The MIPS AUL

• This is material from B.5
• We need an ALU to add and subtract.
• I know you have built a ripple carry adder, but we will take a different approach.
  • They observe that a half adder can be reduced to
    • c_out = a·c_in + b·c_in + a·b
    • sum = abc_in + abc_in+ abc_in + abc_in
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  • This is a bit strange, but they proceed to build a 1 bit ALU.
    • We need an and and an or.
    • Which then requires us to provide a control line to select which output we are interested in.
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    • So we can hook these together to form a (4 bit) ALU.
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    • Show how to and, or , add two numbers with this circuit.
  • This will do for addition but what about subtraction.
    • Subtraction is just addition by a negative of the number.
    • We can take the negative of the number by inverting and adding 1.
    • So the following change is in order:
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    • Do a subtract using this circuit.
      • Set B_Invert to 1
      • Set Carry_In to 1.
    • We need a bitwise NOR, which we can form by
      • a+b = a · b
      • So we need to an an a_invert line, and MUX.
      • 
    • Next we need to deal with the branch conditions.
      • stl $s1, $s2, $s3
      • R[rd] = (R[rs] < R[rt]) ? 0: 1
      • If R[rs] < R[rt] then R[rd] is set to 1.
      • We can just subtract R[rt] form R[rs]
        • -3 -(-2) = -1 (the first is smaller)
        • 7-8 = -1 (the first is smaller)
        • -5 - (-6) = 1 (the second is smaller)
        • 4-3 = 1 (the second is smaller)

        • li	$t0, n
          li	$t1, m
          slt	$t2, $t1, $t2
          

        • So the sign of the result is 1 is the value we wish to store in the destination register.
        • So this is the value we would like to be able to produce from the ALU.
        • We will create a new 1 bit alu, ALU0
        • Add a new operation (3)
        • It is the same as the other ALUs, but
          • Add a new input to ALU0 type called less.
          • This is piped directly to the third output port on the MUX.
          • This needs to be fed the MSB of the sum.
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        • So we need a new 1 bit ALU for the end of the computation as well.
          • Add an output that is the result of the MSB of a sum.
          • This gets wired to the less port of ALU0.
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        • The final is here
    • One last bit... BEQ, BNE
      • This is a subtraction.
      • If A-B = 0, then there is an equal.
      • So we need a way to see if all the bits are 0.
      • We can just do a subtraction and put all the result bits into a big OR.
    • How do we AND two numbers?
      • Place the numbers on A and B
      • Set the Operation to be 0
      • Set A_invert and B_invert to 0
      • Set Carry_In to be 0
    • How do we OR two numbers?
      • Place the numbers on A and B
      • Set the Operation to be 1
      • Set A_invert and B_invert to 0
      • Set Carry_In to be 0
    • How do we NOR two numbers?
      • Place the numbers on A and B
      • Set the Operation to be 0
      • Set A_invert and B_invert to 1
      • Set Carry_In to be 0
    • How do we add two numbers?
      • Place the numbers on A and B
      • Set the Operation to be 2
      • Set A_invert and B_invert to be 0
      • Set Carry_In to be 0
    • How do we subtract two numbers?
      • Place the numbers on A and B
      • Set the Operation to be 2
      • Set A_invert to 0
      • Set B_invert to be 1
      • Set Carry_In to be 1
    • How do we slt?
      • Place the numbers on A and B
      • Set the Operation to be 3
      • Set A_invert to 0
      • Set B_invert to be 1
      • Set Carry_In to be 1