python - Creating a custom sys.stdout class?

Question

What I'm trying to do is simply have the output of some terminal commands print out to a wx.TextCtrl widget. I figured the easiest way to accomplish this is to create a custom stdout class and overload the write function to that of the widget.

stdout class:

class StdOut(sys.stdout):
    def __init__(self,txtctrl):
        sys.stdout.__init__(self)
        self.txtctrl = txtctrl

    def write(self,string):
        self.txtctrl.write(string)

And then I would do something such as:

sys.stdout = StdOut(createdTxtCtrl)    
subprocess.Popen('echo "Hello World!"',stdout=sys.stdout,shell=True)

What results is the following error:

Traceback (most recent call last):
File "mainwindow.py", line 12, in <module>
from systemconsole import SystemConsole
File "systemconsole.py", line 4, in <module>
class StdOut(sys.stdout):
TypeError: Error when calling the metaclass bases
file() argument 2 must be string, not tuple

Any ideas to fix this would be appreciated.

Answer 1

sys.stdout is not a class, it's an instance (of type file).

So, just do:

class StdOut(object):
    def __init__(self,txtctrl):
        self.txtctrl = txtctrl
    def write(self,string):
        self.txtctrl.write(string)

sys.stdout = StdOut(the_text_ctrl)

No need to inherit from file, just make a simple file-like object like this! Duck typing is your friend...

(Note that in Python, like most other OO languages but differently from Javascript, you only ever inherit from classes AKA types, never from instances of classes/types;-).