Cout in While loop Strange behaviour

Question

My code look like below

int i=0;
while(i<10){
cout<<"Hello";
sleep(1);
i++
}

In Windows the code prints on each loop but in Linux it prints everything after exiting while loop . And also if I put an endl at the last of cout then it prints on each loop. Why this happening ?. Can anyone explain this behavior?.

Cout works using a buffer. This thread has more information: http://stackoverflow.com/questions/9274057/c-cout-and-cin-buffers-and-buffers-in-general — bhekman, May 18 '13 at 05:31
borisbn is also right. You should copy&paste your complete code. Some important pieces are missing. — bhekman, May 18 '13 at 05:32

score 4 · Accepted Answer · answered May 18 '13 at 05:29

4

Try to use cout.flush(); maybe the two OS has different policy in term of buffering the stdout.

answered May 18 '13 at 05:29

Felice Pollano

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score 1 · Answer 2 · answered May 18 '13 at 05:30

1

For efficiency reasons, sometimes the standard streams will be implemented with a buffer. Making lots of tiny writes can be slow, so it will store up your writes until it gets a certain amount of data before writing it all out at once.

Endl forces it to write out the current buffer, so you'll see the output immediately.

answered May 18 '13 at 05:30

Antimony

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`endl` forces the output by calling `cout.flush`. – Mats Petersson May 18 '13 at 06:58

score 0 · Answer 3 · answered May 18 '13 at 07:14

0

#include <iostream>
using namespace std;

int main()
{
    int i = 0;
    while(i < 10){
        cout << "Hello" << endl;
        sleep(1);
        ++i;
    }
}

answered May 18 '13 at 07:14

Zera42

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Cout in While loop Strange behaviour

3 Answers3