While loop with unexpected bash exit

Question

Problem statement:

It appears that I have an issue with with a while loop when I utilize set -e:

 1  #!/bin/bash
 2  # This is a script to perform initial configurations of CentOS 7.2
 3  # Written by Robert J.
 4  
 5  # Exit if unexpected condition
 6  set -e;
 7  
 8  # Simple die function
 9  dieaquickdeath(){
10      code=$1;
11      message=$2;
12      printf "$message\n";
13      exit "$code";
14  }
15  
16  #
17  ## There is an initial prompt here that was removed for simplification.
18  #
19  InputPassword1=1
20  InputPassword2=2
21  
22  # Sanity check for password
23  unset n;
24  while [[ "$InputPassword1" != "$InputPassword2" ]];
25  do
26      # Counter
27      ((n++));
28  
29      # Prompt for new password
30      printf "\nPasswords do not match\n";
31      read -sp "Enter password: " InputPassword1;
32      echo;
33      read -sp "Re-enter password: " InputPassword2;
34  
35      # End loop after passwords match.
36      [[ "$InputPassword1" == "$InputPassword2" ]] &&
37          break;
38  
39      # End loop after 3 attempts
40      [[ "$n" == 3 ]] &&
41          dieaquickdeath 1 "Third failed password attempt";
42  
43  done;

Attempted solutions:

I initially thought that this might have just been because of the exit code for test so I hardwired true statements temporary for troubleshooting purposes (This did not work still):

36      [[ "$InputPassword1" == "$InputPassword2" ]] &&
37          break || 
38          true;
39  
40      # End loop after 3 attempts
41      [[ "$n" == 3 ]] &&
42          dieaquickdeath 1 "Third failed password attempt" || 
43          true;

This does not appear to have any issues when running without the set -e;

[root@Wiki ~]# vi testing
[root@Wiki ~]# ./testing
[root@Wiki ~]# grep -v "set -e" testing > test2
[root@Wiki ~]# ./test2
-bash: ./test2: Permission denied
[root@Wiki ~]# chmod +x test2
[root@Wiki ~]# ./test2

Passwords do not match
Enter password: 
Re-enter password: 
Passwords do not match
Enter password: 
Re-enter password: 
Passwords do not match
Enter password: 
Re-enter password: Third failed password attempt
[root@Wiki ~]#

Questions:

Does anyone see where my unexpected exit is?
Is this the cause of some bash caveat that I am unaware of?

I feel like I have been staring at this for far too long.

Thanks in advance for being awesome all.

Thanks man! I didn't know about the bash -x option. This is super helpful. — Robert J
– Robert J, Commented Jun 13, 2016 at 19:17

John1024 · Accepted Answer · 2016-06-13 19:17:04Z

5

To see where it stops, run bash -x testing:

$ bash -x testing
+ set -e
+ InputPassword1=1
+ InputPassword1=2
+ unset n
+ [[ 2 != '' ]]
+ (( n++ ))

It stops because n was (effectively) zero. Observe:

$ unset n; ((n++)); echo code=$?
code=1

An exit code of 1 indicates failure. Hence, set -e terminated execution here.

Here are two possible solutions that will return a zero (success) exit code:

$ unset n; ((++n)); echo code=$?
code=0
$ unset n; ((n++)) || true; echo code=$?
code=0

In general, as discussed here, set -e does strange things. Its use is controversial.

edited Jun 13, 2016 at 19:17

answered Jun 13, 2016 at 19:10

John1024

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3 Comments

Robert J Over a year ago

Awesome man! I bash -x is going to save me a lot of time in the future. I'm kinda a bash noob. Thanks for your time man.

user3439894 Over a year ago

@Robert J, Just like set -e, you can also place it in the script, e.g., set -x for testing purposes and then comment it out or remove it.

shellter Over a year ago

and set +x to "turn it off", and export PS3='$LINENO >' to have it display the line number of the code being executed. Good luck to all.

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