Re: Return code from calling qx

Thanks everyone I found all of your refrences helpful. I'm having trouble understanding shift operators. When I run

qx(dir thisdoesntexist);
print $?;
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I get

File Not Found
256
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Running

qx(dir thisdoesntexist);
print $?>>8;
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returns

 
File Not Found 
1
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What does it mean to be shifting bits by 8 and how did that change the return code to 1?

Thank you,

Jason

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Re^2: Return code from calling qx by johna (Monk) on Oct 27, 2009 at 20:06 UTC
Think in terms of binary: `100000000 (binary) = 256 (decimal)` [download] If you shift it 8 bits to the right (>> 8) you get: `1 (binary) = 1 (decimal)` [download] Hope this helps a little. -John	[reply] [d/l] [select]
Re^3: Return code from calling qx by f50bodykit (Initiate) on Oct 27, 2009 at 20:24 UTC
Yes that helps. Thanks John.	[reply]
Re^4: Return code from calling qx by Discipulus (Canon) on Oct 28, 2009 at 10:31 UTC
..with ends up with this demostrative scripts: `print "the command -->@ARGV<-- returns:\n\n"; qx !@ARGV!; print "\n\n\$? $? (return status)\n", "\$?>>8 ",$?>>8, "(exit value of the subprocess)\n", "\$?&127 ",$?&127,"( which signal, if any, the process died from +)\n", "\$?&128 ",$?&128," (core dump)\n";` [download] Lor*	[reply] [d/l]