GNU make conditional function $(if ...) inside a user-defined function always evaluates to true

Question

I am trying to write a make function to touch/create an empty file and/or set the permissions, user and group, where possible, or warn if not. However, every conditional check within my function seems to evaluate to true.

The essentials of my Makefile are

INSTALL_USER  := fileUser
INSTALL_GROUP := fileGroup

.PHONY: test
test:
    $(call touchFile,~/test.ini)

define touchFile
$(eval fileName := $(strip $(1)))
-touch $(fileName)
-chmod -c 664 $(fileName)

$(info filename info $(fileName))
$(info $(shell stat -c "%a %U:%G" $(fileName)))

$(if ifeq  "foo" "bar", @echo match is broken, @echo match works)
$(if ifneq "foo" "bar", @echo match works, @echo match is broken)

$(if ifneq ($(shell stat -c %a $(fileName)),664), $(warning Error - $(fileName) does not have expected permissions of 664))
-chgrp -c $(INSTALL_GROUP) $(fileName)
$(if ifneq ($(shell stat -c %G $(fileName)),$(INSTALL_GROUP)), $(warning Error - $(fileName) does not belong to $(INSTALL_GROUP) group))
-chown -c $(INSTALL_USER) $(fileName)
$(if ifneq ($(shell stat -c %U $(fileName)),$(INSTALL_USER)), $(warning Error - $(fileName) does not belong to $(INSTALL_USER) user))
endef

Running make test outputs

filename info ~/test.ini
664 myUserName:myGroup
Makefile:7: Error - ~/test.ini does not have expected permissions of 664
Makefile:7: Error - ~/test.ini does not belong to common group
Makefile:7: Error - ~/test.ini does not belong to netserve user
touch ~/test.ini
chmod -c 664 ~/test.ini
match is broken
match works
chgrp -c fileGroup ~/test.ini
changed group of `/home/myUserName/test.ini' to fileGroup
chown -c fileUser ~/test.ini
chown: changing ownership of `/home/myUserName/test.ini': Operation not permitted
make: [test] Error 1 (ignored)

I've considered/tried the following:

$(if ...) is evaluated at "compile-time", before the function is called with a parameter. But, the hard-coded ifeq "foo" "bar" also gives an invalid result. Additionally, $(info ...) correctly evaluates $(fileName) at "compile-time".
The documentation doesn't actually give examples, so in addition to $(if ifeq...), I also tried $(ifeq ...), which seemed to be ignored.
"Non-functional" if (i.e., the ifeq without the $(if...)) inside a function gives /bin/sh: ifeq: command not found.

Can someone help identify why my conditionals aren't behaving as I expect (or why I'm expecting the wrong thing)?

Caveat: I know there are still bugs to be worked out if the file doesn't exist, but that should be trivial compared to this hurdle.

The GNU website, although it looks nice, is so succinct as to be almost pointless, isn't it. They will tell you a function exists, and then not give you any example on how to use it, so as to harass you with it. — Xennex81, Jun 02 '17 at 20:52

Eldar Abusalimov · Accepted Answer · 2016-02-11T12:26:14.437

$(if ...) conditional function evaluates to true when the first argument passed to it is non-empty. In you case the condition is literal text: ifeq "foo" "bar", which is, obviously, non-empty.

ifeq/ifneq conditionals are in fact directives, not functions. They can't be used inside variable definition and in functions.

Back to your example, to test string for equality inside the condition use functions like filter and findstring:

$(if $(filter foo,bar),@echo match is broken,@echo match works)
$(if $(filter-out foo,bar),@echo match works,@echo match is broken)

BTW this could be also turned into an inline form for better readability:

@echo match $(if $(filter foo,bar),is broken,works)
@echo match $(if $(filter-out foo,bar),works,is broken)

score 10 · Answer 2 · answered Jul 16 '15 at 09:18

I faced this problem and I found you can use the result of "filter" function in the "condition" part of "if" function. Here is an example useful for opening a pdf either in Linux (with "evince"), or in OSX with "open")

uname     :=$(shell uname -s)
is_darwin :=$(filter Darwin,$(uname))
viewpdf   :=$(if $(is_darwin), open, evince)

score 3 · Answer 3 · answered Jan 25 '12 at 19:28

You seem to be misunderstanding the way $(if works. From the make info docs:

$(if CONDITION,THEN-PART[,ELSE-PART])' The `if' function provides support for conditional expansion in a functional context

The first argument, CONDITION, first has all preceding and trailing whitespace stripped, then is expanded. If it expands to any non-empty string, then the condition is considered to be true. If it expands to an empty string, the condition is considered to be false.

In all your examples, your condition is something like ifeq SOMETHING OTHERTHING -- which is a non-empty string (from the ifeq irrespective of what the other things are), and so is treated as true.

score 1 · Answer 4 · edited Jun 28 '23 at 04:15

1

If you want to check that your OS version is Linux or something else, and print a message if it is, you can do this:

ifneq ($(TARGETOS), Linux)
    target: server
    @echo
    @echo $(MSG) $(TARGETOS)
else
target: server
    @echo $(MSG) $(TARGET)
endif

Before, I tried with ifeq, but it printed the else part only (while I was running Linux, Ubuntu, to be more precise).

edited Jun 28 '23 at 04:15

Isacc Barker

507
4
15

answered Sep 01 '17 at 11:05

amit pandya

1,384
13
22

GNU make conditional function $(if ...) inside a user-defined function always evaluates to true

4 Answers4