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Re: Test exit status misinterpreted in scripts when buit without job con
Re: Test exit status misinterpreted in scripts when buit without job control
Thu, 4 Aug 2016 14:36:37 -0400
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On 8/4/16 12:05 PM, Dan Cross wrote:
> Bash Version: 4.3
> Patch Level: 30
> Release Status: release
> When bash is built without job control, shell scripts that use
> the 'test' builtin (e.g., via '[') in conditionals may take the
> wrong branch becuase the exit status of the test is lost.
> Configure without job control. Via e.g.,
> ./configure --prefix=/usr --bindir=/bin --without-bash-malloc
> --disable-nls --disable-job-control
> Invoke the resulting shell and run the following sequence of commands:
> $ cat > foo.sh
> if [ $# -lt 2 ]
> echo "$# args is less than 2"
> echo "$# args is not less than 2"
> $ chmod +x ./foo.sh
> $ ./foo.sh 1 2 3 4
> 4 args is less than 2
> Observe the output: '4' is not actually less than '2' yet the
> script incorrectly reports it as such.
Thanks for the report. I took a quick look at this, and it's not disabling
job control that does it: it's disabling both job control and nls.
Disabling either one while leaving the other enabled doesn't produce this
error (which only happens in the case where you run a script with the
execute bit set without a #! line after running an executable that causes
the shell to call waitpid()). It's a strange set of circumstances.
I'll see what I can find.
``The lyf so short, the craft so long to lerne.'' - Chaucer
``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, UTech, CWRU address@hidden http://cnswww.cns.cwru.edu/~chet/