Question

1. How many bytes are allocated in total for the variables (va, vb, vc, vd, and ve)? 2. For each of the instructions highlighted (boldface), derive the effective address of the memory variable (source) if any memory variable is involved, and the contents of the register a or d (whichever one involved) right after the instruction is executed. Answer in hexadecimal. Assume that all bytes in the variables va, vb and vc are initially zero. $0800 RAMStart EQU EQU ROMStart $4000 ORG RAMStart DS.B 1 va vb DS.B 5 DS.W DC.B 4 ve vd $01,810,SAF," 1" $01,810,8800,8806," 1" ," 234" DC.W ve ORG ROMStart daa #$12 #802 $802 ldd ldd daa va ldaa $810 #vd 0,x dx Idaa ldaa 3,х ldaa 2,+x 1,x #2 ldaa ldab daa dx #$802 $14,x] #22 (d,x] ldaa dd ldaa

Answer 1

Solution:

1)

va                DS.B 1
vb                DS.B        5
vc                DS.W       4
vd DC.B $01,$10,$AF,"1"
ve                DC.W      $01,$10,,$800,$806,"1","234"

         each B(byte) takes as 1 byte = 8 bits

         each W(word) taken as 2 byte = 16bits

So

va has(1*1)=1 byte

vb has (5*1)=5 bytes

vc has (4*2)=8 bytes

vd has (4*1)=4 bytes

ve has (2*6)=12 bytes

sorry For 2 i don't have enough knowledge.

Thank you, Have a great day:-)