1. MIPS R3000 Subroutine Calls and Stack
CS 286 – Project #2
MIPS R3000 Subroutine Calls
and Stack
Department of Computer Science
Southern Illinois University Edwardsville
Summer, 2019
Dr. Hiroshi Fujinoki
Subcall/001

2. CS 286 – Project #2 Subroutine Calls in MIPS R3000
This PPT presentations describe how to write an assembly program
for MIPS R3000 processor that uses subroutine calls
The concept of “Stack Frame” in MIPS R3000 processor
Implementation of recursive subroutine calls in MIPS R3000 assembly
program
Subcall/002

3. CS 286 – Project #2 Subroutine Calls in MIPS R3000
the first address
in your memory
the last address
High Address
Low Address
main:
In this presentation, it is assumed that
the low address appears at the top
Each program executes from the
top to the bottom
jr $31
Subcall/003

4. right after the “jar” instruction Go back to the instruction
CS 286 – Project #2
Subroutine Calls in MIPS R3000
main:
In MIPS R3000 assembly language:
jal xyz
(a) Calling a subroutine
Jump to “xyz”
  
“jal” instruction
(b) Returning to a calling routine
right after the “jar” instruction
Go back to the instruction
jr $31
“jr $ra” instruction
xyz:
jr $ra
Subcall/004

5. CS 286 – Project #2 Subroutine Calls in MIPS R3000
“jr $ra” instruction makes sure that the program execution flow
goes back to “the right return point”
main:
jr $31
xyz:
jr $ra
jal xyz
  
jal xyz
  
Subcall/005

6. CS 286 – Project #2 Subroutine Calls in MIPS R3000
 a danger in using subroutine calls in MIPS assembly programming
Destroying register contents during a subroutine call
S0 register as
a loop counter
main:
jr $31
jal xyz
  
li $s0, 10
LOOP:
sub $s0, $s0, 1
jne $s0, $zero, LOOP
xyz:
jr $ra
li $s0, 200
add $s0, $s0, 5
  
Processor
We have one
processor
slt $s0, $t3, $zero
  
Subcall/006

7. CS 286 – Project #2 Subroutine Calls in MIPS R3000
“recursive subroutine call”
main:
xyz:
jr $ra
xyz:
jr $ra
xyz:
jr $ra
jal xyz
li $s0, 10
add $s0, ...
li $s0, 10
add $s0, ...
li $s0, 10
add $s0, ...
  
jal xyz
jal xyz
jal xyz
  
jr $31
Why are recursive subroutine calls problematic
in using processor registers?
It’s guaranteed that the same registers
will be used in every recursive call ...
What should we do this to solve this problem?
Subcall/007

8. CS 286 – Project #2 Concept of stack (1) Stack Area main: li $s0, 10
LOOP:
sub $s0, $s0, 1
jne $s0, $zero, LOOP
xyz:
jr $ra
jal xyz
reserve my stack space
save all registers I modify
$s0
  
li $s0, 200
jal xyz
add $s0, $s0, 5
  
  
slt $s0, $t3, $zero
  
restore all registers I modified
release my stack space
jr $31
xyz:
Stack Area
Subcall/008

9. “recursive subroutine call”
CS 286 – Project #2
Concept of stack (2)
“recursive subroutine call”
main:
xyz:
jr $ra
jal xyz
li $s0, 10
add $s0, ...
xyz:
jr $ra
jal xyz
li $s0, 10
add $s0, ...
save $s0
restore $s0
xyz:
jr $ra
jal xyz
li $s0, 10
add $s0, ...
save $s0
restore $s0
jal xyz
save $s0
restore $s0
  
jal xyz
$s0
  
  
jr $31
Stack
Each called instance of a subroutine
reserves its stack space and saves the
register it modifies
$s0
$s0
$s0
Subcall/009

10. CS 286 – Project #2 Subroutine Calls in MIPS R3000
 “jr $ra” jumps to the address in
$ra register
Resume to “main”
main:
xyz:
jr $ra
xyz:
jr $ra
xyz:
jr $ra
jal xyz
  
jal xyz
$ra
(next instruction)
jal xyz
jal xyz
$ra
next inst.
jr $31
Stack
 Load the address of the next instruction
to “$ra” register
 Jump (set PC register) to the first
instruction in the called subroutine
Subcall/010

11. for a recursive structure
CS 286 – Project #2
Subroutine Calls in MIPS R3000
main:
This structure is
a necessary condition
for a recursive structure
xyz:
xyz:
jr $ra
xyz:
jr $ra
jal xyz
save $s0
save $s0
save $s0
save $ra
save $ra
save $ra
$ra
(next instruction)
jal xyz
jal xyz
$ra
next inst.
$ra
next inst.
restore $ra
restore $ra
restore $ra
restore $s0
restore $s0
restore $s0
jr $31
jr $ra
Stack
$ra
$s0
$ra
$s0
$ra
$s0
Subcall/011

12. CS 286 – Project #2 Subroutine Calls in MIPS R3000 Stack    main:
xyz:
jr $ra
xyz:
jr $ra
xyz:
jr $ra
jal xyz
jal xyz
jal xyz
  
jal xyz
jr $31
Stack
Stack Frame for “xyz(3)”
Stack Frame for “xyz(2)”
Stack Frame for “xyz(1)”
Stack Frame for “main”
Subcall/012

13. CS 286 – Project #2 Example (3) Set up the stack frame
xyz:
# create stack frame
subu $sp, $sp, 32
# save $ra register
sw $ra, 0($sp)
# save other GPR’s
sw $s0, 4($sp)
sw $t0, 8($sp)
   
Set up the stack frame
and save the registers
   
Subcall/013

14. CS 286 – Project #2 Example (4)     Restore the registers
xyz:
   
# restore registers
lw $s0, 4($sp)
lw $t0, 8($sp)
# restore $ra for the caller
lw $ra, 0($sp)
# restore the caller's stack pointer
addu $sp, $sp, 32
Restore the registers
and release the stack
frame
jr $ra
Subcall/014

15. CS 286 – Project #2 Subroutine Calls in MIPS R3000 $sp = $sp-32
High Address
Low Address
$sp = $sp-32
$fp = $sp+28
$sp
-32
+28
$fp
8 registers
$sp
Stack Frame
$fp
4 bytes
Subcall/015