CODEWITHSUNDEEP
bazel
Esben Nielsen
Esben Nielsen

Reputation: 51

Are absolute paths safe to use in Bazel?

I am experimenting with Bazel to be added along with an old, make/shell based build system. I can easily make shell commands which returns an absolute path to some tool or library build by the old build system as early prerequisites. These commands I can use in a genrule(), which copies the needed files (like headers and libs) into Bazel proper to be exposed in form of a cc_library().

I found out that genrule() does not detect a dependency if the command uses a file with absolute path - it is not caught by the sandbox. In a way I am (ab)using that behavior.

It is it safe? Will some future update of Bazel refuse access to files based on absolute path in that way in a command in genrule?

Upvotes: 1

Views: 2972

Answers (1)

Brian Silverman
Brian Silverman

Reputation: 3868

Most of Bazel's sandboxes allow access to most paths outside of the source tree by default. Details depend on which sandbox implementation you're using. The docker sandbox, for example, allows access to all those paths inside of a docker image. It's kind of hard to make promises about future Bazel versions, but I think it's unlikely that a sandbox will prevent accessing /bin/bash (for example), which means other absolute paths will probably continue to work too.

--sandbox_block_path can be used to explicitly block a path if you want.

If you always have the files available on every machine you build on, your setup should work. Keep in mind that Bazel will not recognize when the contents of those files change, so you can easily get stale results in various caches. You can avoid that by ensuring the external paths change whenever their contents do.

new_local_repository might be a better fit to avoid those problems, if you know the paths ahead of time.

If you don't know the paths ahead of time, you can write a custom repository rule which runs arbitrary commands via repository_ctx.execute to retrieve the paths and them symlinks them in with repository_ctx.symlink.

Tensorflow's third_party/sycl/sycl_configure.bzl has an example of doing something similar (you would do something other than looking at environment variables like find_computecpp_root does, and you might symlink entire directories instead of all the files in them):

def _symlink_dir(repository_ctx, src_dir, dest_dir):
    """Symlinks all the files in a directory.
    Args:
      repository_ctx: The repository context.
      src_dir: The source directory.
      dest_dir: The destination directory to create the symlinks in.
    """
    files = repository_ctx.path(src_dir).readdir()
    for src_file in files:
        repository_ctx.symlink(src_file, dest_dir + "/" + src_file.basename)

def find_computecpp_root(repository_ctx):
    """Find ComputeCpp compiler."""
    sycl_name = ""
    if _COMPUTECPP_TOOLKIT_PATH in repository_ctx.os.environ:
        sycl_name = repository_ctx.os.environ[_COMPUTECPP_TOOLKIT_PATH].strip()
    if sycl_name.startswith("/"):
        return sycl_name
    fail("Cannot find SYCL compiler, please correct your path")

def _sycl_autoconf_imp(repository_ctx):
<snip>
            computecpp_root = find_computecpp_root(repository_ctx)
<snip>
            _symlink_dir(repository_ctx, computecpp_root + "/lib", "sycl/lib")
            _symlink_dir(repository_ctx, computecpp_root + "/include", "sycl/include")
            _symlink_dir(repository_ctx, computecpp_root + "/bin", "sycl/bin")

Upvotes: 2

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