In Buck, how do I expand a genrule output directory? - c++

Question

I have a genrule that outputs a directory containing C++ headers. I also have a cxx_library which uses the output of the genrule. The headers field of the cxx_library looks like this:
...
headers = [
':my-headers',
],
...
The problem is that my C++ source files include the headers like this:
#include "my_header.h"
But Buck prepares the staging area like this:
my-cxx-library#default,private-headers/out
Where out is an alias to the folder containing my generated headers.
So in order to use the headers, I would have to include them like this:
#include "out/my_header.h"
The library is not my own, so I would not like to change the source-code. Instead, I would like to use something like a subdir_glob to include everything inside :my-headers/out.
I tried this:
...
headers = subdir_glob([
(':my-headers/out', '**/*.h'),
]),
...
However, it seems that when done this way, the string :my-headers does not get resolved to the output path of :my-headers.
Is there a function in buck that can expand a rule to its output path?

Answer 1

This isn't trivially possible today, but there is a workaround you can use:
genrule(
name = 'headers',
cmd = 'generate_headers.py --outdir=$OUT'
out = 'headers-dir',
)
genrule(
name = 'generated_header_A.h',
cmd = 'cp $(location :headers)/genereated_header_A.h $OUT',
out = 'generated_header_A.h',
)
Then, in your cxx_library, you just need to declare your headers by referencing the location:
headers = {
'some_namespace/header_A.h': ':generated_header_A.h',
},
The above code assumes all of this is in the same build file, but if it isn't, you just need to use a fully-qualified build target instead.