printf in C++

Hi,
I am changing to work in C++ from C. In my C++ code, I have a line like this

Code:

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  printf("%d %d %d %d \n", x0,y0,x1,y1);

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There is nothing printed out on the console. If I call cout next, then I can get the output of both

Code:

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  printf("%d %d %d %d\n", x0,y0,x1,y1);
  cout<<x0<<" "<<y0<<" "<<x1<<" "<<y1<<endl;

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It is like cout push printf to do what it is supposed to. Any idea how to correctly use printf in C++? Thanks!

Update:
There was mistake in my first post.
with "\n" printf works, while without "\n" it doesn't.
Just out of curiosity, why the latter happens?

printf should flush the buffer when you print '\n', but i'm not sure if that's still a rule in c++ like it is in c. you can make sure it gets flushed by adding fflush after printf.

Code:

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printf("%d %d %d %d \n", x0,y0,x1,y1);
fflush(stdout);

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According to Josuttis (The C++ Standard Library: A Tutorial and Reference):

Quote:

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By default, these standard streams are synchronized with the standard streams of C. That is, the C++ standard library ensures that the order of mixed output with C++ streams and C streams is preserved. Before any buffer of standard C++ streams writes data it flushes the buffer of the corresponding C streams and vice versa. Of course, this synchronization takes some time. If it isn't necessary you can turn it off by calling sync_with_stdio(false) before any input or output is done.

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I do not think sync_with_stdio() is the solution here, but it explains why the use of std::cout causes stdout to be flushed. Perhaps you can try fflush(), e.g.,

Code:

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printf("%d %d %d %d \n", x0, y0, x1, y1);
fflush(stdout);

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but I had the impression that writing a new line should trigger the flush anyway since the stream is line buffered.

Does the program continue to run or terminate normally after the print statement. In any case you should see the output at latest when the program stops for next input or when it finishes. What should happen so that you are able to tell that it just doesn't produce output?

Also, endl definitely flushes the output of cout.

Are you getting just one line print out, because if you are it would indicate it's nothing to do with the buffer. Though this may not be relevant, do you have the correct #includes? because I think (though I can't remember at the mo) you need the #include <cstdio> directive.

There was mistake in my first post.
with "\n" printf works, while without "\n" it doesn't.
Just out of curiosity, why the latter happens?

Quote:

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Originally Posted by lehe

with "\n" printf works, while without "\n" it doesn't.
Just out of curiosity, why the latter happens?

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As I noted, stdout is typically line buffered, so it flushes when a new line is printed.

printf is supposed to flush the buffer when it sees '\n'. if it doesn't see '\n', it doesn't have to flush the buffer unless it gets full. and you don't see output until the buffer is flushed.

So why do you need to use printf in C++?

Quote:

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Originally Posted by Elysia

So why do you need to use printf in C++?

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complex formatting can be easier and shorter with printf.

Quote:

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Originally Posted by Meldreth

complex formatting can be easier and shorter with printf.

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That's true, and for the best of both worlds (and something non-standard, heh) there is the Boost format library.

Non-standard perhaps, but it's darn portable, so usually I say boost is worth it :)