Getting an output while running a program without using \n

[aeson25]

Hello,

I was wondering if anyone could help me with a quick problem. I'm trying to run a program that takes a lot of time for computations. I've been wanting to use a clean way to show that the program is still running. e.g. a period "." for every few cycles, and/or that rotating slash "|" -> "/" -> "-" -> "\" that you see in old DOS programs. The thing is, the only way output seems to come out is when I use the new line "\n" escape sequence. And that gives me lines of output that I don't want. Nothing else seems to work in the program. Can someone help?

[Warnis]

For the | -> / -> - -> \
You can use system("CLS") which will clear the screen

[helios]

Bad advice is bad.

char *busy="|/-\\";
char busyindex=0;
/*loop*/{
	//...
	printf("\b%c",busy[busyindex]);
	busyindex=(busyindex+1)%4;
	//...
}

[aeson25]

Thanks for the suggestion, but I think you guys missed my point. Nothing prints during the loop when I do that. It will only show when the program gets to the next "\n" in the program. I'll give and example.

When I do:

/*loop 5 times*/{
	//...stuff that takes 5 seconds to process
	printf(".");
	//...
}
printf("Some other statement\n");

It only prints ".....Some other statement" all at once after 25 seconds. The periods don't come every 5 seconds.

The only thing I know of right now that flushes the buffer onto the screen is "\n". Which I don't want.

[aeson25]

Btw, I don't think this should make a difference, but I'm using cygwin with gcc version 3.4.4 (cygming special, gdc 0.12, using dmd 0.125)

[jsmith]

stdout by default is line buffered which means the output is only flushed to the screen when a newline is output.

What you need is this:

cout << '.' << flush;

The flush says to force the output to be written to the screen immediately.

[aeson25]

Awesome!! That worked perfectly. Thanks!