Read file with timeout in Python

Question

Within Linux, there is a file, /sys/kernel/debug/tracing/trace_pipe, which as the name says, is a pipe. So, let's say I want to read the first 50 bytes from it using Python - and I run the following code:

$sudo python -c 'f=open("/sys/kernel/debug/tracing/trace_pipe","r"); print f; print f.read(50); f.close()<br>
<open file '/sys/kernel/debug/tracing/trace_pipe', mode 'r' at 0xb7757e90>

We can see that opening the file goes fast ( if we have the superuser permissions ) - however, if the trace_pipe file is empty at that moment, it will simply block ( and even if there is content, the content will be dumped until there is no more, and then again the file will block ). Then I have to press Ctrl-C to interrupt the Python script with a KeyboardInterrupt...

How can I have Python 2.7 do a read with timeout?

That is, I want to instruct Python to "try read 50 bytes from this file; if you don't succeed after one second, give up and return"?

Answer 1

Use

os.read(f.fileno(), 50)

instead. That does not wait until the specified amount of bytes has been read but returns when it has read anything (at most the specified amount of bytes).

This does not solve your issue in case you've got nothing to read from that pipe. In that case you should use select from the module select to test whether there is something to read.

EDIT:

Testing for empty input with select:

import select
r, w, e = select.select([ f ], [], [], 0)
if f in r:
  print os.read(f.fileno(), 50)
else:
  print "nothing available!"  # or just ignore that case

Answer 2

f = os.open("/sys/kernel/debug/tracing/trace_pipe", os.O_RDONLY|os.O_NONBLOCK)

Should prevent blocking (works in Unix only).. No need for select here..