Python，为什么我不能覆盖Popen.stdout

Question

Why is possibile to override sys.stdout but not subprocess.Popen.stdout, with my own class. If it is possibile to override please show me how ...

I'm working on some GUI project, and i want to print output of some other program on text_view in "real_time", not when everything is done.

when i am overriding sys.stdout i do somthing like this :

class MyStdOut :
  def __init__(self) : self.text = ""
  def write(self, string) : self.text += string

here is some program 'pyscript.py'

import os

def main() :
  for i in range(10) :
    print i
    os.system('sleep 0.1') ## this is just to make some delay, i did also 3 loops

if __name__ == '__main__' : main()

and here is main program :

import subprocess as sub

def main() :
  popen = sub.Popen('./pyscript.py', stdout=sub.PIPE)
  for line in popen.stdout : print 'Line :', line
  print 'Main done'

if __name__ == '__main__' : main()

Still it is not 'real time', i get everything when it ends.

I also tried communicate, but get same result.

In my real program i have to print output of other program on text_view (GUI GTK+). I suspect that subprocess.stdout must be file, but when i tried to make some class that inherits from IOBase i got errors too.

Everybody say that it is possible, but still I tried many different ways with no success.

Answer 1

You can achieve the same by using:

p = Popen(args, ..., stdout=subprocess.PIPE)
for line in p.stdout:
   print line

where p.stdout behaves like PIPE, so you can read data from it as soon as it arrives. It is not strict realtime, because some buffering comes in play, but still this is a viable alternative.