I have a command line tool in Go, example:
err := doSomething()
if err != nil {
log.Println(err) //fmt.Println(err)
os.Exit(1)
}
In the makefile, I am doing:
V = 0
Q = $(if $(filter1, $V),,#)
.PHONY: dosomething
dosomething: ; $(info $(shell printf "running dosomething")) #
$Q cd $(BASE) && ret=0 \
test -z "$$($(dosomething))" || ret = 1 ; \
exit $$ret
make fails (if error occurs) when using fmt.Println(err), but doesn't print anything. When I use log.Println, it prints the error but make continues. How to fail make as well print the error? Also, what to do in the case of a panic() in golang code?
I don't understand this makefile at all. What is $(info $(shell printf "running dosomething")) supposed to do? Why not just $(info running dosomething)? In general you never want to use make's shell function inside a recipe, it just leads to confusion.
Second, it looks like your formatting is wrong because you've put the # at the end of the line.
Also you need a semicolon after ret=0.
Finally, what is $$($(dosomething)) supposed to do? You've not set the make variable dosomething to any value, so this is basically a no-op.
I can only assume that in your real makefile you set the make variable dosomething to the command you want to run. It's helpful when asking questions, that you provide the actual files that reproduce the issue (creating a small repro case if the real file is too large or complex).
In that case the reason for the confusion is easy; when you run $(dosomething) in the shell it runs the dosomething program and captures its stdout and expands to that string... so the output is not printed. It's captured. For example:
$ echo hi
hi
$ foo=$(echo hi)
$ echo $foo
hi
Note how hi was not printed in the second command because it was captured and stored in the foo variable.
On the other hand, $(...) does not capture stderr, only stdout:
$ echo hi 1>&2
hi
$ foo=$(echo hi 1>&2)
hi
$ echo $foo
Note here that hi was printed (to stdout) and not captured (so foo is now empty).
I don't know much about Go but I can only assume that one of the two functions you show prints to stdout, and the other prints to stderr.
You don't want to capture any output and compare that, instead you want to compare the exit code of the program. You can do it like this:
V = 0
Q = $(if $(filter1, $V),,#)
.PHONY: dosomething
dosomething = <command-to-run>
dosomething:
$(info running dosomething)
$Q cd $(BASE) && $(dosomething)
This allows the output of the dosomething command to run without capturing any output, and it will exit with the exit code of the command, which will be 0 on success and not-0 (1 in your case) on error.
Related
I have the following Makefile target:
target1:
$(eval count_abc := $(shell grep -c "ABC" myFileA))
$(eval count_def := $(shell grep -c "DEF" myFileB))
echo $(count_abc)
echo $(count_def)
ifeq ($(count_abc),$(count_def))
echo "TRUE"
else
echo "FALSE"
endif
But the output is always TRUE, e.g.:
echo 22
22
echo 21
21
echo TRUE
TRUE
What am I doing wrong here? What I want is INSIDE the target do 2 greps and compare their outputs and do something or something else based on the result. Please note that the greps must be done within the target since myFileA and myFileB get created on the target before and don't exist at the beginning when running make.
Thanks,
Amir
The rule file for "make" is declarative in nature - the makefile defines rules and targets, and then the make program evaluate the rules, and decide which action to take based on the target. As a result, execution is not always in the order the lines are entered into the file.
More specifically, the "ifeq" is evaluated at the rule definition stage, but the actions for building the target (eval count_abc ...) are executed when the target is built. As a result, when the ifeq is processed, both count_abc and count_def are still uninitialized, expanded to empty strings.
For the specific case you described - building a target that will compare the grep -c output from the two files, you can try something like below, effectively using shell variables (evaluated when target is evaluated), and not make variables (which are mostly declarative, evaluated when makefile is read)
target1:
count_abc=$(grep -c "ABC" myFileA) ; \
count_def=$(grep -c "DEF" myFileB) ; \
echo $(count_abc) ; \
echo $(count_def) ; \
if [ "$count_abc" -eq "$count_def" ] ; then echo TRUE ; else echo FALSE ; fi
Disclaimer: I did not run the revised makefile, not having access to desktop at this time.
My makefile starts with something as
myVar := $$( etcExecute )
And it prints 10 times the $$ line if I use $(myVar) 10 times in an event. For example this test prints twice:
test:
ls $(myVar)/1
ls $(myVar)/2
How to supress these ugly prints?
There are a specific directive to silent only variable assigments?
I have a bash function inside the makefile command and want to change macros value.
Is it possible?
C_DFLAGS :=
gui :
parse_flags () { echo $$1; for word in $$1; do if [ $${word::2} = -D ] ; then $(eval C_D_FLAGS+=$${word}); fi ; done ; } ; parse_flags "-D/test -D/TEST"
#echo "C_D_FLAGS :$(C_D_FLAGS)"
$(eval) will be interpreted before your actual bash function call. You cannot update make variables from bash - it's a downstream process.
However, the code you try to run is fairly simple to replace with a native syntax, i.e.:
$ cat Makefile
C_D_FLAGS :=
gui: C_D_FLAGS += -D/test -D/TEST
gui:
#echo "C_D_FLAGS: $(C_D_FLAGS)"
$ make gui
C_D_FLAGS: -D/test -D/TEST
If the flags are provided from elsewhere, they can also be filtered, i.e.:
$ cat Makefile
C_D_FLAGS :=
gui: C_D_FLAGS += $(filter -D%,$(EXTRA_FLAGS))
gui:
#echo "C_D_FLAGS: $(C_D_FLAGS)"
$ make gui
C_D_FLAGS:
$ make gui EXTRA_FLAGS="-Isomething -DFOO -m32"
C_D_FLAGS: -DFOO
I got confused with Makefile. I am trying to run a simple command in the Makefile but it gives me the error "/bin/bash: line 3: :=: command not found". I am using shell to run this makefile
This is my part of my Makefile:
all:
vlog Benchmarks/$(NAME)/Syn/*.v
$(eval tux_number := 1)
$(eval range := 1)
$(eval ssh_log := 255)
echo "Start Range: ${range}"
echo "tux-number: ${tux_number}"
while [[ $$range -le 50 ]] ; do \
ssh -l yazdanbakhsh tux-$(tux_number).cae.wisc.edu exit ; \
echo "range: ${range}" ; \
eval $$range := $$((${range}+1)) ; \
done
Thanks
all:
#range=1; \
while [ $$range -le 10 ] ; \
do echo Range: $$range; \
let range=range+1 ; \
done;
Note that the whitespace in front of #range... is the only TAB.
Just to fix your obvious problems with Makefile syntax, here is an attempt at refactoring your attempt into valid code.
tux_number := 1
ssh_log := 255 # not used anywhere
all:
vlog Benchmarks/$(NAME)/Syn/*.v
echo "Start Range: 1" # This is probably no longer very useful output
echo "tux-number: ${tux_number}"
range=1; while [ $$range -le 50 ] ; do \
ssh -l yazdanbakhsh tux-$(tux_number).cae.wisc.edu exit ; \
echo "range: $$range" ; \
range=$$(expr "$$range + 1); \
done
Notice how tux_number and ssh_log are Makefile variables, while range only exists in the shell which executes the while loop. I have avoided the Bashisms in order to make this portable. (If portability is not important, you might want to refactor it back to Bash syntax and use for ((range=1; range<=50; range++)); do... instead.)
Your use of eval is misguided. As you can see, I simply lifted out the Makefile variables outside the recipe where they don't belong. What you were doing was (1) have Make evaluate the expression range := 1 (which evaluates to itself) and (2) use the output as a shell command in a recipe. Since it's not a valid shell command, you got the syntax error from Bash. Without further ado, I'll just take the easy way out here and say that eval is a complex subject, and until you get more experience with Make, it's probably just best to forget that it exists.
In order to properly make use of Make's facilities, I would make this parallelizable, i.e. split it up into 50 individual targets. This is a bit clumsy (there's probably a better way to define range here), but at least it should illustrate a number of differences to your approach. (If you don't insist on having range count up from 1, making it zero-based would make this a little less clumsy. This exploits the fact that the empty string is harmless in a shell snippet, so we can use it instead of a zero prefix. Again, this could be simplifed if you don't care about the human readability of the range index.)
digits := 0 1 2 3 4 5 6 7 8 9
deca := "" 1 2 3 4
range := $(filter-out ""0,$(foreach d,$(deca),$(foreach i,$(digits),$d$i))) 50
# Or, at the expense of an external process,
# range := $(shell perl -le 'print $$_ for 1..50')
.PHONY: all
all: $(patsubst %,ssh-%,$(range))
.PHONY: ssh-%
ssh-%:
ssh -l yazdanbakhsh tux-$(tux_number).cae.wisc.edu exit
echo "range: $*"
This can be run with something like make -j 5 to execute these in parallel batches of five, for example.
Incidentally, the commented-out $(shell ...) call might be the actual answer to your question, if what you really wanted to do was to use Make to drive an external program to calculate something for you.
Here's a shell script:
globvar=0
function myfunc {
let globvar=globvar+1
echo "myfunc: $globvar"
}
myfunc
echo "something" | myfunc
echo "Global: $globvar"
When called, it prints out the following:
$ sh zzz.sh
myfunc: 1
myfunc: 2
Global: 1
$ bash zzz.sh
myfunc: 1
myfunc: 2
Global: 1
$ zsh zzz.sh
myfunc: 1
myfunc: 2
Global: 2
The question is: why this happens and what behavior is correct?
P.S. I have a strange feeling that function behind the pipe is called in a forked shell... So, can there be a simple workaround?
P.P.S. This function is a simple test wrapper. It runs test application and analyzes its output. Then it increments $PASSED or $FAILED variables. Finally, you get a number of passed/failed tests in global variables. The usage is like:
test-util << EOF | myfunc
input for test #1
EOF
test-util << EOF | myfunc
input for test #2
EOF
echo "Passed: $PASSED, failed: $FAILED"
Korn shell gives the same results as zsh, by the way.
Please see BashFAQ/024. Pipes create subshells in Bash and variables are lost when subshells exit.
Based on your example, I would restructure it something like this:
globvar=0
function myfunc {
echo $(($1 + 1))
}
myfunc "$globvar"
globalvar=$(echo "something" | myfunc "$globalvar")
Piping something into myfunc in sh or bash causes a new shell to spawn. You can confirm this by adding a long sleep in myfunc. While it's sleeping call ps and you'll see a subprocess. When the function returns, that sub shell exits without changing the value in the parent process.
If you really need that value to be changed, you'll need to return a value from the function and check $PIPESTATUS after, I guess, like this:
globvar=0
function myfunc {
let globvar=globvar+1
echo "myfunc: $globvar"
return $globvar
}
myfunc
echo "something" | myfunc
globvar=${PIPESTATUS[1]}
echo "Global: $globvar"
The problem is 'which end of a pipeline using built-ins is executed by the original process?'
In zsh, it looks like the last command in the pipeline is executed by the main shell script when the command is a function or built-in.
In Bash (and sh is likely to be a link to Bash if you're on Linux), then either both commands are run in a sub-shell or the first command is run by the main process and the others are run by sub-shells.
Clearly, when the function is run in a sub-shell, it does not affect the variable in the parent shell (only the global in the sub-shell).
Consider adding an extra test:
echo Something | { myfunc; echo $globvar; }
echo $globvar