Hi Stephen,
> From: Stephen Warren
>
> Currently, Spawn.expect() imposes its timeout solely upon receipt of
> new data, not on its overall operation. In theory, this could cause
> the timeout not to fire if U-Boot continually generated output that
> did not match the expected patterns.
>
> Fix the code to additionally impose a timeout on overall operation,
> which is the intended mode of operation.
>
> Signed-off-by: Stephen Warren
> ---
> test/py/u_boot_spawn.py | 7 ++-
> 1 file changed, 6 insertions(+), 1 deletion(-)
>
> diff --git a/test/py/u_boot_spawn.py b/test/py/u_boot_spawn.py
> index 1baee63df25c..df4c67597cab 100644
> --- a/test/py/u_boot_spawn.py
> +++ b/test/py/u_boot_spawn.py
> @@ -122,6 +122,7 @@ class Spawn(object):
> if type(patterns[pi]) == type(''):
> patterns[pi] = re.compile(patterns[pi])
>
> +tstart_s = time.time()
> try:
> while True:
> earliest_m = None
> @@ -142,7 +143,11 @@ class Spawn(object):
> self.after = self.buf[pos:posafter]
> self.buf = self.buf[posafter:]
> return earliest_pi
> -events = self.poll.poll(self.timeout)
> +tnow_s = time.time()
> +tdelta_ms = (tnow_s - tstart_s) * 1000
> +if tdelta_ms > self.timeout:
> +raise Timeout()
> +events = self.poll.poll(self.timeout - tdelta_ms)
> if not events:
> raise Timeout()
> c = os.read(self.fd, 1024)
Reviewed-by: Lukasz Majewski
--
Best regards,
Lukasz Majewski
Samsung R Institute Poland (SRPOL) | Linux Platform Group
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