Suppose that a beq instruction resides at memory address 0x08000000 and that the rightmost 16 bits...

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Suppose that a beq instruction resides at memory address 0x08000000 and that the rightmost 16 bits...

Suppose that a beq instruction resides at memory address 0x08000000 and that the rightmost 16 bits within the machine code for the beq contains the value 0x00BB. To what value (expressed in hex) would the PC be set if the zero flag=1 when the beq instruction is executed?

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Answer 1

Answer #1

Since zero flag =1 this means that content of both register are same hence branching will happen.

Since the rightmost 16 bit contain the offset value by which the branching should happen when offset value is multiplied by 4. Multiply by 4 happen because every address is multiple of 4.

So, present memory address is 0x08000000, then program counter value will be increased by 0x00BB * 4 = 0x02EC .

So, PC will be set to 0x08000000 + 0x02EC = 0x080002EC

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Answer 2

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