CODEWITHSUNDEEP
cassemblymipsstatements
yoshi105
yoshi105

Reputation: 39

Translating MIPS assembly to a C statement

addi $s7, $s7, -4
add  $s7, $s7, $s1
lw   $s0, 4($s7)

Assume integer variables i and j are in registers $s0 and $s1. Assume base address of an integer array X is in register $s7.

So far I have this:

X = X - 4
X = X + j
i = X - 4 + j

Is this correct? I'm not overly sure so just looking for confirmation.

Upvotes: 0

Views: 531

Answers (2)

gusbro
gusbro

Reputation: 22585

It is incorrect. Note that lw instruction reads a word from memory.

In C it would look something like

  //int *x;
  x--;            // addi $s7, $s7, -4 decrements pointer to x one element
  x = (int*)((char*)x + j);  // add $s7, $s7, $s1 increments the address pointed by x j elements
  i = *(x+1);     // lw $s0, 4($s7) reads the next element pointed by x

Addendum after OPs comment:

If j = $s7 is multiple of 4 (note that each integer occupies 4 bytes), then it could be rewritten in C as i = x[j/4].

Upvotes: 4

hims15
hims15

Reputation: 96

lw-->Load word

syntax:lw $t, offset($s) operation: $t = MEM[$s + offset]; advance_pc (4);

Upvotes: 0

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