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If I have the following code in C

A[1] = 2;

Where the starting address A[0] is $s0.

addi $t0, $s0, 1 #t0->A[0]
lw $t1, 0($t0) #t1->A[1]
addi $t1, $t1, 2

This is where I am a bit confused. Is it necessary to add:

sw $t1, 0($t1)

Or is the code fine as is? I think it is fine because, I am adding 2 to the contents of $t1, which is effectively, A[1] = 2.

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    \$\begingroup\$ Isn't this more of a question for Stackoverflow? \$\endgroup\$
    – Oskar Skog
    Mar 5, 2018 at 15:50
  • \$\begingroup\$ In the C code, you are storing one literal into an array. In the assembly code you're adding a literal into a variable whose value is loaded from the array. Without the store instruction, this becomes x = A[1] + 2;, and with the store instruction this becomes A[1] += 2;. Neither of which is the C code you've posted. \$\endgroup\$
    – Oskar Skog
    Mar 5, 2018 at 15:52
  • \$\begingroup\$ The comments in the assembly code seem quite confusing, are you sure they say what you intended them to say? \$\endgroup\$
    – Oskar Skog
    Mar 5, 2018 at 15:59
  • \$\begingroup\$ I also think there is a typo in the store instruction, you're storing to 0($t1) instead of 0($t0) \$\endgroup\$
    – Oskar Skog
    Mar 5, 2018 at 16:02
  • \$\begingroup\$ You must have been quite confused by something. Why did you load the value of A[1] into a register? \$\endgroup\$
    – Oskar Skog
    Mar 5, 2018 at 16:03

1 Answer 1

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If &A is stored in $s0, then A[1] = 2 compiles to

ori $t0, $0, 2      # Move 2 into $t0
sw $t0, 4($s0)      # Store $t0 into $s0 + 4 (because ints are 4 bytes)

As mentioned in the comments, your code doesn't make a lot of sense because you seem to be doing something resembling x = A[1] +2. You have to use a sw because the C code stores a value to the array.

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