[llvm-dev] Nowaday Scalar Evolution's Problem.
Jun Ryung Ju via llvm-dev
llvm-dev at lists.llvm.org
Sun Nov 19 20:57:40 PST 2017
The Problem?
Nowaday, SCEV called "Scalar Evolution" does only evolate instructions that
has predictable operand,
Constant-Based operand. such as that can evolute as a constant.
otherwise we couldn't evolate it as SCEV node, evolated as SCEVUnknown.
important thing that we remember is, we do not use SCEV only for Loop
Deletion,
which that doesn't really needed on nature loops (usually programmers do
not write non-nature loops), it doesn't really need to evolute some
conditional-variables.
but, LLVM uses SCEV for otherall loop-analysis, typically Loop Unroll, etc.
That really does need to anaylze conditional-expressions, lets see
following C/C++ code.
void UnpredictableBackedgeTakenCountFunc1() {
for(unsigned int i = 0; i != 10; ++i) {
// Exception of variable i...
if(i == 4) ++i; // when the variable i is 4, we increase it. so its
5.
if(i == 7) ++i; // when the variable i is 7, we increase it. so its
8.
// do something here....
}
}
Is this loop-count for this code predictable?
yes, it does. its "actually" predictable, also unrollable.
lets see following IR and Assembly that clang(lastest version, 2017-11-17)
emitted.
**IR OUTPUT with (clang.exe -ggdb0 -O3 -S -emit-llvm)**
define void @UnpredictableBackedgeTakenCountFunc1() {
br label %1 ; goto %1
%2 = phi i32 [ 0, %0 ], [ %10, %5 ] ; if(%2 from %0) %2 = 0
; if(%2 from %5) %2 = %10
;
switch i32 %2, label %5 [ ; switch(%2) default: goto %5
i32 10, label %3 ; case 10: goto %3
i32 4, label %4 ; case 4 : goto %4
]
ret void ; return;
br label %5 ; goto %5;
%6 = phi i32 [ 5, %4 ], [ %2, %1 ]
%7 = icmp eq i32 %6, 7 ; %7 = (%6 == 7)
%8 = zext i1 %7 to i32 ; %8 = (i1)%7
%9 = add i32 %6, 1 ; %9 = %6 + 1
%10 = add i32 %9, %8 ; %10 = %9 + %8
br label %1 ; goto %1;
}
**ASSEMBLY OUTPUT (clang.exe -ggdb0 -O3 -S)**
UnpredictableBackedgeTakenCountFunc1():
xor eax, eax ; eax = 0
cmp eax, 4 ; cmpv = (eax == 4)
jne .LBB0_2 ; if(cmpv == false) goto LBB0_2
jmp .LBB0_4 ; goto LBB0_4
.LBB0_5:
xor ecx, ecx ; ecx = 0
cmp eax, 7 ; cmpv = (ecx == 7)
sete cl ; cl = cmpv
lea eax, [rax + rcx] ; eax = *(rax + rcx)
add eax, 1 ; eax++
cmp eax, 4 ; cmpv = (ecx == 4)
je .LBB0_4 ; if(cmpv == true) goto LBB0_4
.LBB0_2:
cmp eax, 10 ; cmpv = (eax == 10)
jne .LBB0_5 ; if(cmpv == false) goto LBB0_5
jmp .LBB0_3 ; goto LBB0_3
.LBB0_4:
mov eax, 5 ; eax = 5
jmp .LBB0_5 ; goto LBB0_5
.LBB0_3:
ret ; return;
.Lfunc_end0:
The loop doesn't even deleted! whats happening to SCEV!
Yes, reason why SCEV cannot delete this loop because SCEV doesn't handle
conditional-variables dynamically when calculating backedge-taken count.
So now, this is why I am suggesting that SCEV to handle
conditional-variables.
Solve Problem?
First, before we solve the problem, we need to remember that SCEV should
NOT include cyclic SCEV node.
this is really important, if we create cyclic SCEV node, it will be
resolved as SCEVUnknown. will not optimized.
Okay. now, lets try to optimized the code that I wrote on top.
void UnpredictableBackedgeTakenCountFunc1() {
for(unsigned int i = 0; i != 10; ++i) {
// Exception of variable i...
if(i == 4) ++i; // when the variable i is 4, we increase it. so its
5.
if(i == 7) ++i; // when the variable i is 7, we increase it. so its
8.
// do something here....
}
}
lets try to evolute "i" as add-recurrence
%0 -> label %0;
%1 -> 0
%1 -> if(%1 == 10)
goto %6;
else
%i = %2;
%2 -> %2 = (%1 == 4) + %3;
%3 -> %3 = (%1 == 7) + %4;
%4 -> %4 = 1;
%5 -> goto %1;
%6 -> label %6;
This is a current add-recurrence SCEV node emmited from SCEV.
for now, %2 and %3 cannot be evoluted.
we cannot calculate the backedge-taken count for this node.
the SCEVAddRec node can only handle a constant variable. so it will be
like..
SCEV:
What is Loop-Latch? : %1
Is The Loop-Latch is only one? : Yes
Is the Loop-Latch is conditional? : Yes
Is The Loop-Latch has Exit? : Yes
What is Loop-Latch's Exit? : %5
What is %1 value? : %2
What is %2 value? : (%1 == 4) + %3
What is first operand value? :
Unknown! We don't know it can be 4! We cannot add it! it
can be 0 or 1!
Stop analyzing! return SCEVUnknown.
(%1 == 4) = SCEVUnknown(%1 == 4)
%2 = SCEVUnknown(%1 == 4) + %3
%1 = SCEVUnknown(%2)
Done Node Analyzing.
SCEV backedge-taken count analyzing:
Initialize :
%1 = 0;
Is Latch condition true? : false.
Count 1 :
%1 is Unknown! We cannot calculate backedge-taken count!.
Its Unpredictable!.
Loop Deletion:
Is loop changes other loop outside variables? : false.
Is loop volatile? : false
Is loop backedge can taken(loop count predictable) : false
Loop Optimized : false.
Printing analysis 'Scalar Evolution Analysis' for function
'UnpredictableBackedgeTakenCountFunc1':
Classifying expressions for: @UnpredictableBackedgeTakenCountFunc1
%2 = phi i32 [ 0, %0 ], [ %10, %5 ]
--> %2 U: full-set S: full-set Exits: <<Unknown>>
LoopDispositions: { %1: Variant }
%6 = phi i32 [ 5, %4 ], [ %2, %1 ]
--> %6 U: full-set S: full-set Exits: <<Unknown>>
LoopDispositions: { %1: Variant }
%8 = zext i1 %7 to i32
--> (zext i1 %7 to i32) U: [0,2) S: [0,2) Exits: <<Unknown>>
LoopDispositions: { %1: Variant }
%9 = add i32 %6, 1
--> (1 + %6) U: full-set S: full-set Exits: <<Unknown>>
LoopDispositions: { %1: Variant }
%10 = add i32 %9, %8
--> (1 + (zext i1 %7 to i32) + %6) U: full-set S: full-set
Exits: <<Unknown>> LoopDispositions: { %1: Variant }
Determining loop execution counts for: @UnpredictableBackedgeTakenCountFunc1
Loop %1: Unpredictable backedge-taken count.
Loop %1: Unpredictable max backedge-taken count.
Loop %1: Unpredictable predicated backedge-taken count.
This is actual optimization dump with opt executable. (opt.exe -S
-scalar-evolution -analyze)
Why? because it only creates a node, %1 doesn't modified yet!
The (%1 == 4) can be 1 or 0. depend on %1 we cannot assume it as constant
variable. its variant!
Now what happens on new one that I suggesting.
Instruction to SCEV node analyzing:
What is Loop-Latch? : %1
Is The Loop-Latch is only one? : Yes
Is the Loop-Latch is conditional? : Yes
Is The Loop-Latch has Exit? : Yes
What is Loop-Latch's Exit? : %5
What is %1 value? : %2
What is %2 value? : (%1 == 4) + %3
What is first operand value? :
Its conditional! do not analyze yet!
Keep analyze it! we are going to calculate later when
calculating backedge-taken count!
(%1 == 4) = SCEVConditional(%1 == 4)
What is second operand value? : %3
What is %3 value? : (%1 == 7) + %4
What is first opernad value? :
Its conditional! do not analyze yet!
Keep analyze it! we are going to calculate later
when calculating backedge-taken count!
(%1 == 7) = SCEVConditional(%1 == 7)
What is second operand value? : %4
What is %4 value? : Constant 1
%4 = SCEVConstant(1)
%3 = SCEVConditional(%1 == 7) + SCEVConstant(1)
%3 = SCEVAddRecExpr[%1, +, SCEVConditional(%1 == 7) +,
SCEVConstant(1))]
%2 = SCEVAddRecExpr[%1, +, SCEVConditional(%1 == 4), +, %3)]
%1 = %2
What is Latch conditional(%5) value? : %1
Done Node Analyzing!
SCEV backedge-taken count analyzing:
Initialize :
%1 = 0;
Is Latch condition true? : false.
Count 1 :
%1 = %2;
%2 = (%1(0) == 4) + %3
%2 = (%1(0) == 4) + %3
%3 = (%1(0) == 7) + %4
%3 = (%1(0) == 7) + %4
%4 = 1
%3 = (%1(0) == 7) + %4(1)
%3 = 1
%2 = (%1(0) == 4) + %3(1)
%2 = 1
%1 += %2(1)
%1 = 1
Current %1 value : 1
Is Latch condition true? : false.
Count 2 :
%1 = %2;
%2 = (%1(1) == 4) + %3
%2 = (%1(1) == 4) + %3
%3 = (%1(1) == 7) + %4
%3 = (%1(1) == 7) + %4
%4 = 1
%3 = (%1(1) == 7) + %4(1)
%3 = 1
%2 = (%1(1) == 4) + %3(1)
%2 = 1
%1 += %2(1)
%1 = 2
Current %1 value : 2
Is Latch condition true? : false.
Count ... :
...
Backedge-Taken Count : 7
Its predictable!
Loop Deletion:
Is loop changes other outside variables? : false.
Is loop volatile? : false
Is loop backedge can taken(loop count predictable) : true
Removing Loop!
Loop Optimized : true.
Okay, now we got a optimized loop!
The way that I resolving this is, visiting conditional-expressions each
step when calculating backedge-taken count.
not only focus on SCEV node predicates, we can still assume it the loop can
predictable.
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