On Sun, Jun 21, 2009 at 1:48 PM, jacques<soliton1...@yahoo.fr> wrote:
> By writing a simple program (program_a), I was able to see some strange
> behaviours of 'ifeq'
>
> program_a:
> ifeq("foo","foo") \
> TOOL = tool \
> endif
>
> start:
> @echo $TOOL
>
>
> whose execution by "make -f program_a" yields a newline.
"make -f program_a" tells make to ignore your Makefile and instead
read the file "program_a" as a Makefile. Did you *really* do that, or
did you instead run
make program_a
?
When I try "make program_a", I get the following:
$ make
make: Nothing to be done for `program_a'.
$
> First question: The gmake on-line manual states that the syntax of a
> conditional test follows the following template (7.2):
>
> conditional-directive
> text-if-true
> else
> text-if-false
> endif
>
> Notice that there is no backslash character in the definition. In
> 'program_a', if I don't put the backslashes, the program gives execution
> errors. So backslashes are compulsory eventhough the manual says otherwise.
> Is it just a bug that will be fixed in later versions or is there a deeper
> significance ( the 'ifdef' and 'ifndef' conditionals behave syntatically
> according to the gmakeon-line manual !!) ?
Where to start..
1) you did not actually write a ifeq conditional: to be an ifeq
conditional, there MUST
be a space after the word "ifeq"!
2) ending a line with a backslash tells make to join the line with the
following line.
By ending those two lines with backslashes, you effectively gave
make the following:
ifeq("foo","foo") TOOL = tool endif
That's a variable assignment! Yes, the name of the variable is
ifeq("foo","foo") TOOL
as can be confirmed by adding this to the 'start' target:
@echo ${ifeq("foo","foo") TOOL}
You'll find that "make start" then echos "tool endif"
So: insert a space after 'ifeq' and remove the backslashes.
You'll then find that it still doesn't work...because you didn't use
the TOOL variable properly. To make, "$TOOL" means "the value of the
'T' variable, followed by 'OOL'". If a variable has more than one
character in it's name, you must enclose the name in either brackets
or parens. Use ${TOOL} or $(TOOL).
I'm not sure what the "program_a" target is supposed to be doing in
your example Makefile. You _do_ understand that the assignment to
TOOL will *not* be specific to that target, don't you?
Philip Guenther
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