I've slightly modified the signal example from the official docs (bottom of page).

I'm calling sleep 10 but I would like an alarm to be raised after 1 second. When I run the following snippet it takes way more than 1 second to trigger the exception (I think it runs the full 10 seconds).

import signal, osdef handler(signum, frame):print 'Interrupted', signumraise IOError("Should after 1 second")signal.signal(signal.SIGALRM, handler)
signal.alarm(1)os.system('sleep 10')signal.alarm(0)

How can I be sure to terminate a function after a timeout in a single-threaded application?

From the docs:

A Python signal handler does not get executed inside the low-level (C)signal handler. Instead, the low-level signal handler sets a flagwhich tells the virtual machine to execute the corresponding Pythonsignal handler at a later point(for example at the next bytecodeinstruction).

Therefore, a signal such as that generated by signal.alarm() can't terminate a function after a timeout in some cases. Either the function should cooperate by allowing other Python code to run (e.g., by calling PyErr_CheckSignals() periodically in C code) or you should use a separate process, to terminate the function in time.

Your case can be fixed if you use subprocess.check_call('sleep 10'.split()) instead of os.system('sleep 10').