The reason is : The stdout stream is buffered.
Please explicitly flush this buffer after the printf. Appending a \n causes
a buffer flush to happen.
Prashast Kumar
ICO
NIIT Ltd.
6203389
-----Original Message-----
From: Anand Singh Bisen [mailto:[EMAIL PROTECTED]]
Sent: Thursday, November 04, 1999 12:24 AM
To: [EMAIL PROTECTED]
Subject: Re: [LI] C Programming problem
type
printf("Loading..\n");
then it'll work properly
anand
Mrinal Kalakrishnan wrote:
> Hi,
>
> This particular piece of code behaves differently when compiled with gcc
> (Linux), when compared to Turbo C or DJGPP (in DOS).
>
> #include <stdio.h>
>
> void main()
> {
> printf("Loading..");
> float i,j;
> j=12345.6789;
> for (i=0.0;i<50000.0;i+=0.01) j=j*j;
> printf("Done\n");
> }
>
> The expected behaviour would be that "Loading" gets printed, then a delay,
and
> then "Done" gets printed. But, that doesn't happen. Instead, there is a
long
> delay initially, then "loading" and "done" get printed immediately one
after
> the other. However, if I add a newline in the "Loading" string, then it
gets
> printed immediately, followed by the delay.
>
> Anybody faced this problem before - any solutions?
>
> ----------------------------------------------------------------
> Mrinal Kalakrishnan
> [EMAIL PROTECTED]
> http://listen.to/mrinal
> ----------------------------------------------------------------
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