Tough Computer Systems question.
Show work for partial credit.
a)
Immediate addressing mode..
Load 3000 will store AX into 3000
AX becomes 3000
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b)
Direct addressing mode...
It stores the data value where at location 3000
So at address 3000, the value is 7000
So AX becomes 7000
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c)
Indirect addressing mode ...
LD AX [3000]
At 3000 address we have value 7000
So accumulator stores value at address 7000
So AX becomes 8000
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d)
Indexed addressing mode ...
Here AX is stored on address + offset
Load AX , 3000 + 5000
Load AX, 8000
So AX becomes 3000
Tough Computer Systems question. Show work for partial credit. Assume the following values are in memory,...
ISA & Addressing Mode The instruction opcodes and formats for a computer system are as follows Format AD AD OP AD SA OP SA SA LDdir LDindir LDrel LDindex ACC ← 씨씨ADn ACC ← OP ACC ← MPC-AD] ACC ← MRtSA].OP] ACC -RISA] 001 010 011 101 110 ·ISA Suppose the Instruction format ts as follows: AD: Address write the Operation for LDimm and LDreg (for immediate and register direct addressing) OP: Constant Operand SA : Register A ACC is...
This question is from Computer architecture. Please show steps and add comments so its easy to understand. Translate the high-level language code below into assembly instructions. The variables A, B, C, D, E and F are located in the memory and can be accessed by their label (e.g., LOAD R1, A will load A from the memory into R1). Minimize the number of instructions in the assembly code that you write. F = (A-B)*(C+D)/(E-D) a) Write the code for an...
pls both ans
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Encode the following instruction in hexadecimal. For partial credit, show your work in binary as sw $to,-16 ($sp) 니3.
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can you show me how you get each one or fill the table?
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