test verify-pylint re-runs pylint on every single file (and that can be expensive/slow).
What is the idiomatic way in bazel to run these steps?
There are more than one solutions.
The cleanest way is to do the verification at build time: you create a genrule for each file (or batch of files) you want to verify, and if verification succeeds, the genrule outputs something, if it fails, then the rule outputs nothing, which automatically fails the build as well.
Since success of verification depends on the file's contents, and the same input should yield the same output, the genrules should produce an output file that's dependent on the contents of the input(s). The most convenient thing is to write the digest of the file(s) to the output if verification succeeded, and no output if verification fails.
To make the verifier reusable, you could create a Skylark macro and use it in all your packages.
To put this all together, you'd write something like the following.
Contents of //tools:py_verify_test.bzl:
def py_verify_test(name, srcs, visibility = None):
rules = {"%s-file%d" % (name, hash(s)): s for s in srcs}
for rulename, src in rules.items():
native.genrule(
name = rulename,
srcs = [s],
outs = ["%s.md5" % rulename],
cmd = "$(location //tools:py_verifier) $< && md5sum $< > $#",
tools = ["//tools:py_verifier"],
visibility = ["//visibility:private"],
)
native.sh_test(
name = name,
srcs = ["//tools:build_test.sh"],
data = rules.keys(),
visibility = visibility,
)
Contents of //tools:build_test.sh:
#!/bin/true
# If the test rule's dependencies could be built,
# then all files were successfully verified at
# build time, so this test can merely return true.
Contents of //tools:BUILD:
# I just use sh_binary as an example, this could
# be a more complicated rule of course.
sh_binary(
name = "py_verifier",
srcs = ["py_verifier.sh"],
visibility = ["//visibility:public"],
)
Contents of any package that wants to verify files:
load("//tools:py_verify_test.bzl", "py_verify_test")
py_verify_test(
name = "verify",
srcs = glob(["**/*.py"]),
)
A simple solution.
In your BUILD file:
load(":gofmt.bzl", "gofmt_test")
gofmt_test(
name = "format_test",
srcs = glob(["*.go"]),
)
In gofmt.bzl:
def gofmt_test(name, srcs):
cmd = """
export TMPDIR=.
out=$$(gofmt -d $(SRCS))
if [ -n "$$out" ]; then
echo "gmfmt failed:"
echo "$$out"
exit 1
fi
touch $#
"""
native.genrule(
name = name,
cmd = cmd,
srcs = srcs,
outs = [name + ".out"],
tools = ["gofmt.sh"],
)
Some remarks:
If your wrapper script grows, you should put it in a separate .sh file.
In the genrule command, we need $$ instead $ due to escaping (see documentation)
gofmt_test is actually not a test and will run with bazel build :all. If you really need a test, see Laszlo's example and call sh_test.
I call touch to create a file because genrule requires an output to succeed.
export TMPDIR=. is needed because by default the sandbox prevents writing in other directories.
To cache results for each file (and avoid rechecking a file that hasn't changed), you'll need to create multiple actions. See Laszlo's for loop.
To simplify the code, we could provide a generic rule. Maybe this is something we should put in a standard library.

Bazel: copy multiple files to binary directory

I need to copy some files to binary directory while preserving their names. What I've got so far:
filegroup(
name = "resources",
srcs = glob(["resources/*.*"]),
)
genrule(
name = "copy_resources",
srcs = ["//some/package:resources"],
outs = [ ],
cmd = "cp $(SRCS) $(#D)",
local = 1,
output_to_bindir = 1,
)
Now I have to specify file names in outs but I can't seem to figure out how to resolve the labels to obtain the actual file names.
To make a filegroup available to a binary (executed using bazel run) or to a test (when executed using bazel test) then one usually lists the filegroup as part of the data of the binary, like so:
cc_binary(
name = "hello-world",
srcs = ["hello-world.cc"],
data = [
"//your_project/other/deeply/nested/resources:other_test_files",
],
)
# known to work at least as of bazel version 0.22.0
Usually the above is sufficient.
However, the executable must then recurse through the directory structure "other/deeply/nested/resources/" in order to find the files from the indicated filegroup.
In other words, when populating the runfiles of an executable, bazel preserves the directory nesting that spans from the WORKSPACE root to all the packages enclosing the given filegroup.
Sometimes, this preserved directory nesting is undesirable.
THE CHALLENGE:
In my case, I had several filegroups located at various points in my project directory tree, and I wanted all the individual files of those groups to end up side-by-side in the runfiles collection of the test binary that would consume them.
My attempts to do this with a genrule were unsuccessful.
In order to copy individual files from multiple filegroups, preserving the basename of each file but flattening the output directory, it was necessary to create a custom rule in a bzl bazel extension.
Thankfully, the custom rule is fairly straightforward.
It uses cp in a shell command much like the unfinished genrule listed in the original question.
The extension file:
# contents of a file you create named: copy_filegroups.bzl
# known to work in bazel version 0.22.0
def _copy_filegroup_impl(ctx):
all_input_files = [
f for t in ctx.attr.targeted_filegroups for f in t.files
]
all_outputs = []
for f in all_input_files:
out = ctx.actions.declare_file(f.basename)
all_outputs += [out]
ctx.actions.run_shell(
outputs=[out],
inputs=depset([f]),
arguments=[f.path, out.path],
# This is what we're all about here. Just a simple 'cp' command.
# Copy the input to CWD/f.basename, where CWD is the package where
# the copy_filegroups_to_this_package rule is invoked.
# (To be clear, the files aren't copied right to where your BUILD
# file sits in source control. They are copied to the 'shadow tree'
# parallel location under `bazel info bazel-bin`)
command="cp $1 $2")
# Small sanity check
if len(all_input_files) != len(all_outputs):
fail("Output count should be 1-to-1 with input count.")
return [
DefaultInfo(
files=depset(all_outputs),
runfiles=ctx.runfiles(files=all_outputs))
]
copy_filegroups_to_this_package = rule(
implementation=_copy_filegroup_impl,
attrs={
"targeted_filegroups": attr.label_list(),
},
)
Using it:
# inside the BUILD file of your exe
load(
"//your_project:copy_filegroups.bzl",
"copy_filegroups_to_this_package",
)
copy_filegroups_to_this_package(
name = "other_files_unnested",
# you can list more than one filegroup:
targeted_filegroups = ["//your_project/other/deeply/nested/library:other_test_files"],
)
cc_binary(
name = "hello-world",
srcs = ["hello-world.cc"],
data = [
":other_files_unnested",
],
)
You can clone a complete working example here.

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