Question

1. Based on the following code segment what values are in the runtime stack immediately after the fourth time the instruction labeled Rec is executed .text li Sa0, 7 jai Fact #assume the address of this instruction is 2000 fother code not shown Fact: addi Ssp, Ssp, -8 sw Sra, 4(Ssp) sw Sa0, 0(Ssp) slti St0, Sa0, 1 beq St0, Szero, Rec li SvO, 1 addi $sp, Ssp, 8 jr Sra Rec: addi $a0, Sa0,-1 Jal Fact #assume the address of this instruction is 5000 lw Sa0, 0(Ssp lw Sra, 4(Ssp) addi Ssp. Ssp. 8 mul StO. Sa0, $0 r Sra

can someone explain how you get the answer to this question?

Answer 1

NOTE :- Final answer is in BOLD format

Explanation :-
Whenever a function call statement is called; the stack pointer is decreased and values are stored such as return address and arguement values.So,accordingly stack is decreased. While returning back to the caller; the stack pointer is increased and return value address is retrieved and control jumps back to the caller.

So, for first time when jal Fact is called from .text

STACK :-
2000 (return address)
7 ($a0) - stack pointer is here now

1st time when beq $t0,$zer0,Rec
Again,Fact is called;
so,now
STACK :-
2000 (return address)
7 ($a0)
5000 (return address)
6 (new $a0) - stack pointer is here now

2nd time when beq $t0,$zer0,Rec
Again,Fact is called;
so,now
STACK :-
2000 (return address)
7 ($a0)
5000 (return address)
6 ($a0) -
5000 (return address)
5 (new $a0) - stack pointer is here now

3rd time when beq $t0,$zer0,Rec
Again,Fact is called;
so,now
STACK :-
2000 (return address)
7 ($a0)
5000 (return address)
6 ($a0) -
5000 (return address)
5 ($a0)
5000 (return address)
4 (new $a0) - stack pointer is here now

4th time when beq $t0,$zer0,Rec
Again,Fact is called;
so,now
STACK :-
2000 (return address)
7 ($a0)
5000 (return address)
6 ($a0) -
5000 (return address)
5 ($a0)
5000 (return address)
4 ($a0)
5000 (return address)
3 (new $a0) - stack pointer is here now