Re: typecasting - explain

[Steve Graegert <graegerts@gmail.com>]

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.

A final example:

	#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?

Regards

	\Steve

--

Steve Graegert <graegerts@gmail.com>
Software Consultancy {C/C++ && Java && .NET}
Mobile: +49 (176)  21248869
Office: +49 (9131) 7126409