a. Suppose the program counter, PC, has the value 0x12345678. What is the value

Question

a. Suppose the program counter, PC, has the value 0x12345678. What is the value of PC after executing the following jump instruction?

j 0x10

Note: The meaning of "0x10" here is that the j instruction has the value 0x10 (= 16 in decimal) in its 26-bit immediate field.

Answer: The value of PC after execution (in hex) = 0x_________ (give all 8 hexits, leading "0x" already given)

b. Suppose the program counter, PC, has the value 0x12345678. What is the value of PC after executing the following instruction?

beq $0, $0, -5

Note: The meaning of "-5" here is that the branch instruction has the value -5 in its 16-bit immediate field.

Answer: The value of PC after execution (in hex) = 0x_________ (give all 8 hexits, leading "0x" already given)

c. Suppose the program counter, PC, has the value 0x12345678. What is the value of PC after executing the following instruction?

bne $0, $0, 5

Note: The meaning of "5" here is that the branch instruction has the value 5 in its 16-bit immediate field.

Answer: The value of PC after execution (in hex) = 0x_________   (give all 8 hexits, leading "0x" already given)

The correct answers are 10000040, 12345668, and 1234567c. Please show me how you got the answers.

Explanation / Answer

a.
j 0x10;

new PC = base address + offset *4 = 0x10000000 + 0x40 = 0x10000040

b.
beq $0,$0,-5

it has -5 in it's immediate field which means the new PC would PC + 4 - offset*4 .Therefore;
NEW PC = PC + 4 - 5*4 = 0x12345678 + 0x4 - 14 = 0x12345668

c.
beq $0,$0,5

it has -5 in it's immediate field which means the new PC would old PC + offset*4 .Therefore;
NEW PC = PC + 4 + 5*4 = 0x12345678 + 0x4 + 14 = 0x12345690

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