Assignment of void pointer variable

[John S. Price]

In article <19390 at grebyn.com> jmbj at grebyn.com (Jim Bittman) writes:
>This works in C, but I'd like to combine the last two lines...
>  void *varptr[10];
>  int  *intptr;
>  int  myint;
>  intptr = varptr[5]; 
>  myint = *intptr;
>Goal:  myint = (int) *varptr[5];  /* doesn't work, it's what I want! */
>Post or mail suggestions, Thanks for the help!
>Jim Bittman
>jmbj at grebyn.com

Umm... I believe this will work.

myint = *(int *)varptr[5];

The reason:
varptr is declared to be a void pointer, so you have
to tell the compiler what you want it to be "looking"
at when you dereference it.  The (int *) cast tells the
compiler that the next pointer is a pointer to an integer,
and the outer dereference takes the integer at that location.

The reason yours didn't work is this:
myint = (int) *varptr[5];

You have a void pointer, defreference it,
and cast the result to an integer.  This,
I assume, isn't what you wanted.

hope this helps, and I might be totally rambling, for
it is quite early in the morning...


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John Price                   |   It infuriates me to be wrong
john at stat.tamu.edu           |   when I know I'm right....
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