Re: Into the Void

[Ron Michael Khu <ronkhu@ntsp.nec.co.jp>]

extra typing?? r u using the term "type" in the same context as "data 
type"(progamming language concept)?
or "type" in the sense "keyboard typing"??

>>So I should use
>> memcpy ((void *) *data_buf, (void *) bp, len) [len] = '\0';

Oops, I think u've misunderstood jeff's reply...
What jeff and the others are trying to say is that the return value of 
memcpy() is of type "void *"

In essence:
void *ptr;
ptr = memcpy(...)

thus, u need to cast the return value to something more appropriate.

And by the way, 
"memcpy ((void *) *data_buf, (void *) bp, len) [len] = '\0';" is WRONG

u still need to cast it properly if ur u persist on doing it the "one-liner way":
( (char*)memcpy ((void *) *data_buf, (void *) bp, len) )[len] = '\0';

*TAKE NOTE of the enclosing parenthesis:
((char*)SOMETHING)[index] ='\0'   

(assuming of course SOMETHING is something that could safely be treated as char *)

and like what George stated earlier... this one-liner thing is a maintenance 
headache(a code readability issue), and thus, not advisable.
better chop that one-liner down to something more readable/maintanable.

-Ron

P.S. Hmmmm... u seem to have trouble understanding what a "void*" is.
Void * could be anything.. thus u cant simply go around indexing a void*
and treating it like an array without properly casting it...



Scott wrote:

>On Wed, Jan 26, 2005 at 08:45:09AM -0500, Jeff.Fellin@rflelect.com wrote:
>  
>
>>The reason you are getting the references to void is the function prototype
>>for memcpy is:
>>
>>      memcpy (void *dest, void *src, size_t n);
>>
>>You are passing char* pointers which are not void * pointers. You have to
>>correct your arguments
>>to remove the error messages. Please use man memcpy() for more details.
>>    
>>
>
>So I should use
>
>   memcpy ((void *) *data_buf, (void *) bp, len) [len] = '\0';
>
>(len is already type size_t) 
>
>Or, I guess, to follow Ron's suggestion too:
>
>   ((char*) memcpy ((void *) *data_buf, (void *) bp, len)) [len] = '\0';
>
>Just curious as to what evil all of this extra typing is supposed to
>be protecting me from?
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