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Not a duplicate of question.
I am trying to make semvar pattern (1.0.0) from array having 1,0,0 as elements and create a git tag git tag $var.
Error:- The above solution is just not working for . only, it is giving 1 0 0, space separated.
4 13 0
fatal: too many params
Error with git tag command.
Approach:-
var=$( IFS=$'.'; echo "${versionArray[*]}" )
echo $var
git tag $var
Please suggest a better way to do it.
My guess is the shell that executes echo $var and git tag $var uses IFS=. as well. Note var=$(IFS=.;…) cannot change the variable value in the current shell, so this unexpected IFS must have been set earlier, probably during some trials and errors.
If I'm right, the variable named var holds the expected value with dots, e.g. 1.0.0. But when you retrieve it like this
echo $var
the current IFS is used to split 1.0.0 into words, so echo receives three separate arguments: 1, 0, 0. And because echo prints its arguments with single spaces in between, you were under impression the value of var was 1 0 0.
This line
echo "$var"
doesn't rely on IFS. I expect it will show you 1.0.0.
A general solution is to double-quote variables. You should have done it anyway:
echo "$var"
git tag "$var"
Also pay attention to IFS. Now it should be clear echo $IFS won't show you what the variable is; echo "$IFS" is better. In addition these methods are useful:
printf '%s' "$IFS" | hexdump -c
printf '%s' "$IFS" | hexdump -C
printf '%s' "$IFS" | xxd
var holds the expected value with dots, I am able to write in another file properly. Thanks for the time and efforts.
– Randeep tomar
Jul 3 '19 at 16:52
IFS active any more than necessary, since (as we're seeing) it can have unexpected effects. var=$(IFS=.;…) is fine, since the change is isolated to a subshell performing one operation. In other cases, when you need to change it, it's best to make the change, do what you need the nonstandard value for, and then immediately set it back to normal.
– Gordon Davisson
Jul 3 '19 at 20:04
$IFSin your main shell? If it's.thenecho $varwill print1 0 0; butecho "$var"should print1.0.0. Note for the same reasonecho $IFSis not a right way to print$IFS. – Kamil Maciorowski Jul 3 '19 at 15:45IFS=.should be enough to set the value..has no special meaning under the ANSI C escaping rules. – choroba Jul 3 '19 at 15:47$IFSis not printing anything. @KamilMaciorowski . – Randeep tomar Jul 3 '19 at 15:54echo "$var"worked for me. Please put it as answer, I'll accept it. – Randeep tomar Jul 3 '19 at 16:00