bash: Unable to run build commands via bash scripts

Question

I am trying to run 'make' command from my bash script to build the code.

I can see that all parameters got assigned and able to display the command that i am trying to run. I could not see any issue with the command. But the issue is when it tries to run the command via bash script it fails.

My command is :- ./build_script LIC=1 DOCUMNETS=1 PROJECTS="cuda bfm"

Script Snippet of parsing all the arguments and constructing make command:-

makeargs=""
for build_flag do
   if [[ "$build_flag " = "PROJECTS="* ]]; then
          apps =`echo $build_flag |sed "s/PROJECTS=//" `
          makeargs="$makeargs PROJECTS=\"$apps \""
   else
        makeargs="$makeargs $build_flag"
    fi
done
echo "make DCOV=1 $makeargs $maketest"
make DCOV=1 $makeargs $maketest

When i run the script, I can see the build command has constructed properly.

Output :-

make DCOV=1 LIC=1 DOCUMNETS=1 PROJECTS="cuda bfm" run_all
GNUmakefile:16: warning: overriding commands for target `/'
GNUmakefile:19: warning: ignoring old commands for target `/'
make: *** No rule to make target `bfm"'.  Stop.

I try to print PROJECTS variable in my 'GNUmakefile' present in build_main folder. I can see the output : PROJECTS is "bfm . It is not taking whole "cuda bfm" as a whole string.

Note:- When i try to run the same build command :- make DCOV=1 LIC=1 DOCUMNETS=1 PROJECTS="cuda bfm" run_all explicitly it works fine.

Seems like issue with Interpreting variables with makefile.

Any solution for this ? Please help. Thanks!

Can't see enough of your script. Either add more code, or paste your code into shellcheck.net or run your script with bash -x build_script ... — Mark Setchell
– Mark Setchell, Commented Feb 20, 2020 at 8:20
Hi Mark, As you can see the command is getting generated properly right? the for loop will parse the arguments and check for PROJECTS string and generates a build command. — santosh
– santosh, Commented Feb 20, 2020 at 9:16
I can't see enough of your code to tell what's wrong. Did you post it in spellcheck.net yet? Because, for example this is incorrect make DCOV=1 $makeargs $maketest as you haven't double quoted "$makeargs" or "$maketest" — Mark Setchell
– Mark Setchell, Commented Feb 20, 2020 at 9:37
No, I haven't posted in speelcheck.net before. We are running it as a build command i.e. make .... So its not necessary to keep double quotes right ? — santosh
– santosh, Commented Feb 20, 2020 at 9:43
Exactly, that's why I made the initial suggestions of running under bash -x and using shellcheck... — Mark Setchell
– Mark Setchell, Commented Feb 20, 2020 at 9:51

user0 · Accepted Answer · 2020-02-20 11:16:09Z

2

Change makeargs string to array before passing it as an arguments group.

eval makeargs_array=( $makeargs )
make UVC=1 "${makeargs_array[@]}" $maketest

Without converting to array, if you enable debug, it shows last line interpretation as

make DCOV=1 LIC=1 DOCUMNETS=1 'PROJECTS="cuda' bfm '"'

Which is clearly ignoring double-quote and considering space as separator. Even double-quote is getting passed as a separate argument in this case.

Explanation:

Word-splitting It says,

The shell scans the results of parameter expansion, command substitution, and arithmetic expansion that did not occur within double quotes for word splitting.

If we use "$makeargs" i.e. surrounded by double-quote, it is not considered by word-splitting and results in LIC=1 DOCUMNETS=1 "PROJECTS=cuda bfm"

But again its a complete string, while requirement is to split the string to use as arguments.

So now using $makeargs.

Word-splitting gets in action as per the default IFS (space, tab, newline), we get result as LIC=1 DOCUMNETS=1 PROJECTS="cuda bfm "

Double-quoted part of string didn't affect the word-splitting since, subject to splitting is complete string here.

Why array worked here?

Array itself expands each element as separate word when using @ and here no further word-splitting requires after expansion.

Arrays

edited Feb 20, 2020 at 11:16

answered Feb 20, 2020 at 9:48

user0

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1 Comment

santosh Over a year ago

May i know why bash is interpreting this way ?

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