Generating list of generated sources (a la foreach) in automake

Question

I am currently working on a project using templates quite extensively, and running into memory constraints during instantiation. I have split up the instantiation into a number of very simple files, which are all three-liners consisting of includes only.

I let these be generated by a rule in Makefile.am. Now I have a quite long list of files that should be generated in my Makefile, and would like to refactor this list with a foreach-like expression. In more specific terms: I have a line like

libfoo_la_SOURCES = a_0.cpp a_1.cpp ... b_0.cpp b_1.cpp ... c_0.cpp c_1.cpp ...

which could be more concisely expressed as

libfoo_la_SOURCES = $(foreach i,a b ...,$(foreach j,0 1 ...,$i_$j.cpp))

However, the second construct is not only warned against by automake, but also does not work: The files given in this manner are neither compiled nor cleaned.

My current workaround is generating this file list by a shell script.

Any ideas how to implement this iteration?

adl adl · Accepted Answer · 2011-09-18T11:29:06

I would forget about making loops: the GNU extension is not standard, and not understood by Automake. One standard (and portable) make construction you can use here is the macro expansion with substitution: $(var:subst1=subst2) will expand to the value of $(var) after replacing any suffix subst1 of a word by subst2. Automake understands this.

If subst1 is empty, as in $(var:=subst2), you are appending subst2 to all files in $(var). You can use this to construct your list of files as follows:

f = a b c d e f
g = $(f:=_0) $(f:=_1) $(f:=_2) $(f:=_3)
all_files = $(g:=.cpp)

echo:
        @echo $(all_files)

Running make echo with the above Makefile will display all files from a_0.cpp to f_3.cpp.

Generating list of generated sources (a la foreach) in automake

2 Answers