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Group of answer choices ch occupies one byte, j occupies 2 bytes and they are both local variables ch occupies one byte, j occupies 2 bytes and they are both global variables ch occupies two bytes, j occupies 1 byte and they are both global variables ch and j are both constants

Group of answer choices ch occupies one byte, j occupies 2 bytes and they are both local variables ch occupies one byte, j occupies 2 bytes and they are both global variables ch occupies two bytes, j occupies 1 byte and they are both global variables ch and j are both constants

Database System Concepts
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
Section: Chapter Questions
Problem 1PE
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Which of the choices best describes the above code?

Group of answer choices
ch occupies one byte, j occupies 2 bytes and they are both local variables
ch occupies one byte, j occupies 2 bytes and they are both global variables
ch occupies two bytes, j occupies 1 byte and they are both global variables
ch and j are both constants
main . BLOCK 1 . BLOCK 2 0000 120006 BR ;global variable #1c ¡global variable #2d 0003 00 ch: 0004 0000 j: 0006 DIFC15 main: ;scanf("%c %d", &ch, &j) charIn,d ch,d j,d j,d 5,i j,d ch,d 1,i ch,d ch,d charOut,d '\n',i charout,d j,d \n',i charout,d LDBA 0009 F10003 STBA 000c 310004 DECI 000F C10004 LDWA ¡j += 5 0012 600005 ADDA 0015 E10004 STWA 0018 001B D10003 LDBA ;ch++ 600001 ADDA 001E F10003 STBA 0021 D10003 LDBA ¡printf("%c\n%d\n", ch, j) 0024 F1FC16 STBA 0027 DO000A LDBA 002A F1FC16 STBA 002D 390004 DECO 0030 DO000A LDBA 0033 F1FC16 STBA 0036 00 STOP 0037 .END
Transcribed Image Text:main . BLOCK 1 . BLOCK 2 0000 120006 BR ;global variable #1c ¡global variable #2d 0003 00 ch: 0004 0000 j: 0006 DIFC15 main: ;scanf("%c %d", &ch, &j) charIn,d ch,d j,d j,d 5,i j,d ch,d 1,i ch,d ch,d charOut,d '\n',i charout,d j,d \n',i charout,d LDBA 0009 F10003 STBA 000c 310004 DECI 000F C10004 LDWA ¡j += 5 0012 600005 ADDA 0015 E10004 STWA 0018 001B D10003 LDBA ;ch++ 600001 ADDA 001E F10003 STBA 0021 D10003 LDBA ¡printf("%c\n%d\n", ch, j) 0024 F1FC16 STBA 0027 DO000A LDBA 002A F1FC16 STBA 002D 390004 DECO 0030 DO000A LDBA 0033 F1FC16 STBA 0036 00 STOP 0037 .END
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