Re: typecasting - explain

[Steve Graegert <graegerts@gmail.com>]

On 9/9/05, _z33 <timid.Gentoo@gmail.com> wrote:
> Steve Graegert wrote:
> > On 9/9/05, _z33 <timid.Gentoo@gmail.com> wrote:
> >
> >>Steve Graegert wrote:
> >>
> >>>      char *s;
> >>>      int  *i;
> >>>
> >>>      s = (int *)malloc(128 * sizeof(char));
> >>>      i = (char *)malloc(128 * sizeof(int));
> >>>
> >>>      s = "malloc is cool!";
> >>>      printf("s: %s\n", s);
> >>>      printf("s: %d\n", s);
> >>>
> >>>      i = "malloc is cool!";
> >>>      printf("i: %s\n", i);
> >>>      printf("i: %d\n", i);
> >>>
> >>>This piece of code prints:
> >>>
> >>>      s: malloc is cool!
> >>>      s: 4350328
> >>>      i: malloc is cool!
> >>>      i: 4350328
> >>>
> >>
> >>  The illustration was too simple and good, for me to understand.
> >>  Thanks :)
> >>
> >>
> >>>You see, casting changes the interpretation of bits when they are
> >>>read.  The assigment of a particular type to a variable of another
> >>>type destroys the original type information (i.e. loss of precision).
> >>>
> >>
> >>  destroys the original type information? Couldn't understand this part.
> >>  You are still able to recover the string, regardless of what kind of
> >>datatype you store in. The only disadvantage I that, this obscures the
> >>logic of the program.
> >
> >
> > Didn't mean this specific example, but when casting a float or double
> > to int then you loose some information which can not be restored
> > afterwards.
> >
> >
> >>>Yes function pointers are legal.  ANSI C99 says:
> >>>
> >>>"J.5.7 Function pointer casts
> >>>1 A pointer to an object or to void may be cast to a pointer to a
> >>>function, allowing data to be invoked as a function (6.5.4).
> >>>2 A pointer to a function may be cast to a pointer to an object or to
> >>>void, allowing a
> >>>function to be inspected or modified (for example, by a debugger) (6.5.4)."
> >>>
> >>>A function name is just a pointer to the memory location where the
> >>>function is found at runtime.  It can be queried, modified and cast to
> >>>other types.  It behaves like a variable.  Take a look at the
> >>>sigaction structure:
> >>>
> >>>      struct sigaction {
> >>>              /* SIG_DFL, SIG_IGN or pointer to function */
> >>>              void (*sa_handler)(int);
> >>>              ... /* some other fields */
> >>>      };
> >>>
> >>>You can define sigaction as follows:
> >>>
> >>>      void handler(int signo) {
> >>>              doneflag = 1;
> >>>      }
> >>>
> >>>      struct sigaction sa;
> >>>      sa.sa_handler = handler;
> >>>      ...
> >>>
> >>>Here, sa_handler is registered as a function to be called when the
> >>>specified signal occurrs.
> >>>
> >>>Just handle function pointers as simple pointer variables.
> >>
> >>  Have still one silly question ---
> >>    since you say function pointers are similar to simple pointer
> >>variables, and that's why the typecast works, what would the following
> >>code mean?
> >>
> >>  void handler1 (int a) {
> >>
> >>    /* body of handler - 1 */
> >>    printf ("HANDLER - 1 \n");
> >>
> >>  }
> >>
> >>  void handler2 (int a) {
> >>
> >>    /* body of handler - 2 */
> >>    printf ("HANDLER - 2 \n");
> >>
> >>  }
> >>
> >>  struct sigaction sa;
> >>  sa.sa_handler = (handler2 *) handler1;
> >>
> >>  Is this possible? if so, what does it mean?
> >
> >
> > In this case you are trying to cast to a __name__ rather than a type.
> 
> 
> 
> > If you write
> >
> >       sa.sa_handler = (void *)handler1;
> >
> > it would be OK, but nothing would happen.  Why?  Because no
> > typecasting takes place (handler1 is of type void already) and
> > therefore nothing changes.  handler1 would be associated to
> > sa_handler.
> >
> 
> hmm.. fine
> 
> > A final example:
> 
>   .."final"..
>   Got scared seeing this :-O
>   my silly questions are annoying you???

No, of course not.  Just hoped that I now can make myself as clear as
prossible to answer all your questions with a "final" example.  No
reason to be scared.  No stupid questions to be asked.

> >
> >       #include <stdio.h>
> >
> >       void func1(int);
> >       void func2(int);
> >
> >       void func1(int i) {
> >               printf("func1: %d\n", i);
> >       }
> >
> >       void func2(int i) {
> >               printf("func2: %d\n", i);
> >       }
> >
> >       int main (void) {
> >               void (*f1)(int) = (void *)func1;
> >               /* the same as above */
> >               void (*f1)(int) = func1;
> >               void (*f2)(int) = func2;
> >               f1(1);
> >               f2(2);
> >
> >               getc(stdin);
> >
> >               return (0);
> >       }
> >
> > Everything fine now?
> 
> yes :)
> 

Cool then.  Don't forget: never stop asking!
 
Regards

	\Steve

--

Steve Graegert <graegerts@gmail.com>
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