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So here's the question that I've been thinking over for a few hours: MIPS to C. Assume $s3 = i, $s4 = j, $s5 = @A. Below is the MIPS code: Loop: addi $s4,$s4,1 # j = j + 1? add $t1,$s3,$s3 # $t1 = 2 * i add $t1,$t1,$t1 …
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