On Mon, Mar 09, 2020 at 07:42:20PM +0100, J.W. Jagersma wrote:
> On 2020-03-09 19:01, Segher Boessenkool wrote:
> > On Mon, Mar 09, 2020 at 01:54:53PM +0100, Richard Biener wrote:
> >> int foo = 0
> >> try
> >> {
> >> asm volatile ("..." : "=r" (foo));
> >> }
> >> catch (...whatever...)
> >> {
> >> foo should be still zero, but SSA doesn't have the correct use
> >> here
> >> }
> >>
> >> that means the compiler really assumes the asm will populate the outputs
> >> even when it throws.
> >
> > How is memory any different here? In both cases you do not know if it
> > is the old value or some new value in foo, after it threw an exception.
>
> If foo were a memory operand, the compiler makes no assumptions about
> its value. When you reference it in the catch block it is always read
> back from memory. Only register operands are clobbered and retain
> their previous value. If you compile such an example with -O3, you'll
> see that the initial "int foo = 0;" is eliminated from the normal code
> path. It is only set to 0 in the catch block.
My question is *why* that is, and/or what in our code makes that so :-)
Segher
