CODEWITHSUNDEEP
cpointersstackmipsstack-pointer
CodeFreak
CodeFreak

Reputation: 110

How to manipulate MIPS code and use stack pointers?

So, I recently made a code to count the number of binary 1's in C-code and in MIPS code. I did so in C by using a remainder value and increment a count_one variable. In MIPS, I did the same program but I shifted the bytes of the number until it counted all of the 1's. Howver, I want to try to learn how to use pointers but I can not seem to grasp the concept. My MIPS code is as follows:

.data
  prompt: .asciiz "Enter a integer: "

.text
  li $v0,4
  la $a0, prompt
  syscall

  li $v0,5
  syscall
  move $s0,$v0
  j count

count:
  beq $s0,0, exit
  andi $t0,$s0,1
  add $t1,$t1,$t0
  srl $s0,$s0,1
  j count

exit:
  move $a0,$t1
  la $v0,1
  syscall
  li $v0,10
  syscall

I get this complete MIPS code but I am unsure on how pointers completely work in MIPS and after reading I still don't understand. Any advice on how to implement pointers?

Upvotes: 1

Views: 2546

Answers (2)

Adam
Adam

Reputation: 846

Here is an example code that translate the flowing C code in MIPS. In order to save and restore the preserved registers it make some place on the stack then using sw and lw to save and restore those registers.

int leaf_example(int g, int h, int i, int j) {
    int f;
    f = (g + h) - (i + j);
    return f;
}    



.text
 main:

        addi $a0,$0,1       #argument 0 = 1
        addi $a1,$0,2       #argument 1 = 2
        addi $a2,$0,3       #argument 2 = 3
        addi $a3,$0,4       #argument 3 = 4
        jal  leaf           # call function leaf
        add  $s0,$v0,$zero  # return value

        li $v0,10
        syscall


    leaf:
        addi $sp, $sp, -12  #adjust stack to make room for 3 items
        sw $s0, 8($sp)      #save register $s0 for use in memory location 8
        sw $t0, 4($sp)      #save register $t0 for use in memory location 4
        sw $t1, 0($sp)      #save register $t1 for use in memory location 0

        add $t0, $a0, $a1   #register $t0 contains $a0 + $a1
        add $t1, $a2, $a3   #register $t1 contains $a2 + $a3
        sub $s0, $t0, $t1   #$t0 = $t0 - $t1 -> $t0 = ($a0 + $a1) - ($a2 + $a3)

        add $v0, $s0, $zero #copy $s0 to return register $v0

        #Before returning, we restore three original values 
        #of registers we pushed onto stack by popping them
        lw $t1, 0($sp)      #restore register $t1 for caller
        lw $t0, 4($sp)      #restore register $t0 for caller
        lw $s0, 8($sp)      #restore register $s0 for caller
        addi $sp, $sp, 12    #adjust stack to delete 3 items

        jr $ra         #jump back to calling routine

Upvotes: 1

oklas
oklas

Reputation: 8220

Most common the command determines the mind of data.

For example in pseudo code

inc $a0

this command increment data in register $a0 work with it as with number

lw $s1, 0($a0)

this command load data from memory pointed by register $a0 work with it as with pointer

Upvotes: 1

Related Questions