Is it possible to use an inline function in a Thread call?

Question

The documentation for threading.Thread(target=...) states that

target is the callable object to be invoked by the run() method. Defaults to None, meaning nothing is called.

I usually use it like this:

import threading

def worker():
    a = 3
    print("bonjour {a}".format(a=a))

threading.Thread(target=worker).start()

Is there a way to chain the function elements in target so that a new one does not need to be defined? Something like (pseudocode obviously)

threading.Thread(target=(a=3;print("bonjour {a}".format(a=a))).start()

I have a bunch of very short calls to make in the Thread call and would like to avoid multiplication of function definitions.

Answer 1

You can use a lambda function in Python 3.x

import threading

threading.Thread(target=lambda a: print("Hello, {}".format(a)), args=(["world"]))

You should probably take a look at this SO question to see why you can't use print in Python 2.x in lambda expressions.

---

Actually, you can fit many function calls into your lambda:

from __future__ import print_function # I'm on Python 2.7
from threading import Thread

Thread(target=(lambda: print('test') == print('hello'))).start()

That will print both test and hello.

Answer 2

I don't really like using exec, but in Python3.x it is a function, so you could do

threading.Thread(target=exec, args=('a=3; print("bonjour {a}".format(a=a))',)