llvm dev - Nov 2011 - [LLVMdev] [LLVM, loop-unswitch tests] preserve-analyses.ll, strange PHI instruction.

Hi all.

Looking at test/Transforms/LoopUnswitch/preserve-analyses.ll, I found 
improper phi instruction  at string #122:
%call.i25.lcssa48 = phi i8* [ %call.i25, %if.then31.i.i ], [ %call.i25, 
%if.then31.i.i ] ; <i8*> [#uses=0]
Is it trick or mistake?

-Stepan.

Stepan Dyatkovskiy wrote:
> Hi all.
>
> Looking at test/Transforms/LoopUnswitch/preserve-analyses.ll, I found
> improper phi instruction  at string #122:
> %call.i25.lcssa48 = phi i8* [ %call.i25, %if.then31.i.i ], [ %call.i25,
> %if.then31.i.i ] ;<i8*>  [#uses=0]
> Is it trick or mistake?
What's improper about it?

The block may have two predecessors, both of which are the same block 
(consider a switch statement with "case 3: case 4: code;"). This is 
called a critical edge. The requirement in the IR is that a PHI node 
have an entry for each pred (including duplicates for critical edges) 
and that the incoming value must be the same.

The best test is to ask the verifier (opt -verify). It knows all the 
special rules, like why this code is legal:

   define i32 @test() {
     unreachable

     %a = add i32 %a, %a
     ret i32 %a
   }

Surprise! That one's my favourite example.

Nick

Hi Stepan,
> Looking at test/Transforms/LoopUnswitch/preserve-analyses.ll, I found
> improper phi instruction  at string #122:
> %call.i25.lcssa48 = phi i8* [ %call.i25, %if.then31.i.i ], [ %call.i25,
> %if.then31.i.i ] ;<i8*>  [#uses=0]
> Is it trick or mistake?
if there are several edges from a basic block A to a basic block B (because for
example A ends in a conditional branch with both branches leading to B, or A
has a switch with several cases leading to B) then A should occur the same
number of times in any phi node in B, with the restriction that the value
associated with A should be the same each time.  In short, the phi node you
mention looks fine to me because there are two edges from the switch in
%if.then31.i.i to this basic block.

Ciao, Duncan.

OK. Thanks. And if I removed "case 4:" for example. Need I scan all
phi
nodes and fix it, by removing one predecessor (if it exists) to switch's 
parent block?

-Stepan.

Nick Lewycky wrote:
> Stepan Dyatkovskiy wrote:
>> Hi all.
>>
>> Looking at test/Transforms/LoopUnswitch/preserve-analyses.ll, I found
>> improper phi instruction at string #122:
>> %call.i25.lcssa48 = phi i8* [ %call.i25, %if.then31.i.i ], [ %call.i25,
>> %if.then31.i.i ] ;<i8*> [#uses=0]
>> Is it trick or mistake?
>
> What's improper about it?
>
> The block may have two predecessors, both of which are the same block
> (consider a switch statement with "case 3: case 4: code;"). This
is
> called a critical edge. The requirement in the IR is that a PHI node
> have an entry for each pred (including duplicates for critical edges)
> and that the incoming value must be the same.
>
> The best test is to ask the verifier (opt -verify). It knows all the
> special rules, like why this code is legal:
>
> define i32 @test() {
> unreachable
>
> %a = add i32 %a, %a
> ret i32 %a
> }
>
> Surprise! That one's my favourite example.
>
> Nick