meson: use library before declaring it - meson-build

Question

Lets say your project looks like this:
meson.build:
project('myproject', 'cpp')
subdir('a')
subdir('b')
subdir('c')
a/meson.build:
mylib_a = library('mylib.cpp')
executable('a', 'main.cpp', link_with : [mylib_a])
with b/meson.build and c/meson.build looking similar.
If you would want to link the 'c' executable against mylib_a or mylib_b, you can simply add them to the array. But if you would want to link the 'a' executable against mylib_b or mylib_c you cannot do this without reordering the "subdir" commands. And If you would want to link 'c' against 'mylib_a' and 'a' against 'mylib_c', the only way, I could make it work is by putting the "executable('a', 'main.cpp', link_with : [mylib_a])" commands into meson.build instead of a/meson.build. This screws up the directory structure.
Is there any way I could link 'a' against mylib_c and 'c' against mylib_a without screwing up my directory structure?
UPDATE:
The meson authors say that there is no way to do this, but they will add one soon:
https://github.com/mesonbuild/meson/issues/8178

Answer 1

You can try placing all executable's in the bottom of the root meson.build file. In this case all mylib_a/b/c variables will be available. You will also need to create in each subdirectory meson.build file file_a/b/c variables, e.g.
file_a = files('main.cpp')
and use them in corresponding executable(...).
This may suffice your needs, but I suggest you to look towards using subprojects concept instead.
UPDATE
To avoid clutter if you have many sub-directories, libraries and executables you can use loop with foreach:
libs = ['a', 'b', 'c']
lib_list = []
foreach l : libs
lib_list += library(l, join_paths(l, 'mylib.cpp'))
endforeach
progs = ['a', 'b', 'c']
foreach p : progs
exe = executable(p, join_paths(p, 'main.cpp'), link_with : lib_list)
endforeach