Why doesn't this bash expr command work?

Question

I am trying to increment a variable in a bash script and it is not working. Here is my code:

#! /bin/bash

COUNTER=0
while [ $COUNTER -lt 5 ]
do
    echo "i will add this line to file mycreation">>./myfile
    COUNTER = `expr $COUNTER + 1`
done

The quotes around the COUNTER assignment are backticks.

I tried replacing COUNTER with $COUNTER like this:

$COUNTER = `expr $COUNTER + 1`

But that did not solve the problem and gave me the following error:

line7: 0: command not found.

The command 0 not found error appears when you use `$COUNTER` on the LHS of the assignment - `$COUNTER` expands to 0 and the shell then tries to execute that command. — Jonathan Leffler, Jan 24 '11 at 18:18

score 34 · Accepted Answer · edited May 23 '17 at 11:53

34

As @Cory rightly pointed out, there should not be spaces around the equal sign or else bash will confuse COUNTER for a command.

COUNTER=$(expr $COUNTER + 1)

going off-topic ...

That said, you could avoid having bash fork a subprocess by using the following alternatives:

Using the bash builtin 'let' command:
```
let COUNTER="COUNTER + 1"
```
or, using bash c-style expression:
```
(( COUNTER++ ))
```

In fact, your while loop can be written as:

for ((COUNTER=0; COUNTER <= 5 ; COUNTER++))
do
    echo "i will add this line to file mycreation">>./myfile
done

Breaking down the error message

When you were met with the error:

line 7:   0:    command not found.
'-----'  '--'  '------------------'
   |       |                 |
location   |            Description of error.
          culprit

my guess is what you had on line 7 was

$COUNTER = `expr $COUNTER + 1`
--------   --------------------
    |                 |
Evaluated to 0        |
                  Evaluated to 1

What bash ends up see is 0 = 1 and since bash statements are generally in the form command arg1 arg1 ..., bash interprets it as run the command 0 with arguments = 1. Thus the error message : 0: command not found.

When you removed the spaces around the equal sign, what bash ends up interpreting is:

0=1

which means run command 0=1 with no arguments, hence the error 0=1: command not found.

Variable assignments should be in the form VAR_NAME=VALUE (without the $), so the syntax you should be using is:

COUNTER=`expr $COUNTER + 1` # or any of the variants above

which bash evaluates and eventually interpret as:

COUNTER=2

edited May 23 '17 at 11:53

Community

1
1

answered Jan 24 '11 at 17:38

Shawn Chin

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Just be sure to shebang with `/bin/bash` if using bashisms, I think some of them actually turn off if bash is invoked as `/bin/sh` (POSIX) or if /bin/sh doesn't point to bash at all (dash on some systems). – Tim Post Jan 24 '11 at 18:29
Hello I removed the spaces around it however all it did is it's now giving me line:7 0=1 command not found. why did it change from 0:0 to 0:1 baffles me. Maybe i shouldn't be using expr at all? I don't understand why it's listed as useful command. Thanks – Marin Jan 24 '11 at 20:01
@Marin when you encountered the error, did you have a $ sign at the beginning? It should be COUNTER=.. not $COUNTER=... – Shawn Chin Jan 24 '11 at 20:14
answer updated, assuming my hunch is correct. – Shawn Chin Jan 24 '11 at 20:44
thanks Shawn! It worked, for some reason the spaces did the trick but I still don't get it why it's so sensitive? Thanks I guess I should just stick to AWK – Marin Jan 25 '11 at 17:14
+++++1 Thanks for explaining "let", "(( COUNT++ ))" and "for (( ))" =) – Viet Feb 16 '12 at 06:40

score 6 · Answer 2 · answered Jan 24 '11 at 17:35

6

Remove the spaces around the equals sign:

COUNTER=`expr $COUNTER + 1`

answered Jan 24 '11 at 17:35

Cory Klein

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score 3 · Answer 3 · answered Jan 24 '11 at 17:59

3

Another way.

COUNTER=$(($COUNTER + 1))

answered Jan 24 '11 at 17:59

webbi

841
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14

3

not well documented, but you don't have to dereference the variables inside the arithmetic expression: `COUNTER=$((COUNTER+1))` or `((COUNTER++))` as has been mentioned – glenn jackman Jan 24 '11 at 19:37

score 2 · Answer 4 · answered Oct 23 '12 at 08:47

2

for i in {0..4}; do 
    echo "i will add this line to file mycreation" >> ./myfile
done

answered Oct 23 '12 at 08:47

August Karlstrom

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Why doesn't this bash expr command work?

4 Answers4

going off-topic ...

Breaking down the error message