The machine code for instruction LDAB $3088 is (7B, 30,88) (Assuming data bus is
ID: 3782660 • Letter: T
Question
The machine code for instruction LDAB $3088 is (7B, 30,88) (Assuming data bus is 16 bit). In which computer component is this instruction stored? How many memory accesses are needed for fetching this instruction to CPU? How many memory accesses are required to execute this instruction? Explain in detail what happens m the computer during each memory access. The machine code for instruction LDAB $3088 is (7B, 30, 88) (Assuming data bus is 8 bit)? How many memory accesses are needed for fetching this instruction to CPU? How many memory accesses are required to execute this instruction? Explain in detail what happens in the computer during each memory access. The machine code for instruction STD #$3088 is (XX, 30, 88) (Assuming data bus is 16 bit). In which computer component is this instruction stored? How many memory accesses are needed for fetching this instruction to CPU? How many memory accesses are required to execute this instruction? Explain in detail what happens in the computer during each memory access. The machine code for instruction STD #$3088 is (XX, 30, 88). (Assuming data bus is 8 bit)? How many memory accesses are needed for fetching this instruction to CPU? How many memory accesses are required to execute this instruction? Explain in detail what happens in the computer during each memory access. The machine code for instruction LDD #$3088 is (XX, 30, 88) (Assuming data bus is 16 bit). In which computer component is this instruction stored? How many memory accesses are needed for fetching this instruction to CPU? How many memory accesses are required to execute this instruction? Explain in detail what happens in the computer during each memory access. The machine code for instruction LDD #$3088 is (XX, 30, 88) (.Assuming data bus is 8 bit)? How many memory accesses are needed for fetching this instruction to CPU? How many memory accesses are required to execute this instruction? Explain in detail what happens in the computer during each memory access.Explanation / Answer
For the instruction LDAB #3088 is (7B,30,88). Here we could see that, most significant bytes are stored in the lower memory address.
1. The instruction is stored in Big Endian Computer
Databus is 16 bits. Opcode and operand size is 16 bits as well.
2. Hence three memory access would be required to fetch the instruction into CPU
$3088 is the address where the data resides. Data read will be loaded into Accumulator. Hence there would be one memory access to read data from the memory address $3088H.
3. One memory access to read data from the memory address $3088H
4. Detailed flow
1. First memory access would be to fetch the instruction and decode it
2. Next the values 30,88H will be read into memory and interpreted as memory address
3. Value present in the memory address 3088H will be read and loaded to accumulator.
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