thread starts running before calling Thread.start

Question

t1=threading.Thread(target=self.read())
print "something"
t2=threading.Thread(target=self.runChecks(), args=(self))

self.read runs indefinitely, so the program won't ever reach the print line. How is this possible without calling t1.start()? (Even if I call that, it shold start running and go on to the next line, shouldn't it?)

Answer 1

You're passing the result of self.read to the target argument of Thread. Thread expects to be passed a function to call, so just remove the parentheses and remember to start the Thread:

t1=threading.Thread(target=self.read)
t1.start()
print "something"

For targets that need arguments, you can use the args and kwargs arguments to threading.Thread, or you can use a lambda. For example, to run f(a, b, x=c) in a thread, you could use

thread = threading.Thread(target=f, args=(a, b), kwargs={'x': c})

or

thread = threading.Thread(target=lambda: f(a, b, x=c))

though watch out if you pick the lambda - the lambda will look up f, a, b, and c at time of use, not when the lambda is defined, so you may get unexpected results if you reassign any of those variables before the thread is scheduled (which could take arbitrarily long, even if you call start immediately).