How to use a with statement to disable printing?

Question

I want to use something like this:

print('Hello')
with dont_print():
    print('some other text')
    print('more text')

print('World')

The output should be:

Hello
World

So everything inside the dont_print won't get printed. How can I get a function like this?

Have you seen the context manager [`redirect_stdout`](https://docs.python.org/3/library/contextlib.html#contextlib.redirect_stdout)? — ikkuh, Aug 07 '19 at 11:28

score 2 · Answer 1 · answered Aug 07 '19 at 11:33

2

from unittest.mock import patch

print('Hello')
with patch('builtins.print'):
    print('some other text')
    print('more text')

print('World')

answered Aug 07 '19 at 11:33

ipaleka

3,745
2
13
33

score 1 · Accepted Answer · answered Aug 07 '19 at 11:33

Overwrite your standard output to some file object. Here is your modified version.

import sys, os

# Disable
def blockPrint():
    sys.stdout = open(os.devnull, 'w')

# Restore
def enablePrint():
    sys.stdout = sys.__stdout__


print('Hello')

blockPrint()
print('some other text')
print('more text')

enablePrint()
print('World')

More on this here

score 0 · Answer 3 · answered Aug 07 '19 at 11:35

You can use redirect_stdout as context manager to redirect stdout to nothingness with os.devnull:

import os
from contextlib import redirect_stdout

print('Hello')
with open(os.devnull, 'w') as f:
    with redirect_stdout(f):
        print('some other text')
        print('more text')
print('World')

How to use a with statement to disable printing?

3 Answers3