The answer for Part(b) and (d) is (b)$A00 $310AFE $142B00 $042B01 $202B01 $FFFFF

Question

The answer for Part(b) and (d) is

(b)$A00 $310AFE $142B00 $042B01 $202B01 $FFFFF

$B00 $001234 $000004 $000004

(d)$000 $20200A $4B2001 $300102 $4C2000 $1B200A $FFF038 $4A2000

$00A $000005   

Can someone show me how to get this answer?

And can someone show me the steps to get answer for (a) and (c)?

Show the assembly output of the foliowing program fragments. The programs are not necessarily meaningful, so don't try to make sense of them. The assembly output should be organized as a table of locations and their contents. Do not use CHASM to solve this problem. Do it manually. EQU ,$100 LDA# ABCD+5 ADA VAR1 STAVAR2 CHA# ABCD-5 EQU , $203 LAB2: LDA VAR1-2 JGE LAB2 EQU DEF, $102 AND# DEF JMP LAB2 .EQU , $200 VAR1:.WORD $1234 VAR2: .BLKH 1 EQU ABCD, $45 END VAR1: .BLKW 4,5 END

Explanation / Answer

LDA-The contents of a memory location, specified by a    16-bit address in the operand, are copied to the accumulator. The contents of the source are not altered.   Example: LDA 2034 or LDA XYZ

CMA-The contents of the accumulator are complemented. No flags are affected.

INC-INCREMENT

JMP-The program sequence is transferred to the memory location specified by the 16-bit address given in the operand.   Example: JMP 2034 or JMP XYZ

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.