On Wednesday 18 October 2006 00:25, Fredrik Lundh wrote:
> |feature. the "exec" system call operates on a lower level than the
> |stdio buffering system.
I did in this manner:
for exe in ('imap4', 'pop3'):
if exe in cfgfil[optsrv + '.protocol']:
exe = exe[:4]; exe = 'cal
If you wanted to make sure stdio was flushed you could always do...
sys.stdout.write("Message\n")
sys.stdout.flush()
Thomas Guettler wrote:
> Hi,
>
> I noticed, that sys.stout does not get flushed before the process is
> replaced. The last print statements (before execvp()) disappear.
>
> It only
Thomas Guettler wrote:
> Is this a bug or feature?
feature. the "exec" system call operates on a lower level than the
stdio buffering system.
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