<div dir="ltr"><div class="gmail_extra"><div class="gmail_quote">On Fri, Jul 31, 2015 at 5:03 PM, Philip Reames <span dir="ltr"><<a href="mailto:listmail@philipreames.com" target="_blank">listmail@philipreames.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<div bgcolor="#FFFFFF" text="#000000"><span class=""><div>On 07/31/2015 04:05 PM, Piotr Padlewski
wrote:</div><blockquote type="cite">
<div dir="ltr">
<div class="gmail_extra">
<div class="gmail_quote">On Fri, Jul 31, 2015 at 3:53 PM,
Philip Reames <span dir="ltr"><<a href="mailto:listmail@philipreames.com" target="_blank">listmail@philipreames.com</a>></span>
wrote:
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<div bgcolor="#FFFFFF" text="#000000"> Quoting from the
google doc: <span style="font-size:14.666666666666666px;font-family:Arial;color:#000000;background-color:transparent;font-weight:normal;font-style:normal;font-variant:normal;text-decoration:none;vertical-align:baseline">"If
we don’t know definition of some function, we assume
that it will not call @llvm.invariant.group.barrier().</span>"<br>
This part really really bugs me. We generally try to
assume minimal knowledge of external functions (i.e.
they can do anything) and this assumption would invert
that. Is there a way we can rephrase the proposal which
avoids the need for this? I'm not quite clear what this
assumption buys us.<br>
<br>
</div>
</blockquote>
<div>This is because without it the optimization will be
useless. For example:</div>
<div>A* a = new A;</div>
<div>a->foo(); //outline virtual</div>
<div>a->foo();</div>
<div><br>
</div>
<div>If we will assume that foo calls
@llvm.invariant.barrier, then we will not be able to
optimize the second call. <br>
</div>
</div>
</div>
</div>
</blockquote></span>
Why not? If foo calls @llvm.invariant.group.barrier, then it would
have to produce a new SSA value to accomplish anything which might
effect the second call. Given the call is on "a", not some return
value from foo or a global variable, we know that any SSA value
created inside foo isn't relevant. We should end up a with two
loads of the vtable using the same SSA value and the same
invariant.group metadata. The later can be forwarded from the
former without issue right?<br>
<br>
%a = ...;<br>
%vtable1 = load %a + Y !invariant.group !0<br>
%foo1 = load %vtable1 + X, !invariant.group !1<br>
call %foo1(%a)<br>
%vtable2 = load %a + Y !invariant.group !0 <-- Per state rules,
this value forwards from previous vtable load<br>
%foo2 = load %vtable2 + X, !invariant.group !1<br>
call %foo2(%a)</div></blockquote><div><br></div><div>Right, you got the intention of the design. The foo call can placement new into %a if it wants to, but it better put things back the way they were before we return, because %a isn't changing.</div><div><br></div><div>We probably missed that sentence when reviewing the proposal. :) </div></div></div></div>