I've got void pointer which I want to do pointer arithmetic on it. to do so, I want to cast it to a char pointer.
( char * ) buf += current_size;
However when I do so, I get the following error:
error: lvalue required as left operand of assignment
I tried also adding parenthesis on the the whole left side, with no success. Why do I get this ?
You can avoid the warning by writing the assignment operator out:
buf = (char *)buf + current_size;
Part of the reason you get the warning is that sizeof(void) is undefined and you cannot do arithmetic on void *. Beware: GCC has an extension whereby it treats the code as if sizeof(void) == 1, but that is not in the C standard.
The cast on the LHS of the assignment has no effect on the assignment.
void"? If it were a pointer with a type, then you could do math on it.void either.Why do I get this
Simply because the language rules state that a cast does not produce an lvalue even if the operand is one.
In a footnote, page 91
A cast does not yield an lvalue.
Because it's invalid, you can't do this. If you want to achieve this, either
declare buf as a char pointer; or use a temporary variable:
char *tmp = buf; // note you don't need the casting, void * works with everything
tmp += current_size;
buf = tmp;
Edit: as others also suggested, you can even remove that ugly tmp:
buf = (char *)buf + current_size;