compiler improvement of register usage part 1 - chapter 8, through section 8.4 anastasia braginsky

Compiler Compiler Improvement of Improvement of Register UsageRegister Usage

Part 1 - Chapter 8, through Section 8.4

Anastasia Braginsky

Roadmap

Introduction

Scalar Replacement

Unroll-and-Jam

Introduction

Processor cycle time is decreasing; while memory time remains almost the same.

Better usage of registers set (especially for RISC architectures)

No cache discussion here

Register Allocation Algorithms

1. Defining live range per variable.

2. Building interference graph to model which pairs of live ranges can not be assigned to the same register.

3. Using a fast heuristic coloring algorithm.

4. If coloring fails take at least one live range from registers and repeat from step 3.

Register Allocation Algorithms – so what’s the problem? Only non-array variables are assigned to

registers.

Almost no optimization for floating-point registers, typically used to hold temporarily individual elements of array variables.

7 9 4 05 6 1 8 32Array A:Register RFor Array A:


registers.


7 9 4 05 6 1 8 32Array A:


registers.


7 9 4 05 6 1 8 32Array A:

We want to eliminate the unneeded loads and stores.

Introduction – example:

DO I=1,N

DO J=1,M

A(I)=A(I)+B(J)

ENDDO

ENDDO

1. Load from some memory location A(I) to register RA.

2. Load from some memory location B(J) to register RB.

3. RA = RA + RB

4. Store RA to memory – A(I)

1. Load from some memory location A(I) to register RA.

2. Load from some memory location B(J+1) to register RB.

3. RA = RA + RB

4. Store RA to memory – A(I)

Introduction – example (cont.)

DO I = 1, N

T = A(I)

DO J = 1, M

T = T + B(J)

ENDDO

A(I) = T

ENDDO

All loads and stores to A in the inner loop have been saved

High chance of T being allocated a register by the coloring algorithm

Source-to-source transformation

Data Dependence for Register Reuse Two application of dependence:

To determine the correctness of different transformations

To determine the transformations to improve the performance of particular memory accesses

Types of dependences

A true or flow dependence Assignment to variable and then usage of it

S1: V = … Store the register representing V to V’s

memory location

S2: … = V Load V to register representing V

Load can be saved (store after the usage). Cache miss can be saved

R1 R2

R1

R2


An antidepencence dependence

Usage of variable and then assignment to it



memory location

Nothing can be done to improve the registers usage (if used once), but cache miss can be saved.

R1 R2

1


An output dependence

Assignment to variable repeats


memory location


memory location

First store is not needed.

R1 R2

o

Types of dependences - example

S1: A(I) = …

S2: … = A(I)

S3: A(I) = …

Store

Load

Store

True dependence between S1 and S2

Output dependence between S1 and S3


An input dependence

Usage of variable repeats



Second load is not needed.

R1 R2

i


Loop-independent dependence

Loop-carried dependence:

Consistent dependence – a loop-carried dependence with a constant dependence distance throughout the loop.

Non consistent dependence – the opposite.

Dependence graph modification

S1: A(I)=…

S2: …=A(I)

S3: …=A(I) load from A(I)

Two True dependences between S1 and S2

)Load can be saved(

and between S1 and S3

)Load can be saved(

Input dependence from S2 to S3

)Load can be saved(

It is not possible to save three memory references the dependence graph should be pruned.

load from A(I)

store to A(I)

Sum reduction example (again)

DO I = 1, N

DO J = 1, M

A(I) = A(I) + B(J)

ENDDO

ENDDO Save second load

…=V

…=V

Input

Save first store

V=…

V=…

Output

Save nothing

…=V

V=…

Anti- dependence

Save load

V = …

… = V

True

1

o

i

1J

J

oJ i

I

iJ

Sum reduction example (again)

DO I = 1, N

DO J = 1, M

A(I) = A(I) + B(J)

ENDDO

ENDDO

1J

J

oJ i

I

Load from A(I) to R1

…

Load from B(J) to R2

R1 = R1 + R2

Store R1 to A(I)

Load from A(I) to R1

Load from B(J+1) to R2

R1 = R1 + R2

Store R1 to A(I)

…Load from A(I) to R1

iJ

Scalar Register Allocation

DO I = 1, N

T = A(I)

DO J = 1, M

T = T + B(J)

ENDDO

A(I) = T

ENDDO

Roadmap

Introduction

Scalar Replacement

Unroll-and-Jam

Loop Independent Dependence

Simple replacement

DO I=1,N

A(I) = B(I) + C

X(I) = A(I) + Q

ENDDO

DO I=1,N

t = B(I) + C

X(I) = t + Q

A(I) = t

ENDDO

Loop Carried Dependences

Spanning single iteration

DO I=1,N

A(I) = B(I-1)

B(I) = A(I) + C(I)

ENDDO

I=1

A(1) = B(0)

B(1) = A(1)+C(1)

I=2

A(2) = B(1)

B(2) = A(2)+C(2)

I=3

A(3) = B(2)

B(3) = A(3)+C(3)

…

Loop Carried True dependence

Loop Carried Dependences

Spanning single iteration

DO I=1,N

A(I) = B(I-1)

B(I) = A(I) + C(I)

ENDDO

I=1

A(1) = B(0) or T

B(1) or T = A(1)+C(1)

I=2

A(2) = B(1) or T

B(2) or T = A(2)+C(2)

I=3

A(3) = B(2) or T

B(3) or T = A(3)+C(3)

…

T = B(0)

DO I=1,N

A(I) = T

T = A(I) + C(I)

B(I) = T

ENDDO

Dependences Spanning Multiple Iterations Distance for the dependence is 2 iterations

DO I=1, N

A(I) = B(I-1) + B(I+1)

ENDDO

Use 3 different scalar temporaries defined as follows: t1 = B(I-1) t2 = B(I) t3 = B(I+1)

Input dependenceiI

Dependences Spanning Multiple IterationsDO I=1, N

A(I) = B(I-1) + B(I+1)

ENDDO

t1 = B(I-1)

t2 = B(I)

t3 = B(I+1)

DO I=1, N t1 = B(I-1)

t2 = B(I)

t3 = B(I+1)

A(I) = t1 + t3

t1 = t2

t2 = t3

ENDDO

B[0]

B[1]

B[2]

B[3]

B[4]

t1

t2

t3

t1

t2

t3


A(I) = B(I-1) + B(I+1)

ENDDO

t1 = B(I-1)

t2 = B(I)

t3 = B(I+1)

t1 = B(0)

t2 = B(1)

DO I=1, N t3 = B(I+1)

A(I) = t1 + t3

t1 = t2

t2 = t3

ENDDO

2 pipeline cycles/slots wasted for each loop!


A(I) = B(I-1) + B(I+1)

ENDDO

t1 = B(I-1)

t2 = B(I)

t3 = B(I+1)

t1 = B(0)

t2 = B(1)

DO I=1, N t3 = B(I+1)

A(I) = t1 + t3

t1 = t2

t2 = t3

ENDDO

2 pipeline cycles/slots wasted for each loop!

B[0]

B[1]

B[2]

B[3]

B[4]

t1

t2

t3

t1

t2

t3

B[6]

B[5]

…

Dependences Spanning Multiple Iterations t1 = B(0) t2 = B(1) DO I=1, N

t3 = B(I+1) A(I) = t1 + t3 t1 = B(I+2) A(I+1) = + t2 = B(I+3) A(I+2) = t3 + t2

ENDDO

I = 1 A(1)=B(0)+B(2) I = 2 A(2)=B(1)+B(3) I = 3 A(3)=B(2)+B(4) I = 4 A(4)=B(3)+B(5) …

B[0]

B[1]

B[2]

B[3]

B[4]

t1

t2

t3

B[6]

B[5]

…

t2 t1t1

,3

Eliminating Scalar Copies

t1 = B(0)

t2 = B(1)

mN3 = MOD(N,3)

DO I = 1, mN3

t3 = B(I+1)

A(I) = t1 + t3

t1 = t2

t2 = t3

ENDDO

DO I = mN3 + 1, N, 3

t3 = B(I+1)

A(I) = t1 + t3

t1 = B(I+2)

A(I+1) = t2 + t1

t2 = B(I+3)

A(I+2) = t3 + t2

ENDDO

Pre-loop the same as the previous solution

Main loop - three sums in one iteration

t1 = B(0)

t2 = B(1)

DO I=1, N

t3 = B(I+1)

A(I) = t1 + t3

t1 = t2

t2 = t3

ENDDO

Scalar Replacement

1. Loop-independent dependence – simple replacement

2. Handling loop-carried dependence – spanning one iteration

3. Dependences spanning multiple iterations

4. Eliminating scalar copies

5. Pruning the dependence graph

6. Moderating register pressure

Break?

Pruning the dependence graph

DO I = 1, N

A(I+1) = A(I-1) + B(I-1)

A(I) = A(I) + B(I) + B(I+1)

ENDDO

I = 1A(2)=A(0)+B(0)A(1)=A(1)+B(1)+B(2)

I = 2A(3)=A(1)+B(1)A(2)=A(2)+B(2)+B(3)

I = 3A(4)=A(2)+B(2)A(3)=A(3)+B(3)+B(4)

…

Save

last

load

…=V

…=V

Input

Save

first

store

V=…

V=…

Output

Save

nothing

…=V

V=…

Anti-dependence

Save

load

V=…

…=V

True


Two kinds of edges to be pruned:

True and input dependence edges which are killed by an intervening assignment to the same location.

Input dependence edges that are redundant.

Three-phase algorithm for pruning edges Phase 1: Eliminate killed dependences

Can be true or input dependences

Looking for the store to the location involved in the dependence between the endpoints of the dependence.


True dependence

Looking for output dependence from the source to the assignment before the sink (there should be true dependence from the assignment to the sink).

S1: A(I+1) = …

S2: A(I) = …

S3: … = A(I)True

True

Output


Input dependence

Looking for anti-dependence from the source to the assignment before the sink (there should be true dependence from the assignment to the sink).

S1: …=A(I+1)

S2: A(I) = …

S3: … = A(I)Input

True

Anti

Three-phase algorithm for pruning edges Phase 2: Identify generators

Generator is a reference with at least one true or input dependence starting

from it

AND

with no input or true dependence into it

Actually if we are looking on the graph represented only by input and true dependences, generators are the sources.

Three-phase algorithm for pruning edges Phase 3: Find name partitions and eliminate

input dependences

A name partition is a set of references that can be replaced by a reference to a single scalar variable.

A(I+2)

A(I)

B(I)

A(I+1)B(I)

A(I)

B(I)

A(I-1)

2 2

Three-phase algorithm for pruning edges Phase 3: Find name partitions and eliminate

input dependences A name partition is a set of references that can be

replaced by a reference to a single scalar variable.

Starting at each generator mark each reference reachable from the generator by a flow or input dependence as part of the name partition for that variable. (similar to the typed fusion problem)

Eliminate input dependences within same name partition, unless source is generator.

Three-phase algorithm for pruning edges Phase 3 and a half: Anti-dependence

Note that anti-dependence can’t directly give a rise to register reuse and are always pruned as the last step in the graph pruning.


DO I = 1, N

A(I+1) = A(I-1) + B(I-1)

A(I) = A(I) + B(I) + B(I+1)

ENDDO

Phase 1 Try to eliminate true

dependences

Try to eliminate input dependences

Phase 2 Identify generators

Candidates are sources of true and input dependences

ii

io

i

1


DO I = 1, N

A(I+1) = A(I-1) + B(I-1)

A(I) = A(I) + B(I) + B(I+1)

ENDDO

Phase 3 Starting each

generator mark each reference reachable from the generator by input or true dependence as part of the name partition for that variable.

Phase 3.5

i

i

io

1


DO I = 1, N

A(I+1) = A(I-1) + B(I-1)

A(I) = A(I) + B(I) + B(I+1)

ENDDO

ii

io

i

1

How to start scalar replacement from here?

1. Generators: A(I+1) A(I) B(I+1)

2. The number of scalars for each generator = how many iterations are spanned by the dependence.

A(I-1) B(I-1) B(I)


DO I = 1, N

A(I+1) = A(I-1) + B(I-1)

A(I) = A(I) + B(I) + B(I+1)

ENDDO

i

i

o

References: A(I-1) = tA1

A(I) = tA2

A(I+1) = tA3

B(I-1) = tB1

B(I) = tB2

B(I+1) = tB3


DO I = 1, N

A(I+1) = A(I-1) + B(I-1)

A(I) = A(I) + B(I) + B(I+1)

ENDDO

tA1 = A(I-1), tA2=A(I), tA3=A(I+1)

tB1 = B(I-1), tB2=B(I), tB3=B(I+1)

tA1 = A(0)

tA2 = A(1)

tB1 = B(0)

tB2 = B(1)

DO I = 1, N

tA3 = tA1 + tB1

tB3 = B(I+1)

tA1 = tA2 + tB2 + tB3

// Above: tA1=tA2

A(I) = tA1

tA2 = tA3

tB1 = tB2

tB2 = tB3

ENDDO

A(N+1) = tA3

One Load

One Store


DO I = 1, N

A(I+1) = A(I-1) + B(I-1)

A(I) = A(I) + B(I) + B(I+1)

ENDDO

1. Load A(I-1) to R1

2. Load B(I-1) to R2 R1 = R1 + R2

3. Store R1 to A(I+1)

4. Load A(I) to R1

5. Load B(I) to R2 R1 = R1 + R2

6. Load B(I+1) to R2 R1 = R1 + R2

7. Store R1 to A(I)


Special cases: References in the Loop

DO I=1, N

= B(I) + C(I,J)

C(I,J) = + D(I)

ENDDO

A(J)

A(J)

1

i

otA

tA

A(J) = tA Assumption: N>1


Special cases: Forcing stores and loads

DO I = 1, N

A(I) = A(I-1) + B(I)

A(J) = A(J) + A(I)

ENDDO

I=1

A(1) = A(0) + B(1)

A(J) = A(J) + A(1)

I=2 A(2) = A(1) + B(2) A(J) = A(J) + A(2)

I=3…



DO I = 1, N

tAI = A(I-1) + B(I)

A(I) = tAI

A(J) = A(J) + tAI

ENDDO

I=1

A(1) = A(0) + B(1)

A(J) = A(J) + A(1)

I=2 A(2) = A(1) + B(2) A(J) = A(J) + A(2)

I=3…



tAI = A(0)

DO I = 1, N

tAI = tAI(was A(I-1))+B(I)

A(J) = A(J) + tAI

A(I) = tAI

ENDDO

J ?= I

I=1

A(1) = A(0) + B(1)

A(J) = A(J) + A(1)

I=2 A(2) = A(1) + B(2) A(J) = A(J) + A(2)

I=3…


tAI = A(0)

DO I = 1, N

tAI = tAI + B(I)

A(J) = A(J) + tAI

A(I) = tAI

ENDDO

A(J) ?= A(I)

DO I = 1, N

tAI = A(I-1) + B(I)

A(J) = A(J) + tAI

A(I) = tAI

ENDDO

tAI = A(0)

tAJ = A(J)

JU = MAX(J-1,0)

DO I = 1, JU

tAI = tAI + B(I)

tAJ = tAJ + tAI

A(I) = tAI

ENDDO

// Here I = J

IF(J.GT.0.AND.J.LE.N) THEN

tAI = tAI + B(J)

tAJ = tAJ + tAI

A(J) = tAI

tAI = tAJ

ENDIF

DO I = JU+2, N //I starting from J+1

tAI = tAI + B(I)

tAJ = tAJ + tAI

A(I) = tAI

ENDDO

A(J) = tAJ


Special cases: Inconsistent dependence

DO I = 1, N

A(2*I) = A(I) + B(I)

ENDDO

Bag edge in the typed fusion framework above.

Moderating of Register Pressure

Scalar replacement may produce to many scalar quantities.

Have to choose name partitions for scalar replacement to maximize register usage


M name partitions R: {R1, R2, R3,…,Rm}

The value of the name partition: v(R) – the number of memory loads and stores saved by replacing

The cost of the name partition: c(R) – the number of registers needed to hold all the scalar values.


The desired solution, given the limit of register-resident scalars to use - n:

The sub-collection of name partitions:

or {R1, R2,…,Rm} where Ri=0/1

Such that:

And

1

( )m

i ii

c R R n

1

max[ ( ) ]m

i ii

v R R

1 2{ , ,..., }

ki i iR R R

0/1 knapsack problem solutions

Dynamic programming solution – O(nm)

Heuristic solution:

Order the name partition set in decreasing order of the ratio v(R)/c(R)

Select elements from the beginning of the list until the registers are full.

Scalar replacement algorithm

For the loops that do not include conditional flow of control:

1. Prune dependence graph. (Apply typed fusion.) Get set of name partitions as the part of the result.

2. Select a set of name partitions using register pressure moderation



3. For each selected partition, replace it with a reference to scalars:

A) If non-cyclic, replace using set of temporaries

True dependence is replaced by set of temporaries. The number of temporaries is defined by how many iterations are spanned by the dependence.

Output dependence – move stores after the end of loop.

Input dependence – move loads before the beginning of loop.



B) If cyclic replace reference with single temporary

C) For each inconsistent dependence

Use index set splitting or insert loads and stores

4. Unroll loop to eliminate scalar copies

Experimental Data

Speedup = running time of the original program divided by running time of the versionafter scalar replacementLL = Livermore loops

Roadmap

Introduction

Scalar Replacement

Unroll-and-Jam

Unroll-and-Jam

Recall introduction example:

DO I=1,N

DO J=1,M

A(I)=A(I)+B(J)

ENDDO

ENDDO

Unroll-and-Jam

After transformation called unroll-and-jam

Assume Nmod2=0

DO I=1,N,2

DO J=1,M

A(I)=A(I)+B(J)

A(I+1)=A(I+1)+B(J)

ENDDO

ENDDO

This is unroll-and-jam to factor 2

Improving the efficiency of pipelined functional unit

DO J=1, 2M

DO I=1, N

A(I,J) = A(I+1,J) + A(I-1,J)

ENDDO

ENDDO

A(1,J) = A(2,J) + A(0,J)

A(2,J) = A(3,J) + A(1,J)

A(3,J) = A(4,J) + A(2,J)

…


InstructionFetch

InstructionDecode

OrRegister

Fetch

RegisterFile

ExecuteMemoryAccess

WriteBackTo

RegisterFile


InstructionFetch

InstructionDecode

OrRegister

Fetch

RegisterFile

ExecuteMemoryAccess

WriteBackTo

RegisterFile

Forward Unit


InstructionFetch

InstructionDecode

OrRegister

Fetch

RegisterFile

MemoryAccess

WriteBackTo

RegisterFile

Forward Unit

Execute


DO J=1, 2M

DO I=1, N

A(I,J) = A(I+1,J) + A(I-1,J)

ENDDO

ENDDO

A(1,J) = A(2,J) + A(0,J)

A(2,J) = A(3,J) + A(1,J)

A(3,J) = A(4,J) + A(2,J)

…

A(1,J+1) = A(2,J+1) + A(0,J+1)

A(2,J+1) = A(3,J+1) + A(1,J+1)

A(3,J+1) = A(4,J+1) + A(2,J+1)

…

Improving the efficiency of pipelined functional unitDO J=1, 2M, 2

DO I=1, N

A(I,J) = A(I+1,J) + A(I-1,J)

A(I,J+1)=A(I+1,J+1)+A(I-1,J+1)

ENDDO

ENDDO

Legality of Unroll-and-Jam

DO I=1, 2N

DO J=1, M

A(I+1, J-1) = A(I,J) + B(I,J)

A(I+2, J-1) = A(I+1,J) + B(I+1,J)

ENDDO

ENDDO

I=1, J=1A(2, 0) = A(1, 1) + B(1, 1)I=1, J=2A(2, 1) = A(1, 2) + B(1, 2)I=1, J=3A(2, 2) = A(1, 3) + B(1, 3)…I=2, J=1A(3, 0) = A(2, 1) + B(2, 1)I=2, J=2A(3, 1) = A(2, 2) + B(2, 2)I=2, J=3A(3, 2) = A(2, 3) + B(2, 3)…

,2


I=1, J=1A(2, 0) = A(1, 1) + B(1, 1)I=1, J=2A(2, 1) = A(1, 2) + B(1, 2)I=1, J=3A(2, 2) = A(1, 3) + B(1, 3)…I=2, J=1A(3, 0) = A(2, 1) + B(2, 1)I=2, J=2A(3, 1) = A(2, 2) + B(2, 2)I=2, J=3A(3, 2) = A(2, 3) + B(2, 3)…


DO I=1, 2N

DO J=1, M

A(I+1, J-1) = A(I,J) + B(I,J)

ENDDO

ENDDO

The direction vector of the loop: [<, >]

Swap [>, <] illegal

Should we assume that loop unroll-and-jam is illegal whenever loop interchange is illegal?


DO I=1, 2N

DO J=1, M

A(I+2, J-1) = A(I,J) + B(I,J)

ENDDO

ENDDO

Direction vector is (<,>); still unroll-and-jam possible


Definition:

An unroll-and-jam to factor n consist of:

Unrolling the outer loop n-1 times to create n copies of the inner loop

And fusing those copies together

Unroll-and-Jam to factor n=2


DO I=1,2N

DO J=1,M

A(I)=A(I)+B(J)

ENDDO

ENDDO

Unrolling the outer loop n-1=1 times to create n=2 copies of the inner loop

DO I=1,2N

DO J=1,M

A(I)=A(I)+B(J)

ENDDO

DO J=1,M

A(I)=A(I)+B(J)

ENDDO

ENDDO



DO I=1,2N, 2

DO J=1,M

A(I)=A(I)+B(J)

ENDDO

DO J=1,M

A(I+1)=A(I+1)+B(J)

ENDDO

ENDDO


DO I=1,2N

DO J=1,M

A(I)=A(I)+B(J)

ENDDO

ENDDO



DO I=1,2N, 2

DO J=1,M

A(I)=A(I)+B(J)

ENDDO

DO J=1,M

A(I+1)=A(I+1)+B(J)

ENDDO

ENDDO



DO I=1,2N, 2

DO J=1,M

A(I)=A(I)+B(J)

ENDDO

DO J=1,M

A(I+1)=A(I+1)+B(J)

ENDDO

ENDDO

And fusing those copies together

DO I=1,2N,2

DO J=1,M

A(I)=A(I)+B(J)

A(I+1)=A(I+1)+B(J)

ENDDO

ENDDO


Theorem:

An unroll-and-jam to factor n is legal if and only if there exist no dependence with direction vector (<, >) such that the distance for the outer loop is < n.

To check this note to use the full dependence graph and not the pruned one.


What happens if there exist dependence with direction vector (<, >) but that the distance for the outer loop is n.

Note that an unroll-and-jam was to factor n

Unroll-and-Jam to factor m Algorithm

1. Create preloop

2. Unroll main loop m times

3. Apply typed fusion to loops within the body of the unrolled loop

4. Apply unroll-and-jam recursively to the inner nested loop

Unroll-and-Jam example

DO I = 1, N

DO K = 1, N

A(I) = A(I) + X(I,K)

ENDDO

DO J = 1, M

DO K = 1, N

B(J,K) = B(J,K) + A(I)

ENDDO

ENDDO

DO J = 1, M

C(J,I) = B(J,N)/A(I)

ENDDO

ENDDO

DO I = Nmod2+1, N, 2

DO K = 1, N

A(I) = A(I) + X(I,K)

A(I+1) = A(I+1) + X(I+1,K)

ENDDO

DO J = 1, M

DO K = 1, N

B(J,K) = B(J,K) + A(I)

B(J,K) = B(J,K) + A(I+1)

ENDDO


C(J,I+1) = B(J,N)/A(I+1)

ENDDO

ENDDO


DO K = 1, N

A(I) = A(I) + X(I,K)

A(I+1) = A(I+1) + X(I+1,K)

ENDDO

DO J = 1, M

DO K = 1, N

B(J,K) = B(J,K) + A(I)

B(J,K) = B(J,K) + A(I+1)

ENDDO


C(J,I+1) = B(J,N)/A(I+1)

ENDDO

ENDDO


DO K = 1, N

A(I) = A(I) + X(I,K)

A(I+1) = A(I+1) + X(I+1,K)

ENDDO

DO J = 1, Mmod2

DO K = 1, N

B(J,K) = B(J,K) + A(I)

B(J,K) = B(J,K) + A(I+1)

ENDDO


C(J,I+1) = B(J,N)/A(I+1)

ENDDO

DO J = Mmod2+1, M, 2

DO K = 1, N

B(J,K) = B(J,K) + A(I)

B(J,K) = B(J,K) + A(I+1)

B(J+1,K) = B(J+1,K) + A(I)

B(J+1,K) = B(J+1,K) + A(I+1)

ENDDO


C(J,I+1) = B(J,N)/A(I+1)

C(J+1,I) = B(J+1,N)/A(I)

C(J+1,I+1) = B(J+1,N)/A(I+1)

ENDDO

ENDDO

Effectiveness of Unroll-and-Jam


InstructionFetch

InstructionDecode

OrRegister

Fetch

RegisterFile

ExecuteMemoryAccess

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RegisterFile

compiler improvement of register usage part 1 - chapter 8, through section 8.4 anastasia braginsky

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