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MIPS code examples Making Use of a Stack Outline Assumptions for example Implementation: sprod/prod/add2 Example: factorial Assumptions for Example We suppose the ... – PowerPoint PPT presentation

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Title: Procedures%202a%20MIPS%20code%20examples


1
Procedures 2aMIPS code examples
  • Making Use of a Stack

2
Outline
  • Assumptions for example
  • Implementation sprod/prod/add2
  • Example factorial

3
Assumptions for Example
  • We suppose the following
  • x1,y1,x2,y2 stored in memory starting at location
    X
  • Procedures
  • add2 - adds together 4, 5 puts result in 6
  • prod - multiplies 4, 5 puts result in 6
  • sprod (x1 y1) (x2 y2), puts result in 8
  • 29 is the stack pointer (address of top of
    stack)
  • The stack in MIPS grows downwards
  • so that subtracting 4 from 29 makes room for a
    push
  • and adding 4 to 29 adjusts for a pop

4
  • Strategy for example
  • First examine code for simple procedures
  • Prod
  • Add2
  • Then examine code for more complex procedure
  • sprod

5
Prod
  • prod addi 29,29,-4 step 1 of push
  • sw 31,0(29) step 2 of push 31
  • mult 6,4,5
  • lw 31,0(29) step 1 of pop
  • restore return address
  • addi 29,29,4 step 2 of pop
  • adjust top of stack
  • jr 31 return

6
add2
  • add2 addi 29,29,-4 step 1 of push
  • sw 31,0(29) step 2 of push 31
  • add 6,4,5
  • lw 31,0(29) step 1 of pop
  • restore return address
  • addi 29,29,4 adjust top of stack
  • jr 31 return

7
Implementation of sprod
  • sprod lw 4,X(0) x1 into 4
  • li 5, 4
  • lw 5,X(5) y1 into 5
  • addi 29,29,-4 step 1 for push
  • sw 31,0(29) step 2 for push 31
  • return address now saved,
  • to be popped at end sprod
  • jal prod call prod, result in 6
  • add 20,0,6 copy result into 20
  • li 4, 8
  • lw 4,X(4) x2 into 4
  • li 5, 12
  • lw 5,X(5) y2 into 5
  • jal prod call prod, result in 6
  • continued

8
sprod continued
  • add 4,0,20 first result into 4
  • add 5,0,6 second result into 5
  • jal add2 puts result in 6
  • lw 31,0(29) step 1 of pop
  • restores return address
  • addi 29,29,4 step 2 of pop
  • adjust top of stack
  • jr 31 return to caller of sprod

9
Example Factorial
  • n! n?(n-1)?...?2?1
  • 4! 4?3?2?1 24
  • 4! 4?(3?2?1) 4?3!
  • n! n?(n-1)! (unless n1, in which case
    n!1)
  • Function to implement factorial
  • fact(n)
  • if n1 then return 1
  • else return nfact(n-1)

10
Factorial in MIPS (Assumptions)
  • The argument (n) is in 4
  • 8 is already initialised to 1
  • 8 will hold the result
  • We will use the pretend instructions
  • push
  • pop
  • print
  • for sake of simplicity.

11
Nearly a MIPS representation
  • MAIN li 4,4 the argument n 4
  • li 11,1 set 11 to have the value 1
  • li 8,1 initialize 8 to have the value 1
  • jal fact call fact procedure
  • print 8 no such instruction!
  • exit no such instruction - halt altogether
  • fact beq 4,11,RETURN if n1 goto RETURN
  • push(31) save on stack
  • push(4) save argument
  • addi 4,4,-1 n becomes n-1
  • jal fact recursive call to fact
  • RETURN 4 pop() no such instruction!
  • 31 pop()
  • mult 8,8,4 8 84
  • jr 31 return

12
Summary
  • jal
  • jump and link
  • saves return address in 31
  • jr 31
  • jumps to return address in register 31
  • Stack pointer used to store nested return
    addresses (29)

13
Next Time
  • How to handle arguments to procedures
  • How to write programs for the MIPS simulator SPIM
  • Answering your questions about the Coursework.
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