Suppose pep/8 contains following four hexadecimal values: A= 19AC, X=FE20, Mem[0
ID: 3677534 • Letter: S
Question
Suppose pep/8 contains following four hexadecimal values: A= 19AC, X=FE20, Mem[0A3F]=FF00, Mem[0A41]=103D
If it has these values before running instruction C10A3F, what are the values after running the instruction? Select all correct values.
Question 1 options:
A=FF00
A=19FF
X=20EF
Mem[0A3F]=19AC
X=FF00
Question 2
Suppose pep/8 contains following four hexadecimal values: A= 19AC, X=FE20, Mem[0A3F]=FF00, Mem[0A41]=103D
If it has these values before running instruction D10A3F, what are the values after running the instruction? Select all correct values.
Question 2 options:
A=19FF
A=1900
X=FFAC
X=19FF
X=20EF
Question 3
Suppose pep/8 contains following four hexadecimal values: A= 19AC, X=FE20, Mem[0A3F]=FF00, Mem[0A41]=103D
If it has these values before running instruction D90A3F, what is the value of X after running the instruction?
Question 3 options:
X=FE20
X=FF20
X=FEFF
X=EF00
Question 4
Suppose pep/8 contains following four hexadecimal values: A= 19AC, X=FE20, Mem[0A3F]=FF00, Mem[0A41]=103D
If it has these values before running instruction F10A41, what is the value of Mem[0A41] after running the instruction?
Question 4 options:
Mem[0A41]=10AC
Mem[0A41]=AC3D
Mem[0A41]=1019
Mem[0A41]=193D
Question 5
Suppose pep/8 contains following four hexadecimal values: A= 19AC, X=FE20, Mem[0A3F]=FF00, Mem[0A41]=103D
If it has these values before running instruction E90A3F, what is the value of Mem[0A3F] after running the instruction?
Question 5 options:
Mem[0A3F]=19AC
Mem[0A3F]=FF20
Mem[0A3F]=FE20
Mem[0A3F]=1920
A=FF00
A=19FF
X=20EF
Mem[0A3F]=19AC
X=FF00
Explanation / Answer
1Ans) Answer is A= FF00
The load instruction has instruction specifier 1100 raaa. This instruction loads one word (two bytes) froma memory location into either the accumulator or the index register depending on the value of r. It affects the N and Z bits. If the operand is negative, it sets the N bit to 1; otherwise it clears the N bit to 0. If the operand consists of 16 0's, itsets the Z bit to 1; otherwiseitclears the Z bit to 0. Theregister transfer language(RTL) specification oftheload instruction is
r Oprnd; N r<0, Zßr=0
The above question is initially Accumulater have this value A= 19AC,
X=FE20,
Mem[0A3F]=FF00,
Mem[0A41]=103D
C10A3F instruction means load.
Opcode r aaa operand specifiers
1 1 0 0 0 001 0000 1010 0011 1111
C 1 0 A 3 F
After excute this C10A3F load instruction
Load the value of 0A3F memory lacation into Accumulator. Mem[0A3F]=FF00 to Accumulator A.
Answer is A= FF00
2Ans) Answer is A= 19FF
The load instruction has instruction specifier 1101 raaa. This instruction loads one byte from a memory location into either the accumulator or the index register depending on the value of r. It affects the N and Z bits. If the operand is negative, it sets the N bit to 1; otherwise it clears the N bit to 0. If the operand consists of 16 0's, itsets the Z bit to 1; otherwiseitclears the Z bit to 0. Theregister transfer language(RTL) specification oftheload instruction is
r Oprnd; N r<0, Zßr=0
The above question is initially Accumulater have this value A= 19AC,
X=FE20,
Mem[0A3F]=FF00,
Mem[0A41]=103D
C10A3F instruction means load.
Opcode r aaa operand specifiers
1 1 0 0 0 001 0000 1010 0011 1111
C 1 0 A 3 F
After excute this D10A3F load instruction
Load the one byte of 0A3F memory lacation into Accumulator. Mem[0A3F]=FF00 to Accumulator A.
Answer is A= 19FF
3Ans) Answer is X= FEFF
The load instruction has instruction specifier 1101 raaa. This instruction loads one byte from a memory location into either the accumulator or the index register depending on the value of r. It affects the N and Z bits. If the operand is negative, it sets the N bit to 1; otherwise it clears the N bit to 0. If the operand consists of 16 0's, itsets the Z bit to 1; otherwiseitclears the Z bit to 0. Theregister transfer language(RTL) specification oftheload instruction is
r Oprnd; N r<0, Zßr=0
The above question is initially Accumulater have this value A= 19AC,
X=FE20,
Mem[0A3F]=FF00,
Mem[0A41]=103D
C10A3F instruction means load.
Opcode r aaa operand specifiers
1 1 0 0 1 001 0000 1010 0011 1111
C 1 0 A 3 F
After excute this D90A3F load instruction
Load the one byte of 0A3F memory lacation into index register because r=1. Mem[0A3F]=FF00 to Accumulator A.
Answer is X= FEFF
4Ans) Answer is Mem[0A41]=10AC
The store instruction has instruction specifier 1111 raaa. This instruction stores one byte fromeither the accumulator or the index register to a memory location. With directaddressing, the operand specifies the memory location inwhich theinformation is stored. The RTL specification for thestoreinstruction is
Oprnd r
The above question is initially A= 19AC,
X=FE20,
Mem[0A3F]=FF00,
Mem[0A41]=103D
F10A41 instruction means store byte.
Opcode r aaa operand specifiers
1 1 1 1 0 001 0000 1010 0100 0001
F 1 0 A 4 1
After excute this F10A41 store byte instruction
Store the one byte of 0A41 memory lacation from Accumulator because r=0.
Answer is Mem[0A41]=10AC
5Ans) Answer is Mem[0A3F]= FE20
The store instruction has instruction specifier 1110 raaa. This instruction stores one word (two bytes) fromeither the accumulator or the index register to a memory location. With directaddressing, the operand specifies the memory location inwhich theinformation is stored. The RTL specification for thestoreinstruction is Oprnd r
The above question is initially A= 19AC,
X=FE20,
Mem[0A3F]=FF00,
Mem[0A41]=103D
E90A3F instruction means store.
Opcode r aaa operand specifiers
1 1 1 0 1 001 0000 1010 0100 0001
F 1 0 A 4 1
After excute this E90A3F store instruction
Store the one byte of 0A3F memory lacation from index register because r=1.
Answer is Mem[0A3F]= FE20
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