hail linux
my freind thats becuse printf uses buffered output and
incidentally the buffer got cleared sooner when u used the \n character.
try flushing the buffer
using fflush call
ashurazdan
On Thu, 4 Nov 1999, Anand Singh Bisen wrote:
> 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
> > ----------------------------------------------------------------
> > --------------------------------------------------------------------
> > The Linux India Mailing List Archives are now available. Please search
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>
> --------------------------------------------------------------------
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