Silence the stdout of a function in Python without

Question

Is there a way in Python to silence stdout without wrapping a function call like following?

Original Broken Code:

from sys import stdout
from copy import copy
save_stdout = copy(stdout)
stdout = open('trash','w')
foo()
stdout = save_stdout

Edit: Corrected code from Alex Martelli

import sys
save_stdout = sys.stdout
sys.stdout = open('trash', 'w')
foo()
sys.stdout = save_stdout

That way works but appears to be terribly inefficient. There has to be a better way. Any ideas?

Answer 1

Chiming in very late to this with what I thought was a cleaner solution to this problem.

import sys, traceback

class Suppressor(object):

    def __enter__(self):
        self.stdout = sys.stdout
        sys.stdout = self

    def __exit__(self, type, value, traceback):
        sys.stdout = self.stdout
        if type is not None:
            # Do normal exception handling

    def write(self, x): pass

Usage:

with Suppressor():
    DoMyFunction(*args,**kwargs)

Answer 2

redirect_stdout() has been added to contextlib since python 3.4

For python >= 3.4, this should do it:

import contextlib
import io

with contextlib.redirect_stdout(io.StringIO()):
    foo()