Question

Stack Operation 10. (10 pts.) Show the contents of the stack and affected registers at the two marked points in the execution of the followin code. Assume RO-0, R1-1, R2-2, R3-3, R4-4, R5-5, and R6-6. The initial value for stack pointer (prior to executing this code block) is given as SP-0x20001000 PUSH R2,R3) ADD R4, R1, Re ;<---A POP R5, R6) ADD R5, R5, R4 ADD R6, R6, R5 PUSH (R4-R6); SUBS Re, RO,R1-B a) Show the contents of Stack, SP and affected registers at execution point A Memory Memory Register Register Value Affected Contents Address 0x20000FF4 0x20000FF8 R1 0x20000FFC R2 0x20001000 R3 0x20001004 R4 0x20001008 R5 0x2000100C R6 SP b) Show the contents of Stack, SP and affected registers at execution point B: Memory Register Register Value Contents Address Affected 0x20000FF4 RO 0x20000FF8 R1 0x20000FFC R2 0x20001000 R3 0x20001004 R4 0x20001008 R5 0x20001000C R6 SP

Answer 1

Initial Values for registers are :

When First instruction (PUSH{R2,R3} is executed :

Value of SP(stack pointer) is decreased by 4

now SP have value 0x20000FFC then value in Register R2 is pushed into stack at the Memory Address 0x20000FCC(value represented by SP)

after that value of SP is decreased by 4 and became 0x20000FF8 now value of R3 register is pushed at stack at Memory Address 0x20000FF8.

When Second instruction(ADD R4,R1,R0;) is executed it adds value of R1 and R0 to R4.

what ADD instruction do is take value from second place(here R1) and third place(here R0) to do arithmetic(adding) then store the result at first place(here R4). after execution of this instruction value of R1 and R0 remains same as before 1 and 0 respectively but value of R4 is changed to sum of 1 and 0. so value of R4 became 1 from 4.

At the point A values are :

What POP instruction do is read the value from Memory Address to the Register and then increase SP value.

When third instruction is executed(POP{R5,R6}):

value at memory address 0x20000FF8(represented by SP) is copied to R5 then value of SP is increased by 4 and became 0x20000FFC now value at memory address 0x20000FFC is copied to R6 and value of SP is increased by 4. now value of R5 is 3, R6 is 2 and SP is 0x20001000.

Forth instruction ADD R5,R5,R4 stores the value got from sum of R5(value at this point 3) and R4(value at this point 1) at R5. now value of R5 becomes 4.

After Fifth instruction ADD R6,R6,R5 value stored by R6 became 6 by adding value from R6(before instruction executes 2) and R5(value 4)

Sixth instruction PUSH{R4-R6} pushes value of registers R4,R5andR6 to the Stack.(to the Memory Address by decreasing the value of SP) first value of SP is decrease to 0x20000FFC and value of R4(1) is copied to Memory Address 0x20000FFC, then value of SP is decreased to 0x20000FF8 and value of R5(4) is copied to Memory Address 0x20000FF8, later value of SP is decrease to 0x20000FF4 and value of R6(6) is copied to Memory Address 0x20000FF4.

The last instruction SUBS R0,R0,R1 : subtracts the value of R1(1) from value of R0(0) and stores the result into R0(-1).

what SUBS do is stores the result at first place(R0) got from subtracting value of third place(R1) from second place (R0).

SUBS               R0                    R0               R1           means

operator     destination           operand1        operand2

At the point B values are :

Program only explores first 3 memory address so the remaining value of memory address is not known. it can contain any possible value.