Re^3: ztk-webcam snapshot-grabber

Would you explain what "$|++" meaning is in the code? Thanks

Its like $foo++ except working on the variable $| , a reall shell-ish (or hack-ish) way to turn autoflush on, as in

use IO::Handle;
print 1; # doesn't print, puts 1 in the buffer
STDOUT->autoflush(1)
print 1; # prints 1 immediately (and the previously buffered 1)
print "\n2\n"; # \n forces a flush
[download]

Tutorials: Input and Output: Suffering from Buffering?

Comment on Re^3: ztk-webcam snapshot-grabber Select or Download Code

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Re^4: ztk-webcam snapshot-grabber by PhillipHuang (Beadle) on Jul 22, 2011 at 08:35 UTC
Just think me as a kid. When i run the below script, it really print 1(not as your comment). `phillip@athens:/tmp$ perl -e 'use IO::Handle;print 1;'` And thanks for the reply(Re^3: ztk-webcam snapshot-grabber), I find "$_" mean in http://perldoc.perl.org/perlvar.html. Is its printing "1" caused by there's nothing to be flush in buffer?	[reply] [d/l]
Re^5: ztk-webcam snapshot-grabber by Anonymous Monk on Jul 22, 2011 at 09:08 UTC
Just think me as a kid. When i run the below script, it really print 1(not as your comment). Well, that is not my code ;) buffering means printing doesn't happen immediately, it is delayed, barring a crash of some sort, it will happen eventually, like when a program exits, when all buffers are flushed . Try this instead `use IO::Handle; print "\n2\n"; # \n forces a flush, 2 printed immediately print 1; # doesn't print, puts 1 in the buffer sleep 1; # still nothing printed STDOUT->autoflush(1); # buffered 1 printed print "\n"; # prints immediately sleep 1; print 1; # prints 1 immediately sleep 1; print "\n";` [download] Without sleep, you wouldn't notice the buffering, and without sleep, I forget to add sleep or test my original code:) but its ok since I linked to the tutorial for you to read, and run its code, and believe what it says is true :) because it is :) I did, I did taw a puddy tat Anonymous Tweety Pie	[reply] [d/l]


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