Re: AW: optimizer discards sign information

[Xi Ruoyao <xry111@xry111.site>]

On Wed, 2024-04-10 at 12:03 +0200, stefan@franke.ms wrote:
> Yes, there is an overflow when the value gets assigned to x
> 
> 	u32 x = *a * *b;
> 
> And after that line of code, x is a valid unsigned int, no matter what
> value was assigned. And the compiler must not throw away that
> unsignedness.

If *a * *b does not overflow, x is a valid unsigned int.  But if *a * *b
overflows, x is not valid, at all.  Its type does not matter.

> Also an add can overflow:
> 
> u64 faa(int a, int b) {
>     u32 x = a + b;
>     u64 r = x;
> 
> And in this case the optimizer doesn't discard the variable x

Only an overflow on *arithmetic operation* is undefined behavior.  An
overflow on *conversion* is not.  In fact an "overflow on conversion" is
even not referred as "overflow" in the standard.

So in this case if a and b are both -1, x *must* be 0xfffffffdU, r
*must* be 0xfffffffdULL, and there's no undefined behavior.  So it will
be incorrect (i.e. violating the standard, not "different from what a
person thinks") to use a signed extension here, and the compiler does
not do that.

But for

u16 a, b;
u32 x = (int)a * (int)b;

(int)a and (int)b must be non-negative, and since an overflow on
multiplication is UB, (int)a * (int)b must be non-negative too.  So it's
valid (i.e. allowed by the standard, not "doing exactly what a person
thinks") to use a signed extension (though maybe it's not optimal, and
we may have a missed-optimization here).

-- 
Xi Ruoyao <xry111@xry111.site>
School of Aerospace Science and Technology, Xidian University