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Home/ Questions/Q 783165
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T20:30:15+00:00 2026-05-14T20:30:15+00:00

What I wanted is printing out 5 dots that a dot printed per a

  • 0

What I wanted is printing out 5 dots that a dot printed per a second using time.sleep(), but the result was 5 dots were printed at once after 5 seconds delay.
Tried both print and sys.stdout.write, same result.

Thanks for any advices.

import time
import sys

def wait_for(n):
    """Wait for {n} seconds. {n} should be an integer greater than 0."""
    if not isinstance(n, int):
        print 'n in wait_for(n) should be an integer.'
        return
    elif n < 1:
        print 'n in wait_for(n) should be greater than 0.'
        return
    for i in range(0, n):
        sys.stdout.write('.')
        time.sleep(1)
    sys.stdout.write('\n')

def main():
    wait_for(5)    # FIXME: doesn't work as expected

if __name__ == '__main__':
    try:
        main()
    except KeyboardInterrupt:
        print '\nAborted.'
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  1. Editorial Team
    Editorial Team
    2026-05-14T20:30:16+00:00Added an answer on May 14, 2026 at 8:30 pm

    You need to flush after writing.

    sys.stdout.write('foo')
    sys.stdout.flush()
    wastetime()
    sys.stdout.write('bar')
    sys.stdout.flush()
    
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