打印功能最近打印耗时的东西

Question

i am trying to understand, why start... is not print first. but got nothing...

from __future__ import print_function
# from time import sleep

def sleeping():
    # sleep(2)
    for i in range(60000000):  # or xrange(60000000)
        pass

print('start...', end='')
sleeping()
print('stop.')

same thing happen in this

from __future__ import print_function
from time import sleep

print('this will print lately')

sleep(2)

print('and this one')

i have tried this on both python 3 and 2. on python 2 it works fine.but causes problem on python 3.

EDIT :

different output on sublime text and terminal (may be configuration problem on sublime). on terminal, second one works well in both version and first one causes problem in both version.

Answer 1

sys.stdout (which print prints to by default) is line-buffered when running interactively.

Since print('start...', end='') makes the line not end, the output isn't automatically flushed out.

To force the buffer to be flushed, you can specify flush=True when calling print .

In your case: print('start...', end='', flush=True)

Another option is to flush the stream manually by using sys.stdout.flush() .

Further reading:

• Documentation of print
• Documentation of sys.stdout

Answer 2

The print statements are in most of the cases buffered. In order to force print to stdout you have to flush the buffer (using sys):

from __future__ import print_function
# from time import sleep
import sys


def sleeping():
    # sleep(2)
    for i in range(60000000):  # or xrange(60000000)
        pass

print('start...', end='')
sys.stdout.flush()
sleeping()
print('stop.')