C Newbie Question: Explain (textBuffer *) in buf = (textBuffer *)XtMalloc(sizeof(textBuffer));

[Randy Kramer]

On Saturday 17 November 2007 08:11 am, Tony Balinski wrote:
> Quoting Randy Kramer <rhkramer at gmail.com>:
> > Can anybody help me understand the following statement, or more
> > specifically,
> > the "(textBuffer *)" part:
> >
> > buf = (textBuffer *)XtMalloc(sizeof(textBuffer));
 
> Pretty close. buf is already a pointer to a textBuffer structure, but
> is not pointing to anything in particular before the assignment (I suppose).
> So it's got to point at some bit of memory which holds/can hold that
> structure. XtMalloc here returns a pointer to such memory, freshly 
allocated.
> The trouble is that the XtMalloc function returns a pointer to characters
> (char *) - that makes no difference to the pointer value, but the compiler
> only wants to allow buf to take a value which is of type textBuffer *.
> 
> The (type)value construct is a cast: when applied to pointers, it tells the
> compiler to treat the operand - value - as a pointer of type type. In our
> we interpret the returned value of XtMalloc as a pointer to a textBuffer
> structure rather than one to a character buffer. The assignment to buf then
> becomes an assignment of type-compatible values. (No actual machine code is
> generated for this. In C, the return value of a function returning a void *
> pointer can be assigned directly to a pointer to any other type as a special
> case, to avoid writing these casts - as malloc does; but here XtMalloc 
returns
> char * so the cast is necessary.) In short, you're bending C's type system,
> but telling the compiler that you know what you're doing.

Tony,

Thanks very much!

I might have recognized something like:

(textBuffer)XtMalloc(sizeof(textBuffer))

... as a cast (probably not immediately, but eventually ;-)  What I have to 
absorb now is that the syntax (<type> *) is a cast for a pointer (and not 
some new fangled multiplication ;-)

thanks,
Randy Kramer