BRANCHING IN FORTRAN

J. G. Sullivan

Abstract

The execution time of Fortran programs can be improved by properly considering both the methd used to cause a p r o g w

I. branch and fhe position of the branch point. The discussion \--.. points out some general rules which can be applied to Fortran programs which wit l improve execution time.

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Branching i n Fortran

There a re f i v e methods available t o the Fortran programmer fo r

choosing among paths fo r program execution. Four of them are , essen t ia l ly ,

performing a t e s t of some variable and, depending on i t s v a u e , branching t o

a l t e rna te port ions of t he code. One, the GO TO n type statement where n i s

a statement number (e .g . , GO TO 273)) i s a branch with no t e s t . The burden

of t h i s discussion w i l l be the implications of statemerts which t e s t and

branch and t he considerations necessary fo r the FortraK programmer t o

e f f i c i en t l y choose among the abundance of possibilities.

The four methods are:

1 ) " ~ o m ~ u t e d " GO TO (nl, n> ,..., n.), I; e .g . , GO ~0(1,2 ,3 ,43,7) , 1 J

2) "Assigned" GO TO; e.g., GO TO I, where I has been previously

defined by an ASSIGN statement.

3 Arithmetic.IF; e.g., IF(A-E) 1, 2, 3

4 ) Logical IF; e.g., IF(A.GT.B).W TO 100

As methods 1 and 2 are undoubtedly r a r e r ,in occurrence than methods

3 and 4, a few remarks w i l l be made about them before devoting t he bulk ,of t h e

discussion t o methods 3 and 4.

GO TO STATEMENTS

F i r s t of all, t he "assigned" GO TO does not d i f f e r l og i ca l l y from

the "conputed" GO TO statement.

Example l a :

ASSIGN 200 TO ITRA

IF (x- 5 . ) 10, 20, 30

1 0 CONTINUE

- FORTRAN STATEMENTS-

ASEIGN 50 TO ITRA -

CQ TO 40

20 CONTINUE

- FORTRAg STArnENTS -

ASSIGN 73 to ITRA

Go 70 '0

30 m(x-6.) 31, 40, 40

f l C3I!JTINLX

- F - 3 R M STATEWNTS - ASSIGN 371 I1Y) ITRA

kl G O Y O r n

where t h e e=ecutix~ of statement 40 will cause a transfer of execution to

one of hhe follob-ing statements : 203, 50, 73, or 371.

The sm-e cz~ding, but x i n g the "ccmpnted" GO TO, is illustrated

in the .?ollcwing coding:

Example. 1t :

' I m = i

IF(X-5.1 10, 20, 30

10 CONTINUE

- F0R'nA.N STXTErnJTS -

- FORTRAN STA'TEMENTS -

= :,

0 TC* 4c

3 I F ( x - ~ . ) 31, 40, 40

31 CONTINUE

- FORTRAN STATrnIENTS - ITRA = 4

40 GO TO (200,50,73,371), ITRA

There i s no log ica l difference i n t h i s usage but t he use of t he "computedtt

GO TO probably makes t he coding a l i t t l e eas ie r t o follow. One advantage . .

fo r choosing t h e "computed" form over t h e "assigned" Porm i s t h a t , during t h e

debugging phase, t he "assigned" GO TO can give pecul isr (and nonreproduzible)

e r ro rs i f the re i s a GO TO ITRA and t h i s statement i s executed p r io r t o an

ASSIGN TO ITRA. Also, some "programmers" a re prone t o wri te ar i thmet ic

assignment statements (ITRA = 200) when they should have wri t ten ASSIGN;

200 TO ITIW. One disadvantage with using t h e "computed" GO TO during

debugging i s t h a t while t he action taken at statement 40 i n Example l b

i s c lea r ly defined f o r ITRA = 1,2,3, or 4 various corrpilers react i n

d i f fe r ing ways f o r ITRA < 1 and ITRA > 4. One's predi lect ions must be t h e

determining fac tor here. So f a r as speed of execution i s ' concerned t h e use

of t he "assigned" GO TO i s definit 'ely f a s t e r . This i s evident from cozsidering

t h a t the "computedt1 GO TO must e i t he r do IF-l ike t e s t s , e.g. :

or e l s e t r e a t t h e addresses 'of the .b rmch points as i f they were an array

subscripted by ITRA. There .a lso may be s a e "mickey-mouse" i n taking -are of

values of ITRA outside t he range of t rans fe rs . The "assigned" GO TO i s never

more t h m a load of ITRA and a branch t o the 'address specified b y ' t h e value of

ITRA. One might a lso note t ha t i f there a re only th ree branch points: GO TO

(200,50,73), ITRn the IF(ITRA-2)200,50,73 w a ~ r of wri t ing it i s probably a

shade f a s t e r i n execution. A l l i n all, use of t he "computed" GO TO i s

4

possibly b e t t e r during t h e debugging phase and t h e "assigned1' GO TO i s b e t t e r

i n the execution phase. (This, of course, only applies when methods 1 and 2

a re used interchangeably as i n Example 1.) I f one, say, , s t a r t s off a program

with:

READ 1, ITRA .

GOT0 (1, 2, 3, 4), ITRA

then use of t h e "assigned" GO TO i s , of course, not applicable.

I F STATEDENTS

The l o g i c d IF and tk-e arfthmetic IF can, i n many instances, be

ra ther f r2ely interchanged :

1 C O I ' r I W 1 CONTINUE

However, because 3f t h e greater complexity of the l og i ca l IF each method w i l l

be taken up s epar 5 te ly .

Method .3. ARITHMETIC! IF

In t h i s method there %re, of course, two items f o r consideration:

1) the e q r e s s i c n t 3 be evaluated t e s t ed and 2) t he possible branch points.+

Whenever my mithmet ic operations a r e . ca l l ed f o r i n t h e evaluation of an

expressior (2.g., A-B) t h e action takan i s e s sen t i a l l y t he same as t h a t

performed on t h e right-hand s ide of ?he ari thmetic replacement statement except

t h a t no s tor ing of t he r e s u l t occurs and t he r e su l t i s t e s t ed i n whatever way

t h e computer hardware permits. When no ari thmetic operation appears, then t h e

manner i n whicn t he t e s t of t h e var iable i n storage i s made i s a function of

ccmputer hazdware. (some computers c a ~ t e s t t h e var iable i n memory d i rec t ly ;

o thers Eay have t c load t he var iable i n t o an accumulator or r e g i s t s r of some

kind and then 2er fom t h e t e s t ) . For i l l u s t r a t i v e purposes t h e 360 hardware

operations are used t o describe the action taken and the mnemonics t c be

used are defined as follcws:

MNEMONIC, Var Table 'Action

LOAD,X The variable, X, i s loaced int.0 the r eg i s t e r

ADD, X The variable, X, i s added t o the r eg i s t e r

SUB,X ' , The variable, X, i s subtracted from the r eg l s t e r

. . B, 1 Branch t o statement 1

B Z , l Branch ' i f the reg is te r i s i e ro t o statement 1 otherwise, proceed t o the next ins t ruct ion

BP, 1 Branch i f the reg is te r i s greater than 0 t o statement 1, ctherwise proczed.

BM, 1 , . Branch on minus,' etc.

BPZ,1 ~ r m c h on greater than or equal t o zero, e tc .

BM, 1 Branch on minus or zero, e tc .

BPM, 1 Branch on greater than zero or l e s s than zero, e tc .

TEST .Test the reg is te r , .

It i s necessary t o note tha t a f t e r an addit5on or subtraction the reg is te r

can be tes ted and 'the branches made without the necessity for a TEST instruct ion.

After a LOAD, a TEST instruct ion must be given before the branch cm be made.

(statement numbers w i l l appear a t the l e f t of the mnemonic code).

E Y W U : LOAD,X Explanation: Y i s s7ubtracte3 from X.

SUB,Y , . . and i f the r e su l t i s zero 3,

BZ, 1 branch t o stztement 1 i s taken.

EXAMPLF:: LOAD,X Explanation: X i s loaded, tes ted, and

TEST i f X i s l e s s than zero a branzh t o

BM, 1 statement 1 i s taken.

EXAMFU: LOAD.,X Explanation: WRONG. The TEST must be

BM, 1 given a f t l r the LOAD s s in the

preceding example.

Example 2

TDEIYGS i n psec for 360175 I

.9 :LOAD,A 7

I3 7

BM, 1 1.0

3z,2 1 .0

3p,: . 1.0

Finding .the amount of computer time t o execute ststement 9 i s a s follows:

Case . . Time (psec)

A < B 2.4

A = ' B , 3.4

A > B 4.4

Thus, ~ons iderab le diff,erence i n exes~ t ion . t ime occur. In fac t , t he

e x e c u t l o ~ time for ;he case where A > B i s almost double t ha t fo r the case

where -< < E. Coxpilers have no information upon which t o base the order i n

which the t s s t s -A-ould be "most optimally" performed and w i l l usually t rans la te

t h i s 13' i n 3 neckanical way (BN, BZ, EP) or (BP, EZ, BM), e tc . E n n , i f the

prograrmer lcnows t h e t A i s usuallf greater than B, t e must a lso k?ow

the order i n uhich the compiler w i l l s s t up th? t e s t s . There i s n3 log ica l

difference :n the fcllou3ng rearrangem~nt of statement 9 :

5: IF(B--~) 3, 2, 1

yet t h i s arrangement might take about half the time of the statement as

or ig ina l ly written! . .

Take the most extreme example:

Example 3

Timing' (p sec )

TEST

BZ, 2 . 1.0

ere the time taken fo r A < 0. i s 2 .1 psec and the time taken

f o r A > 0 i s 4 .1 psec, a considerable difference.

A complication (e i ther helpful or harmf'ul so f a r a s the t o t a l

execution time) appears when' one of the branch points immediately follows

the IF. For example :

Example 4 . .

9 IF@-B)I, 2 , 3

r" CONTINUE

This could be . t rans la ted :

Example 4a.

r n , B - 7

BMj 1 1.. 0 . . .

BZ, 2 1 . 0

BP, 3

2: e tc .

SUE, B

2: e tc .

Timings CPsec)

Exanple : 4.3. 4b.

In moet case;, compile-s make aoEe use of <he f a c t tha t the following

s ta te ren t na;r be one ci' the I F stat.ement branch ~ o i n t s , so z gain i n

I execution ;ine ,c,r, a t . least , no lcsss) i s possible i f a brzrc'n point of an

IF imcnedia-,ely follows the I F statemeat.

In o r e r t o Lee how c i f f i c u l t l i f e can .be, make the following

assumption the A > B branch I s most l i ke ly . If we write:

Example 5

9 IF:S-4) 2, 2, 1

2 CmTm

which t ranslz ted, might be :

LOAD, I3 .7'

SUB, A 7 BZ,2 1 .0

BP, 1 1.0

3 et.2.

The time t o reach stctement 3 i s 3.4 , ~ s e c .

I f we were t o move statement 3 away from t h e IF t e s t :

Example 5a 9. IF(B-A) 3, 2, 1

4 CONTINUE

and, i f t he compiler makes t h e minus t e s t f i r s t , t h e most frequently

followed path i s performed i n 2.4 psec . In many cases two of t he branch points of t he ari thmetic

'IF may be iden t ica l :

IF(A-B) 1, 2, 1

This i s generally t rans la ted as:

LOAD, A

SUB, B

BPM, 1

BZ, 2

The same s o r t of consideration holds as i n the discussion above where t h e

branch points were d i f fe ren t . One t e s t has s t i l l t o be performed f i r s t and

a proper asrangement of statement locat ions (e i ther following t h e I F o r not

following t h e IF) must be undertaken f o r maximum efficiency.

S m q : I n order t o determine how t o minimize execution time

t h e following must be considered:

1) Should 'the t e s t be reversed?

2) Should one of t h e branch points follow the, IF?

3 ) If the answer t o 2 ) i s yes, which brmch point should

be chosen?

4) 'Look a t t he machine language coding actual ly produced and

discover whether t he compiler has "outsmarted" you!

Method 4. LOGICfL IF

Tne glneral form of t h i s IF d l o v s several l o . ~ i c a l expressions t o

appear i n what laoks l i k e one great t e s t but the yrograrrwr should bear care-

f u l l y in mind th;%t the order i n which the expressions amzar may be qui te

important :

EWU: IF(A.I;T.B.GR.C.LT.D;: G O T O 100

This stat?men% i q l i e s t ha t i f e i t he r A > B o r C < D the branch to; statement

100 shculd 5e performed. It i s a b i t simpler f o r discussion purposes t o break

t h i s I F i n to two I F statements perfcrming.exactly the s m e log i ca l function:

This i s e s s e n t i a 1 . l ~ the way i n -&ich most cc~mpilers w i l l hreak .up the o r ig ina l .

IF befc-e t r a r s l a ~ i n g t o machin? 1an.guage. There i s no s t r i c t Fortran ru l e

t ha t the -bests are performed i n order f'rom l e f t t o r i gh t so some compilers,

i n the t rans la t ing process, coulri jus t a s wel l reverse t h? order:

IF(C.LT.D) GO TO 100

IF(A.GT.B) GO TO 100

Naturally, i f fhe resul t of one of t he two t e s t s , say A > 3, i s expected t o be

t r ue a large percentage of the time, it behooves the programmer t o have t h i s

t .est performed f i r s t . Unllss the c h ~ r a c t e r i s t i c s of the compiler a r e known

(a process 05' compiling a few sample I F ' S and h o k i n g st the machinz language

r e su l t ) , it i s gmera l l y wiser to brsak complicated IF 'S i n t o t h e i r simpler

form t o cc8ntrcl t h e order of t es t ing .

Fc.r example :

IF(A.GT.B.AND.C.LT.D.OR. I .GE.O) GO TO 100

This i s , logical ly , r e a l l y three I F statements

I F ( A . ~ . B ) GO TO 1

IF(C.LT.D) GO TO 100

Obviously, i f I i s going t o be zero or pos i t ive a large percentage of the

time the t e s t a s wri t ten could take on the order of three times as long a s

i t needs t o take i n ac tua l s i tuat ions . While the s tmpl ic i ty of s t r ing ing

the various log ica l expressions .has same advantage i n i n ~ t a n t a n e ~ u s recognition

of what the code i s doing, the application of E, l i t t l e knowledge concerning

the probable outcome of t he t e s t s could provide overriding advantages i n the

reduction of execution time.

The remarks made' i n the discussion' of Method 3 'concerning e i t h e r

simplifying or removing IF t e s t s from innermost'D0 loops apply exact ly the

same a s fo r Method 4.

One question which aprogrammer shodd ask i s :

How do t he following s e t s 'of coding compare? . .

a) . . IF@-B) 1, 100, ,100

1 CONTINllE

b ) IF(A.GE.B) GO TO 100

1 CONTINUE

In most cases, the resu l t ing code would be comparable but a look a t a

few examples of the compiler output would point out my spec i s l cases

where a time advantage could be achieved.

12

One d e t z i l concerns e i t k e r type of I F statement imbedded ins ide

nested 30 loops: t h e LOAD &KC TEET ill take 1.1 psec and t h e LOAD and SUB

w i l l t a i e 1 . 4 psec. T h e follcwing example i l l u s t r a t e s poss ible gains:

Example 6

BEFORE DO 1 1=1,100

DO 1 J=1,100

DO 1 K=1,100

IF(FLAG-1. ) 10,?0,1o

1 0 COIflINUE

- FORTRAN STATE?rIEN;rS - GO TO 1

20 COmTINJE

- FORTw 3TATEMERTS - 1 CON'TINJE

AFTER :

ExampLe 6~

FLAGA=FLZG-1.

DO 1 I=1,100

DO 1 J = l , l G O

DO 1 K=1,100

IF (FLAGA) 10, 20, 10

e tc .

In f a c t , depending on the complexity of t he 1oop.and 03 the r e l a t i v e

mount of time spent in performing t h e t e s t as contrasted t o t h e

remainder of t he calculations in t he innermost loop, i t i s en t i r e ly

possible t h a t the following can be much f a s t e r :

. Example 6~

100

IF(FLAG-1. ) loo, 200, 100

DO 101 I=1,100

DO 101 J=1,100

DO 101 K=1,100

CONTINUE

FORTRAN STATEMENTS - CONTINUE

GO TO 7

DO 201 I=1,100

DO 201 J=1,100

DO 201 K=1,100

CONTINUE

FORTRAN STATEMENTS - CONTINUE

CONTINUE

Jus t as constant sub-expressions i n ari thmetic replacement

statements should be removed from inner loops, so a l so should IF

expressions be simplif ied.

DO 1 I=1,10

DO 1 J=1,20

z = A(I,J)+(I:J)

DO 1 K=1,100

IF(Z-x(I,J,K~+~.) 2, 3, 4

e t c .

should b? rewrit ten as:

1'0 1 I=1,10

DO 1 J=1,20

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

"EETZ = Z + 1 .. DO 1 K=1,100

IF(TESTZ-x(I,J,K)) 2, 3, 4

2tc.

a1 timir-.gs given &re app-oximate m d depend on verious conditions

and assumptions as t o average use ~ . f t h e com~uter . The memory tirnings used

here a re fron - IBM SYSTEM/~~O NODEL FUNCTIONAL CHARACTERISTICS, Form

A22-6889-0. Repeated attempt a t o c'btain more deta i led information from

IBM concerning t h e assumptions used have met with f a i l u r e . No t i n ing

given here may be taken as ccrrect but t he general cons ide r~ t i ons are correct .

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