🚀 go-pugleaf

dat@thinkage.on.ca (David Adrien Tanguay) writes:
>How is this any different from the array example?  p may point
>outside a[1], but it is still within the a object.  Similarly, ...
>is outside the f->s object but within the malloc object.

The point is, "int a[5][5];" doesn't give you 25 ints, it gives you five arrays
of five ints each.  "(int(*)[5])malloc(5*sizeof(int[5]))" gives you
sizeof(int)*25 bytes which are ALSO five arrays of five ints.

The first looks like this:
	int[5][5]:
	    int[5]:
		int
		int
		int
		int
		int
	    int[5]:
		...
	    ...

And the second looks like this:
	char
	char
	char
	char
	...

Now that cast magically makes the second be close enough to the first in
outward appearance for all intents and purposes, but it's still the second
underneath.  Obviously in almost all compilers the memory layout will be the
same.  However, an environment such as Saber-C should be free to keep track of
the details and that's why I've always thought that this is a good
interpretation ruling.

I'm not sure what you get when you take a pointer to that array five of array
five of int, cast it to a (char *), then cast it to a (int(*)[5]).