how to store in array in MIPS assembly language?

Question

A[k++] = A[k] + A[k]
# i-> $s0, k-> $s1 , base of A[] ->$s2

i tried the following but can't figure out how to store back in array A[k++]....

sll $t0 , $s1 , 2
add $t0 , $t0 , $s2
lw $t1, 0($t0)
add $s1, $t1 , $t1 # i added A[k] + A[k]

How about using sw to store it since you use lw to load it, or is there some other problem? — Sami Kuhmonen
– Sami Kuhmonen, Commented Feb 12, 2017 at 7:01
# shift left logical sll $t1 , $s1 , 2 # add 1 addi $t1 , $t1 , 1 # now add with the base add $t1, $t1 , $s2 # finally store back in array sw $s1 , 0($t1) will it work ? — Mahadi
– Mahadi, Commented Feb 12, 2017 at 7:47
You already have the address in $t0, no need to calculate it again. Just store. — Sami Kuhmonen
– Sami Kuhmonen, Commented Feb 12, 2017 at 7:54
the element memory location are calculated as follow: array variable base address + 4bytes index in case you have an array with or words... otherwise you should index with the element size, for string will be byte. — A. Herlas
– A. Herlas, Commented Oct 12, 2022 at 9:58

Gustavo Mello · Accepted Answer · 2022-09-25 17:59:58Z

1

In order to iterate throught arrays in MIPS, you need to add 4 bytes for each position of the array. Example:

.data
  myArray: .store 12 # array with length of 3 (3 x 4 bytes)
.text
  li $t0, 0
  li $s0, 1
  sw $s0, myArray($t0) # myArray[0] = 1
  addi $t0, 4
  sw $s0, myArray($t0) # myArray[1] = 1
  addi $t0, 4
  sw $s0, myArray($t0) # myArray[2] = 1

edited Sep 25, 2022 at 17:59

answered Sep 25, 2022 at 17:59

Gustavo Mello

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