Question

1. Convert the following pseudo code routine into MIPS assembler language:

- Set register $t0 = 1

- In four separate instructions add 2, then 3, then 4, then 5 into register $t0

$t0 = $t0 + 2 + 3 + 4 + 5

- Copy the result from register $t0 into register $t1

2. Convert the following into MIPS assembler language:

- Set register $t0 = 0

- Initialize the register $t1 = 10

- Use register $t2 as a counter, and create a loop that repeats 5 times

- Each time through the loop, add register $t1 to $t0

- And add 10 (decimal) to register $t1

- As a result, $t0 should have the sum of the equation: 10 + 20 + 30 + 40 + 50

3. Generate an instruction sequence that will result in an arithmetic overflow. Do not use an endless loop, but rather create a specific code sequence.

Please add comments!!

Answer 1

1.
addi $t0 $zero 0x1 //set register $t0 =1
addi $t0 $t0 0x2 //add 2 to register $t0
addi $t0 $t0 0x3 //add 3 to register $t0
addi $t0 $t0 0x4 //add 4 to register $t0
addi $t0 $t0 0x5 //add 5 to register $t0
add $t1 $t0 $zero //$t1 = $t0+0

2.
addi $t0 $zero 0x0 //set register $t0 =0
addi $t1 $zero 0xA //set register $t1 =10
addi $t2 $zero 0x0 //set register $t2 =0

Loop:
   slti $t3 $t2 0x5 //set $t3 = 1 if $t2<5
   beq $t3 $zero end //if $t3 is not set branch to end
   add $t0 $t0 $t1 //$t0 = $t0+$t1
   addi $t1 $t1 0xA //$t1 = $t1+10
   addi $t2 $t2 0x1 //$t2 = $t2+1
j Loop //jump to Loop

end:

3.
MIPS integers are 32-bit, and since you'll be using signed integers, the maximum value is 2^31-1 . Thus any addition which results in a number larger than this should produce an overflow , e.g if you try to add 1 to 2147483647:

# Load 2147483647 into $s1
lui $s0, 32767
ori $s1, $s0, 65535

# Add 1 to $s1 and store in $s2. This should produce an overflow
addi $s2, $s1, 1