Re: typecasting - explain

[_z33 <timid.Gentoo@gmail.com>]

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???

> 
> 	#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 :)

_z33
-- 
I love TUX; well... that's an understatement :)