match.pd: Fix up ((C << A) & D) != 0 simplification [PR126476]

This simplification for power of two @1 and @2 folds to false (resp.
to true for the == version) if @1 is larger than @2 (in unsigned
comparison), because @1 & @2 is known to be zero (i.e. for shift count 0)
and for shift count larger than that it will be zero too, either because
@1 << @0 is even larger, or if @0 is too large @1 << @0 overflows to zero.
This is the case of e.g. ((4 << x) & 2) != 0, which is always false.
Now, this PR is about a different problem, if @1 is smaller than @2, say
((1 << x) & 256) != 0, but x has a very narrow type, say unsigned _BitInt(3),
then the largest possible value of x is 7 and ((1 << 7) & 256) is
still 0, 1 << 7 is 128 and so still smaller than 256.
So, if c1 - c2 doesn't fit into the shift count type
(resp. for the other case c2 - c1), it will be also always false (resp.
true).
Trying to improve it and using range of x (aka @0) is not needed,
this simplification folds it into @0 != (c1 - c2) and so will be folded
later.  Just the case where c1 - c2 overflows is problematic because
we've lost the details (unless we'd promote both operands or something).
Another possible way to do this would be build_int_cst and check for
the overflow flags, but I think this is shorter.

2026-07-31  Jakub Jelinek  <jakub@redhat.com>

	PR tree-optimization/126476
	* match.pd (((C << A) & D) != 0 -> A == 0,
	((C << A) & D) == 0 -> A != 0): Fold to false/true if
	c1 - c2 resp. c2 - c1 doesn't fit into TREE_TYPE (@0).

	* gcc.dg/torture/bitint-103.c: New test.

Reviewed-by: Richard Biener <rguenth@suse.de>
(cherry picked from commit 5c62a2771f2c7d5301ee6456f1817098c2fea113)