Modify global variable from loop

Question

Why do I get

/tmp/test: line 4: 0=Done: command not found

from the below

a="0"

while [ true ]; do
  $a="Done"
  exit
done

echo $a

I were expecting it would output Done.

You should use break to exit a loop. exit will exit the whole shell. — Alfe
– Alfe, Commented May 8, 2018 at 11:10
Why you need a while [ true ] loop at all if you have an unconditional break or exit? — oliv
– oliv, Commented May 8, 2018 at 11:13
Why test the string "true" when you could just execute the true command (i.e. while true; do ...; done). Not that there's any point in a loop with unconditional break, but... — Toby Speight
– Toby Speight, Commented May 8, 2018 at 13:02

Lix · Accepted Answer · 2018-05-08 11:08:00Z

3

You don't need to use the $ when defining a variable, only when you are accessing it.

You'll need to change the line defining the variable a to:

a="Done"

As to an explanation, what I believe is happening here is that $a is being resolved to 0 and then the shell is seeing the entire 0=Done as a single (unfound) command. Accessing undefined variables still returns a 0 exit code.

answered May 8, 2018 at 11:08

Lix

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2 Comments

Sandra Schlichting Over a year ago

If I do that, then it doesn't print anything. @Alfe's comment is the solition.

Lix Over a year ago

Yes this would have been the next issue :) My answer handles the error you reported in your original post.

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