En Mon, 09 Feb 2009 07:43:36 -0200, John O'Hagan <resea...@johnohagan.com>
escribió:
I'm using the socket module (python 2.5) like this (where 'options'
refers to
an optparse object) to connect to the Fluidsynth program:
host = "localhost"
port = 9800
fluid = socket(AF_INET, SOCK_STREAM)
try:
fluid.connect((host, port)) #Connect if fluidsynth is
running
except BaseException:
print "Connecting to fluidsynth..." #Or start fluidsynth
soundfont = options.soundfont
driver = options.driver
Popen(["fluidsynth", "-i", "-s", "-g", "0.5",
"-C", "1", "-R", "1", "-l", "-a", driver, "-j",
soundfont])
timeout = 50
while 1:
timeout -= 1
if timeout == 0:
print "Problem with fluidsynth: switching to
synth."
play_method = "synth"
break
try:
fluid.connect((host, port))
except BaseException:
sleep(0.05)
continue
else:
break
(I'm using BaseException because I haven't been able to discover what
exception class[es] socket uses).
Usually socket.error, which is a subclass of IOError, but others might
happen too I think. In any case, the most generic except clause you should
use is
try:
except Exception: ...
(because you usually don't want to catch KeyboardInterrupt nor SystemExit,
that is, let Ctrl-C and sys.exit() do their work)
The problem is that this fails to connect ( the error is "111: Connection
refused") the first time I run it after booting if fluidsynth is not
already
running, no matter how long the timeout is; after Ctrl-C'ing out of the
program, all subsequent attempts succeed. Note that fluidsynth need not
be
running for a success to occur.
Always the same exception? In both lines?
--
Gabriel Genellina
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