Hi everybody,
it appears that a $(foreach) statement appearing in any recipe line is
evaluated before all remaining lines regardless where in the recipe it
appears, then all remaining recipe lines are executed. I would expect the
lines to be executed in the order of appearance.
A simple culprit example:
$ make --version
GNU Make 4.2.1
$ cat -n makefile
1 .PHONY: all
2 all: list.txt
3 list.txt: $(MAKEFILE_LIST)
4 $(RM) $@
5 $(foreach x,$^,$(shell echo $x >> $@))
6 test -f $@
# for copying the code
$ cat makefile
.PHONY: all
all: list.txt
list.txt: $(MAKEFILE_LIST)
$(RM) $@
$(foreach x,$^,$(shell echo $x >> $@))
test -f $@
$ make
rm -f list.txt
test -f list.txt
make: *** [Makefile:6: list.txt] Error 1
The test -f fails, i.e. list.txt does not exist, since the $(foreach) loop
is evaluated first writing to list.txt, then the remaining recipe lines in
appearance order are executed: list.txt is removed by the recipe first line
such that the test in the third recipe line fails.
Is it intended that the recipe line order is not honored for $(foreach)? I
think this behavior is counterintuitive and for me it smells like a bug.
As a side remark, a workaround for this sample issue is to move $(foreach)
into the **first** line with adjusted recipe code logic (I am well aware
that there are other make techniques to do this, but this is about the
$(foreach) precedence issue in a recipe block):
.PHONY: all
all: list.txt
list.txt: $(MAKEFILE_LIST)
$(foreach x,$^,$(shell echo $x >> $@.tmp))
mv -f $@.tmp $@
Any input highly appreciated.
Many thanks,
J.
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