BASH Syntax Checking Debug Mode Malfunction?

Question

We can use bash -n script.sh to validate the syntax of a shell script. However, when I was trying to test this function, I noticed not all the syntax errors could be found by this option.

For example:

root@ubuntu:~/testenv# cat test 
#!/bin/bash
SEND=1
if [ "$SEND" -eq 0 ]
        echo no
fi

Now, let's test the script:

root@ubuntu:~/testenv# bash -n test 
test: line 5: syntax error near unexpected token `fi'
test: line 5: `fi'

It works fine. However, if I just remove one of the bracket:

root@ubuntu:~/testenv# cat test     
#!/bin/bash
SEND=1
if [ "$SEND" -eq 0 
then 
        echo no
fi
root@ubuntu:~/testenv# bash -n test 
root@ubuntu:~/testenv#

Nothing happened!

I also checked the man page of bash, it describes the "-n" is:

 -n               Read  commands  but  do not execute them.  This may be used to check a
                  shell script for  syntax  errors.   This  is  ignored  by  interactive
                  shells.

It is a script file, so it shouldn't be an "interactive shell" right? So,how could this happen?

@NeilWang: Looks valid! Do highlight the same, people may mis-interpret it — Inian
– Inian, Commented Jan 3, 2017 at 8:05
@Inian In the second example, I missed a close bracket "]", but the syntax checker didn't find! Please execute them on your own box, you will find it will throw an error. — NeilWang
– NeilWang, Commented Jan 3, 2017 at 8:08
@NeilWang: You can't expect to catch that error, it ain't a syntax violation, [ is a shell built-in — Inian
– Inian, Commented Jan 3, 2017 at 8:09
This is a good question, why do people want to close it and downvoted it? — gniourf_gniourf
– gniourf_gniourf, Commented Jan 3, 2017 at 9:13

David Z · Accepted Answer · 2017-01-03 08:57:44Z

I'm guessing you have run into a very strange quirk of the way the shell implements single-bracketed conditionals: [ is a command, not a special character. Look in your system executable directory (probably /usr/bin) and you will find an executable file literally named [ which implements this command. When you write something like

[ "$SEND" -eq 0 ]

then you're actually invoking the command [ with four arguments:

The value of $SEND
The string -eq
The string 0
The string ]

The command [ checks that the last argument is ] (because it would look weird otherwise), then puts the remaining arguments together to form a condition and return the result of testing the condition.

Now, because [ is a command, it's not a syntax error to invoke that command with any set of arguments you like. Sure, if you leave off the trailing ], you will get an error, but that error comes from the command [, not from the shell. That means you have to actually run the script to get the error - the syntax checker won't see anything wrong with it. As far as bash is concerned, [ is just a command name, no different from, say, my_custom_conditional_test, and if you were to write

my_custom_conditional_test "$SEND" -eq 0

it would be obvious that this is fine, right? Bash thinks of [ the same way.

I should note that for efficiency, bash doesn't actually use the executable file /usr/bin/[; it has its own builtin implementation of [. But people expect [ to act the same way regardless of whether it's built in to the shell or not, so the Bash syntax checker can't give its own [ special treatment. Since it wouldn't be a syntax error to invoke /usr/bin/[ with no trailing ], it can't be a syntax error to invoke the builtin [ without a ].

You can contrast this with [[, which does more or less the same thing (testing a condition) but is given special meaning by the shell. [[ is a special token in shell syntax, not a command. If you write [[ instead of [, and you omit the corresponding trailing ]], you bet Bash is going to complain about a syntax error.

You are exactly damn right! I had never known there is a [ under /usr/bin. Thank you for solving my concerns.

Collectives™ on Stack Overflow

BASH Syntax Checking Debug Mode Malfunction?

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related