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Re: local keyword hides return code of command substitution

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On Tue, Sep 22, 2015 at 10:19:56AM -0400, idallen@home.idallen.ca wrote:
> Description:
> 	Adding a "local" keyword to a variable assignment hides the
> 	return code of a command substitution.	Same problem in both
> 	bash and dash shells.

Yes, this is how it works.  If you care about the return value of a
command used to initialize a local variable, you have to write it in
two steps:

foo() {
    local foo
    foo=$(bar) || return
}

http://mywiki.wooledge.org/BashPitfalls has this and many more.

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Re: local keyword hides return code of command substitution Greg Wooledge <wooledg@eeg.ccf.org> - 2015-09-22 11:45 -0400

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