Makefile macro modifier - linux

Question

I am totally confused about the following macro modifier in a Makefile I have come onto,
TOOLS = $(TOOL_ROOTS:%=$(OBJDIR)%$(TOOL_SUFFIX))
Here
TOOL_ROOTS=some filename prefixes
OBJDIR=$HOME/obj/
TOOL_SUFFIX=.so
Can someone tell me what this line actually means?

Answer 1

TOOL_ROOTS must be getting assigned some value other than the empty string at some point or that does nothing (which I'll show in a moment).
So first things first just expanding the variables takes us from:
TOOLS = $(TOOL_ROOTS:%=$(OBJDIR)%$(TOOL_SUFFIX))
to:
TOOLS = $(:%=~/obj%.so)
(which we can immediately see doesn't look right, and as I'll explain in a moment doesn't do anything)
So lets pretend it has a value instead.
TOOL_ROOTS = shovel axe hammer
And try the expansion again:
TOOLS = $(shovel axe hammer:%=~/obj%.so)
(That OBJDIR definition looks odd also. I would expect it to be ~/obj/ or something... and that's ignoring that ~ is a bad choice here and that $HOME would be much better.)
The next thing we need to know is what that syntax is all about. Well it is a Substitution Reference.
A substitution reference substitutes the value of a variable with alterations that you specify. It has the form ‘$(var:a=b)’ (or ‘${var:a=b}’) and its meaning is to take the value of the variable var, replace every a at the end of a word with b in that value, and substitute the resulting string.
When we say “at the end of a word”, we mean that a must appear either followed by whitespace or at the end of the value in order to be replaced; other occurrences of a in the value are unaltered. For example:
foo := a.o b.o c.o
bar := $(foo:.o=.c)
sets ‘bar’ to ‘a.c b.c c.c’. See Setting Variables.
A substitution reference is actually an abbreviation for use of the patsubst expansion function (see Functions for String Substitution and Analysis). We provide substitution references as well as patsubst for compatibility with other implementations of make.
Another type of substitution reference lets you use the full power of the patsubst function. It has the same form ‘$(var:a=b)’ described above, except that now a must contain a single ‘%’ character. This case is equivalent to ‘$(patsubst a,b,$(var))’. See Functions for String Substitution and Analysis, for a description of the patsubst function.
For example:
foo := a.o b.o c.o
bar := $(foo:%.o=%.c)
sets ‘bar’ to ‘a.c b.c c.c’.
So, the first % there is matching the entirety of every word in the value of the variable (here shovel axe hammer) and then replacing each value with the expansion of the second part.
So shovel becomes ~/objshovel.so, etc. and we end up with:
TOOLS = ~/objshovel.so ~/objaxe.so ~/objhammer.so
See what I meant about OBJDIR being odd before? OBJDIR=~/obj/ would have left us with this instead:
TOOLS = ~/obj/shovel.so ~/obj/axe.so ~/obj/hammer.so
which makes a lot more sense to me.