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Re: % stdout?

From: Terry (beebumgmail.com)
Date: Thu Nov 09 2006 - 11:07:57 CST

Also, I think you mean:

    fprintf(stdout, foo);

not

    fprintf(stdout, bar);

right?

Terry

On Thu, Nov 09, 2006 at 04:49:20PM +0000, Andreas Kahari wrote:
> Have a look in your C code book. The you will need to printf "%%" to get a
> '%'.
>
> Andreas
>
>
> On 09/11/06, Cassio B. Caporal <cassioostec.com.br> wrote:
> > Hey,
> >
> > I have problems to print '%' in stdout... Suppose code below:
> >
> > #include <stdio.h>
> >
> > main() {
> > char foo[] = "bar=30%\n";
> > fprintf(stdout, bar);
> > }
> >
> > OpenBSD returns : bar=30
> > Linux returns : bar=30%
> >
> > How can I solve this? Thanks,
> >
> >
>
>
> --
> Andreas Kahari
> Somewhere in the general Cambridge area, UK