The code for the figure above has a CPWA at 000C to test the value of j. Because the program branches to that instruction from the bottom of the loop, why doesn’t the compiler generate an LDWA j, d at that point before the compare instruction? High-Order Languageinclude «etdio.h>int main() (int jifor (j 0: j 3; j {printf ( sd\n". j):return 0;Assembly Language0000 120003jiBRnain.EQUATE Oilocal variable #2a0003 580002 main:SUDSP2,1ipush #j0006 Co0000ifor (- 0LDWA0,10009 E30000STKA1,8000c ADD003 for:3,1i < 3CPKA000P 1C002ABRGEendPor0012 49002ESTROneg.diprintt( sdin", j)0015 3B000DDECO0018 DO00GALODA001B FIFC16001E C30000STBAcharout,dLOMA0021 600001ADDA1,i0024 E30000STWA1.80027 12000CBRfor002A 500002 endPor: ADDSP2,1pop002D 00STOP0028 GA203D mag:.ASCII j \x0020000033. END

The reason is that the present value of “i” will be in the ACCUMULATOR regardless whether control…

Answered: The code for the figure above has a…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

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The code for the figure above has a CPWA at 000C to test the value of j. Because the program branches to that instruction from the bottom of the loop, why doesn’t the compiler generate an LDWA j, d at that point before the compare instruction?

High-Order Language
include «etdio.h>
int main() (
int ji
for (j 0: j 3; j {
printf ( sd\n". j):
return 0;
Assembly Language
0000 120003
ji
BR
nain
.EQUATE O
ilocal variable #2a
0003 580002 main:
SUDSP
2,1
ipush #j
0006 Co0000
ifor (- 0
LDWA
0,1
0009 E30000
STKA
1,8
000c ADD003 for:
3,1
i < 3
CPKA
000P 1C002A
BRGE
endPor
0012 49002E
STRO
neg.d
iprintt( sdin", j)
0015 3B000D
DECO
0018 DO00GA
LODA
001B FIFC16
001E C30000
STBA
charout,d
LOMA
0021 600001
ADDA
1,i
0024 E30000
STWA
1.8
0027 12000C
BR
for
002A 500002 endPor: ADDSP
2,1
pop
002D 00
STOP
0028 GA203D mag:
.ASCII j \x00
2000
0033
. END

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