how to convert ads to iar?

"lzh08" <cppcpldfpga@gmail.com> wrote in message 
news:1179732737.322113.259380@b40g2000prd.googlegroups.com...
> who could tell me?thx

Without knowing what you want to convert then no-one can help you. If 
you've written ANSI C then everything will run just dandy. However, if 
you've used compiler extensions then you need to find the appropriate 
extensions under IAR. Their support people are very helpful (at a price) 
and respond quickly.

If you write code that conforms to standards and liberally document the 
places where it doesn't then you won't get this problem in the future. 

On 21/05/2007 Chris Hills wrote:

> In article <f2sl6q.1o4.1@stefan.msgid.phost.de>, Stefan Reuther
> <stefan.news@arcor.de> writes
> > Chris Hills wrote:
> > > /* "keil" to "iar" will need some thinking about */
> > 
> > No problem.
> > 
> > #include <stdio.h>
> > int main() {
> >  char x[6];
> >  fgets(x, 6, stdin);
> >  printf("%c%c%c\n", x[2], x[1] - 4, x[0]+x[3]-x[1]);
> >  return 0;
> > }
> > 
> > Even works for upper-case :-)
> 
> 
> Any one want to try "windriver" to" iar"? :-)

#include <stdio.h>
int main() {
  char x[11];
  char *p = &x[1];
  fgets(x, 11, stdin);
  printf("%c", *p);
  p += 2;
  printf("%c", *p - 3);
  p += 5;
  printf("%c\n", *p);
  return 0;
}


-- 
John B

Chris Hills wrote:
>
... snip ...
> 
> /* how to convert ads to iar? */
> 
> #include <stdio.h>
> 
> int main(void )
> {
>   char one = 'a';
>   char two  ='d';
>   char three ='s';
>   char four  = 8;
>   char five = 3;
>   char six =1;
>   printf("%c %c %c \n", one, two, three);
> 
>   one = one + four;
>   two = two -five;
>   three = three - six;
>   printf("%c %c %c \n", one, two, three);
> 
>   return 0;
> }

Fails if the char. set is ebcdic (or some others) :-)

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