CODEWITHSUNDEEP
c++buck
sdgfsdh
sdgfsdh

Reputation: 37121

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

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 that can expand a rule to its output path?

Upvotes: 5

Views: 1123

Answers (1)

sdwilsh
sdwilsh

Reputation: 4682

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.

Upvotes: 5

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