Thank you for your reply. I understood I'm misusing ` `, actually I
don't need ` ` at all in my example.
Em 26-02-2018 14:58, G.raud escreveu:
On Mon, Feb 26, 2018 at 02:30:15PM -0300, Lauro Costa wrote:
Subject: Unexpected(?) return value inside if-else
The "if" on b is of course false, so it goes to the else and returns the
return value of a(). Dash will print "b() = 1" and bash will print "b() =
0".
If you comment out lines 6,7,8 and 10 bash and dash will print "b() = 0".
This is just a sample code and doesn't do anything useful. And it would be
easy to workaround this issue changing the code. And I'm very likely doing
something non POSIX-compliant with return `something`. But I don't know what
it is and I'm very curious to know what is going on. Please take a look,
thank you.
________________________
a(){
return `echo x |grep -q x`
}
b(){
if echo x |grep -q y ;then
return 100
else
return `a`
fi
}
a
echo "a() = $?"
b
echo "b() = $?"
_________________________
You probably misunderstood what process substitution (``) is: a simple
string substitution. In your examples all it does is execute a command
but substitute nothing.
Your examples reduce to:
$ dash -c 'b() { false; `true` return; }; b; echo $?'
1
$ bash -c 'b() { false; `true` return; }; b; echo $?'
0
Without process substitution both shells return 1:
$ dash -c 'b() { false; return; }; b; echo $?'
1
$ bash -c 'b() { false; return; }; b; echo $?'
1
What happens is that bash returns the return value of the process
substitution on the return command and dash the return value of the last
command. I do not know if there is a specification mandating one or the
other behaviour but that's bash's behaviour that I find surprising.
--
Lauro Costa
Analista de Sistemas
Phone: +55 (41) 3045-9016
<http://www.polilinux.com.br>
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