Question

1. Translate the following code into MIPS code.

Test (int i, int j)

                        {

                        int k;

                        k = i +j +10;

                        k = Sub(k+1) + Sub (k)

                        return k;

                        }

Sub (int m)

                        {

                        int g;

                        g = m + m;

                        return g;

                        }

Assume variables k and g are assigned to registers $s0 and $s1, respectively. Note: your code should be complete.

Answer 1

Solution:

Given, variables k and g are assigned to registers $s0 and $s1, respectively.

Let us assume i, j and m are assigned to $a0, $a1, $a2 respectively.

Translation of C code to MIPS:

Test: addi $sp, $sp, -4 // adjust stack for 2 items

sw $ra, 0($sp) //save the return address

sw $a0, 0($sp) //save the argument i

sw $a1, 0($sp) //save the argument j

sw $a2, 0($sp) //save the argument m

addi $s0, $a0, $a1, 10 //store k equal to i + j + 10

add $t0, $s0, 1 //add K + 1 and store in temporary register

sub $t0, $t0, $s0 //subtract k+1 and store temporary register

add $s0, $s0, $t0 //addition of sub(K+1) and $s0

jr $s0 // return k

Exit

Sub: sw $a2, 0($sp) //save the argument m

sw $s1, 0($sp) //save the variable g

add $s1, $a2, $a2 //g = m + m

jr $s1 // return g

Please give thumbsup, if you like it. Thanks.