Strange behaviour of std::cout in Linux

Question

I am trying to print results in 2 nested for cycles using std::cout. However, the results are not printed to the console immediately, but with delay (after both for cycles or the program have been finished).

I do not consider such behavior normal, under Windows printing works OK. The program does not use threads.

Where could be the problem? (Ubuntu 10.10 + NetBeans 6.9).

No code, common printing in two nested for cycles. There is nothing to analyze in code :-)) — johny, Mar 25 '11 at 20:18
As Cameron pointed out, it sounds like you're not flushing the buffer by using std::endl (write new line and flush buffer) or std::flush (just flush buffer). — Peter Huene, Mar 25 '11 at 20:19
@Cameron: The problem arises when printing dot representing % of processed data. — johny, Mar 25 '11 at 20:21
@johny: We wouldn't have to ask all these questions if you posted the code. "Common printing" is hardly specific enough for us to help, except by guessing — Cameron, Mar 25 '11 at 20:21

MacGucky · Answer 1 · 2011-03-25T20:41:00.183

9

std::cout is an stream, and it is buffered. You can flush it by several ways:

std::cout.flush();

std::cout << std::flush;

std::cout << std::endl;  // same as:  std::cout << "\n" << std::flush`

johny:

I am flushing the buffer before the cycle using std::endl. The problem arises when printing dot representing % of processed data inside the cycle.

If you flush the buffer before the cycle, that does not affect the output in the cycle. You have to flush in or after the cycle, to see the output.

edited Mar 25 '11 at 20:41

answered Mar 25 '11 at 20:20

MacGucky

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I am flushing the buffer before the cycle using std::endl. The problem arises when printing dot representing % of processed data inside the cycle. – johny Mar 25 '11 at 20:31
@johny: You need to flush inside the cycle, or else `cout` can delay printing as long as it wants. If you don't want newlines, use `std::flush` instead of `std::endl`. – aschepler Mar 25 '11 at 20:39
@aschepler - I would like to add (while it isn't important for your case) that `::std::cout` cannot delay 'as long as it wants'. First, if you read data using `::std::cin` a flush will happen. Secondly, it must flush before your program ends if it ends by a normal return from `main` (even if that return is non-zero). Also, buffer flushes are expensive, don't do them unless you have to (including by using `::std::endl`). – Omnifarious Mar 25 '11 at 21:54

score 1 · Answer 2 · answered Mar 25 '11 at 20:27

If you don't flush your output, your output is not guaranteed to be visible outside your program. The fact that it is not printing in your terminal is just a consequence of the default behavior in linux to do line buffering when the output is a tty. If you run your program on linux with its output piped to another thing, like

 ./your_program | cat

Then the default buffer will be MUCH larger, most likely it will be at least 4096 bytes. So nothing will get displayed until the big buffer gets full. but really, the behaviour is OS specific unless you flush std::cout yourself.

To flush std::cout, use :

std::cout << std::flush;

also, using

std::cout << std::endl;

is a shortcut to

std::cout << '\n' << std::flush;

Strange behaviour of std::cout in Linux

2 Answers2

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