I am trying to make a following setup run with Bazel. With calling “bazel build” a Python script should generate unknown number of *.cc files with random names, and then compile these into single static library (.a file), all within one Bazel call. I have tried following: one generated file has fixed name, this one is referenced in outs of genrule() and srcs of cc_library rule. The problem is I need all generated files to be built as a library, not only the file with fixed name. Any ideas how to do this? My BUILD file:
py_binary(
name = "sample_script",
srcs = ["sample_script.py"],
)
genrule(
name = "sample_genrule",
tools = [":sample_script"],
cmd = "$(location :sample_script)",
outs = ["cpp_output_fixed.cc"], #or shall also the files with random names be defined here?
)
cc_library(
name = "autolib",
srcs = ["cpp_output_fixed.cc"],
#srcs = glob([ #here should all generated .cc files be named
# "./*.cc",
# "./**/*.cc",
# ])+["cpp_output_fixed.cc"],
)
Python file sample_script.py:
#!/usr/bin/env python
import hashlib
import time
time_stamp = time.time()
time_1 = str(time_stamp)
time_2 = str(time_stamp + 1)
random_part_1 = hashlib.sha1(time_1).hexdigest()[-4:]
random_part_2 = hashlib.sha1(time_1).hexdigest()[-4:]
fixed_file = "cpp_output_fixed" + ".cc"
file_1 = "cpp_output_" + random_part_1 + ".cc"
file_2 = "cpp_output3_" + random_part_2 + ".cc"
with open(fixed_file, "w") as outfile:
outfile.write("#include <iostream>"
"int main() {"
" std::cout <<'''Hello_world''' <<std::endl;"
" return 0"
"}")
with open(file_1, "w") as outfile:
outfile.write("#include <iostream>"
"int main() {"
" std::cout <<'''Hello_world''' <<std::endl;"
" return 0"
"}")
with open(file_2, "w") as outfile_new:
outfile_new.write("#include <iostream>"
"int main() {"
" std::cout <<'''Hello_world''' <<std::endl;"
" return 0"
"}")
print ".cc generation DONE"
[big edit, since I found a way to make it work :)]
If you really need to emit files that are unknown at the analysis phase, your only way is what we internally call tree artifacts. You can think of it as a directory that contains files that will only be inspected at the execution phase. You can declare a tree artifact from Skylark using ctx.actions.declare_directory.
Here is a working example. Note 3 things:
genccs.bzl:
def _impl(ctx):
tree = ctx.actions.declare_directory(ctx.attr.name + ".cc")
ctx.actions.run(
inputs = [],
outputs = [ tree ],
arguments = [ tree.path ],
progress_message = "Generating cc files into '%s'" % tree.path,
executable = ctx.executable._tool,
)
return [ DefaultInfo(files = depset([ tree ])) ]
genccs = rule(
implementation = _impl,
attrs = {
"_tool": attr.label(
executable = True,
cfg = "host",
allow_files = True,
default = Label("//:genccs"),
)
}
)
BUILD:
load(":genccs.bzl", "genccs")
genccs(
name = "gen_tree",
)
cc_library(
name = "main",
srcs = [ "gen_tree" ]
)
cc_binary(
name = "genccs",
srcs = [ "genccs.cpp" ],
)
genccs.cpp
#include <fstream>
#include <sys/stat.h>
using namespace std;
int main (int argc, char *argv[]) {
mkdir(argv[1], S_IRWXU);
ofstream myfile;
myfile.open(string(argv[1]) + string("/foo.cpp"));
myfile << "int main() { return 42; }";
return 0;
}
1) List all output files. 2) Use the genrule as a dependency to the library.
genrule(
name = "sample_genrule",
tools = [":sample_script"],
cmd = "$(location :sample_script)",
outs = ["cpp_output_fixed.cc", "cpp_output_0.cc", ...]
)
cc_library(
name = "autolib",
srcs = [":sample_genrule"],
)
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