From 7405c16533f30acf6c0b1f2035898666871449c3 Mon Sep 17 00:00:00 2001
From: Berker Peksag <berker.peksag@gmail.com>
Date: Thu, 21 Jan 2016 23:59:49 +0200
Subject: [PATCH] Issue #18620: Improve Pool examples in multiprocessing
 documentation

A single call to Pool.apply_async() will create only one process. To use all
of the pool's processes, it should be invoked multiple times:

    with Pool(processes=4) as pool:
        results = [pool.apply_async(func, ()) for i in range(4)]

Patch by Davin Potts.
---
 Doc/library/multiprocessing.rst | 37 +++++++++++++++++++++++----------
 1 file changed, 26 insertions(+), 11 deletions(-)

diff --git a/Doc/library/multiprocessing.rst b/Doc/library/multiprocessing.rst
index 5fdd0e238ee..8209ae9ac8d 100644
--- a/Doc/library/multiprocessing.rst
+++ b/Doc/library/multiprocessing.rst
@@ -361,8 +361,9 @@ processes in a few different ways.
 
 For example::
 
-   from multiprocessing import Pool
-   from time import sleep
+   from multiprocessing import Pool, TimeoutError
+   import time
+   import os
 
    def f(x):
        return x*x
@@ -378,15 +379,29 @@ For example::
            for i in pool.imap_unordered(f, range(10)):
                print(i)
 
-           # evaluate "f(10)" asynchronously
-           res = pool.apply_async(f, [10])
-           print(res.get(timeout=1))             # prints "100"
+           # evaluate "f(20)" asynchronously
+           res = pool.apply_async(f, (20,))      # runs in *only* one process
+           print(res.get(timeout=1))             # prints "400"
 
-           # make worker sleep for 10 secs
-           res = pool.apply_async(sleep, [10])
-           print(res.get(timeout=1))             # raises multiprocessing.TimeoutError
+           # evaluate "os.getpid()" asynchronously
+           res = pool.apply_async(os.getpid, ()) # runs in *only* one process
+           print(res.get(timeout=1))             # prints the PID of that process
+
+           # launching multiple evaluations asynchronously *may* use more processes
+           multiple_results = [pool.apply_async(os.getpid, ()) for i in range(4)]
+           print([res.get(timeout=1) for res in multiple_results])
+
+           # make a single worker sleep for 10 secs
+           res = pool.apply_async(time.sleep, (10,))
+           try:
+               print(res.get(timeout=1))
+           except TimeoutError:
+               print("We lacked patience and got a multiprocessing.TimeoutError")
+
+           print("For the moment, the pool remains available for more work")
 
        # exiting the 'with'-block has stopped the pool
+       print("Now the pool is closed and no longer available")
 
 Note that the methods of a pool should only ever be used by the
 process which created it.
@@ -2171,13 +2186,14 @@ with the :class:`Pool` class.
 The following example demonstrates the use of a pool::
 
    from multiprocessing import Pool
+   import time
 
    def f(x):
        return x*x
 
    if __name__ == '__main__':
        with Pool(processes=4) as pool:         # start 4 worker processes
-           result = pool.apply_async(f, (10,)) # evaluate "f(10)" asynchronously
+           result = pool.apply_async(f, (10,)) # evaluate "f(10)" asynchronously in a single process
            print(result.get(timeout=1))        # prints "100" unless your computer is *very* slow
 
            print(pool.map(f, range(10)))       # prints "[0, 1, 4,..., 81]"
@@ -2187,9 +2203,8 @@ The following example demonstrates the use of a pool::
            print(next(it))                     # prints "1"
            print(it.next(timeout=1))           # prints "4" unless your computer is *very* slow
 
-           import time
            result = pool.apply_async(time.sleep, (10,))
-           print(result.get(timeout=1))        # raises TimeoutError
+           print(result.get(timeout=1))        # raises multiprocessing.TimeoutError
 
 
 .. _multiprocessing-listeners-clients: