tee() with your splice()?
Posted Apr 13, 2006 16:36 UTC (Thu) by RobSeace
In reply to: tee() with your splice()?
Parent article: tee() with your splice()?
I'm not sure I understand why it would have to be any different internally...
I understand what tee() is doing, but I was just saying why not have a new
flag to splice() specify doing exactly that, rather than add a new syscall
to do it... Then, your example code that does tee() followed by splice(),
would instead just do two separate splice() calls... But, internally,
splice() could do the same thing tee() does (when the hypothetical new flag
is set), couldn't it? Or, is there some internal implementation trickery
that I'm completely missing? (Or, is it just that it seems unclean to add
such behavior to splice()? Ie: splice() should always be expected to
consume the input, and you don't want break that assumption, even with a
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