RE: why is indirect assignment inside eval, not working??

[Paul Smith]

Please don't top-post to the mailing list; thanks.

> From: Philip Guenther [mailto:address@hidden 
> 
> On Tuesday, February 9, 2010, Mark Galeck (CW) <address@hidden> wrote:
> > define FOOBAR
> > FOO:=foo
> > BAR:=$(FOO)
> > endef
> >
> > $(eval $(call FOOBAR))
> 
> Conside the output if you instead write
> $(info $(call FOOBAR))
> 
> That shows you what $(eval) is seeing.  Make more sense? 

On Tue, 2010-02-09 at 21:06 -0800, Mark Galeck (CW) wrote:
> Thank you Philip!  But... I already knew that...  that is, the BAR
> gets nothing assigned, that's obvious, and info just confirms that.
> The question is, why doesn't it get anything assigned??  It is only
> inside eval that this happens...  if I just do
> 
> FOO:=foo
> BAR:=$(FOO)
> 
> bar:
>         echo $(BAR)
> 
> everything is fine...  that is my question, what is the
> difference???  

You aren't thinking deeply enough.  What Philip is trying to point out
is that this doesn't have anything to do with the $(eval).  How can it,
when you see the same thing when you don't use $(eval) at all?

Without the $(eval), what's left?

The $(call).  What does $(call) do?  It evaluates the value of the
variable it's given, with some arguments assigned to $1, $2, etc.

What does evaluation do?  It expands the value.  What is the value?  The
value of FOOBAR is the string:

FOO:=foo
BAR:=$(FOO)

Remember, to $(call) this is just a string.  $(call) doesn't think of
this as makefile syntax, it's expanding this value exactly as if it were
something like "$(CC) $(CFLAGS)" or whatever.

At this point, what's the value of the variable FOO?  It doesn't have
any value.  So, what's the result of the expansion?  Just as you saw
from the $(info) output, it's:

FOO:=foo
BAR:=

And so when you pass that expanded string to $(eval), that's exactly
what you get.

If you want to defer the expansion of the $(FOO) variable so it's
expanded by the $(eval) and not by $(call), you have to escape it so
that $(call) doesn't expand it:

define FOOBAR
FOO:=foo
BAR:=$$(FOO)
endef

Of course in this case, the $(call) is completely useless since you're
not assigning any arguments like $1, $2.  You might as well just use:

$(eval $(FOOBAR))

directly.  And, since you don't want to expand any part of the value of
FOOBAR before $(eval) gets it, you can avoid escaping $(FOO) by using
the $(value) function:

$(eval $(value FOOBAR))

However, in real life this isn't often very useful since if there were
no variable values in FOOBAR that you wanted to replace, you might as
well not bother to use $(eval) anyway.

-- 
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