On Tue, Apr 15, 2008 at 2:47 PM, Tony Strauss
<[EMAIL PROTECTED]> wrote:
> Calling $(eval) within $(eval) can lead to very unintuitive (at least to me
> :) behavior. In particular, the inner $(eval) is expanded *and* executed
> during the expansion of the outer $(eval).
Right, because make needs to know what the inner $(eval) expands to so
that it knows what the outer $(eval) is getting as its argument.
> For instance (I've also attached
> this example, since my email client mangled the tabs in the code below):
>
> define func2
> MY_FUNC2_VAR := $$(SOME_VALUE)
> all::
> @echo "Here2: $$(MY_FUNC2_VAR)"
> endef
>
> define func3
> MY_FUNC3_VAR := $$(SOME_VALUE)
> all::
> @echo "Here3: $$(MY_FUNC2_VAR)"
> endef
>
> define func1
> SOME_VALUE := 1
> MY_FUNC1_VAR := $$(SOME_VALUE)
> all::
> @echo "Here1: $$(MY_FUNC1_VAR)"
> $(eval $(call func2))
> ifndef SOME_VALUE
> $(eval $(call func3))
> endif
> endef
>
> $(eval $(call func1))
> all::
Umm, why are you using $(call funcN)? Since you're not passing any
arguments and they aren't recursive, that's the same as $(funcN).
(This appears to be a common misconception, that $(call) is needed
with $(eval). It's not.)
> The execution of the inner $(eval)s before the execution of the outer
> $(eval) means that:
> 1.) SOME_VALUE is not defined when either func2 or func3 is executed.
> 2.) func3 is expanded and executed despite being within a conditional that
> never will be true.
Right. If you don't want those effects, then don't use inner
$(eval)s! If you leave out the $(eval)s and just reference $(func2)
and $(func3) inside the definition of func1, those issues don't arise:
define func1
SOME_VALUE := 1
MY_FUNC1_VAR := $$(SOME_VALUE)
all::
@echo "Here1: $$(MY_FUNC1_VAR)"
$(func2)
ifndef SOME_VALUE
$(func3)
endif
endef
$ make
Here1: 1
Here2: 1
$
Philip Guenther
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