In article <[EMAIL PROTECTED]>,
Anthony Greene <[EMAIL PROTECTED]> wrote:
> On Sat, 01 Apr 2006 14:56:02 -0500, Roy Smith wrote:
>
> > I've got a bunch of strings in a list:
> >
> > vector = []
> > vector.append ("foo")
> > vector.append ("bar")
> > vector.append ("baz")
> >
> > I want to send all of them out a socket in a single send() call, so
> > they end up in a single packet (assuming the MTU is large enough). I
> > can do:
> >
> > mySocket.send ("".join (vector))
> >
> > but that involves creating an intermediate string. Is there a more
> > efficient way, that doesn't involve that extra data copy?
>
> Is sendall() what you're looking for?
No. Sendall() is actually what I'm using now. It handles the other side
of the issue; issuing repeated send() calls if the system fragments your
buffer. I'm trying to aggregate lots of small buffers into one large one.
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