POSIX shell trivia

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Nov. 28th, 2007 at 10:06 PM

#!/bin/sh
false
if [ $? = 1 ]
then
	echo $?
fi

The above code block outputs "0". If $? is defined as, "Expands to the status of the most recently executed foreground pipeline," can you define exactly why the output is "0"?

hint: ~~expr~~ test (updated)

( 6 comments — Leave a comment )

ivazquez

Nov. 29th, 2007 05:01 am (UTC)

Simple. "[" is a command, and it has to evaluate to true (i.e., 0) in order for the body of the if statement to be executed.

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jasondclinton

Nov. 29th, 2007 05:13 am (UTC)

[ is expr, hence the hint.

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(Deleted comment)

jasondclinton

Nov. 29th, 2007 06:03 am (UTC)

I guess I should check more than one shell's manpage before I post trivia. grrr...

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(Deleted comment)

(Anonymous)

Nov. 29th, 2007 08:35 am (UTC)

Re: this reminds me of another people...

$? in shell is just like errno in C and all those weird built-in variables in Perl: If you need it, save it to a different variable first. Always.
GCC should have a -Werrno switch that complains if errno is used for anything but assigning it to something else.

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fragglet

Nov. 29th, 2007 09:40 am (UTC)

Shell scripts

Yet another reason to never use shell scripts for anything serious, ever.

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(Anonymous)

Nov. 2nd, 2009 12:48 pm (UTC)

its because of the false. it evaluates to a non zero exit status 1. So the return code $? is set to the value of 1 which tests as true in the [ $? = 1 ]. and hence prints 0. Check the debug mode of execution using a '#!/bin/sh -x'

+ false
+ '[' 1 = 1 ']'
+ echo 0
0

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( 6 comments — Leave a comment )

September 2011