Re: pointer to integer to pointer conversions

vipps...@[EMAIL PROTECTED]
 wrote:
> vippstar wrote:
> > That's possible. Using `intptr_t' or `uintptr_t'.
> > You have to include <stdint.h> to use it.
> > There is also uintptr_t, which you will use to store the void pointer
> > does not matter.
> > The type only matters if you use it in arithmetic.
> > ie, uintptr_t foo = malloc(123); foo = ~foo; free((void*)~free);
>                      ^                                      ^^^^
> > With intptr_t undefined behavior might be invoked in this
> > example (if, for example, malloc() returns NULL)
>
> Ignore the small error I made (typing free instead of foo), and
> that I did not cast malloc(123) to (uintptr_t), which might be
> necessary.

It is.

> The point of this snip is to show that ~ in signed integer with
> value 0 might be a trap representation, which you can avoid
> by using an unsigned integer. I cannot think of an example
> that signed is preferred to unsigned, but there has to be at
> least one.

There are ha****ng techniques that involve negative values.
[Note that Java has no unsigned types.]

--
Peter