Arithmetic%20I%20CPSC%20350

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Arithmetic ICPSC 350

• E. J. Kim

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Any Questions?
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What happened so far?

• We learned the basics of the MIPS assembly
  language
• We briefly touched upon the translation to
  machine language
• We formulated our goal, namely the implementation
  of a MIPS processor.

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Pipelined MIPS Processor
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Welcome to the Future!

• The execution of machine instructions can follow,
  for example, the steps
• Instruction fetch
• Instruction decode and register read
• Execute opn. or calculate an address
• Access operand in data memory
• Write the result into a register

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Pipelined MIPS Processor
We concentrate first on the arithmetic-logic unit
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The Arithmetic-Logic Unit

• Arithmetic (addition and subtraction)
• we need to know number representations
• there exist various interesting algorithms for
  addition and subtraction
• integer and floating point arithmetic
• Logical operations (and, or, not)

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Computer Arithmetic
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Unsigned Numbers

• 32 bits are available
• Range 0..232 -1
• 11012 232220 1310
• Upper bound 232 1 4,294,967,295

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Unsigned Numbers

• If we have n bit unsigned integers, then addition
  really means
• ab mod 2n
• For example, if n4, then
• 11012 10012 13 9 22 6 mod 16
• 11012 10012 01102

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Number representations

• What signed integer number
• representations do you know?

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Signed Numbers

• Sign-magnitude representation
• MSB represents sign, 31bits for magnitude
• Ones complement
• Use 0..231-1 for non-negative range
• Invert all bits for negative numbers
• Twos complement
• Same as ones complement except
• negative numbers are obtained by inverting all
  bits and adding 1

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Ones Complement

• Suppose we want to express -30 as an 8bit integer
  in ones complement representation.
• 30 0001 11102
• Invert the bits to obtain the negative number
• -30 1110 00012

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Twos Complement

• Suppose we want to express -30 as an 8bit integer
  in twos complement representation.
• 30 0001 11102
• Invert the bits to obtain the negative number
• 1110 00012
• Add one
• -30 1110 00102

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Advantages and Disadvantages

• sign-magnitude representation
• ones complement representation
• twos complement representation

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Signed Numbers (3bits)
sign magnitude ones complement twos complement
0002 0 0002 0 0002 0
0012 1 0012 1 0012 1
0102 2 0102 2 0102 2
0112 3 0112 3 0112 3
1002 -0 1002 -3 1002 -4
1012 -1 1012 -2 1012 -3
1102 -2 1102 -1 1102 -2
1112 -3 1112 -0 1112 -1
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Twos complement

• The unsigned sum of an n-bit number and its
  negative yields?
• Example with 3 bits
• 0112
• 1012
• 10002 2n gt negate(x) 2n-x
• Explain ones complement

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MIPS 32bit signed numbers

• 0000 0000 0000 0000 0000 0000 0000 0000two 0ten
• 0000 0000 0000 0000 0000 0000 0000 0001two
  1ten
• 0000 0000 0000 0000 0000 0000 0000 0010two
  2ten
• ...
• 0111 1111 1111 1111 1111 1111 1111 1110two
  2,147,483,646ten
• 0111 1111 1111 1111 1111 1111 1111 1111two
  2,147,483,647ten
• 1000 0000 0000 0000 0000 0000 0000 0000two
  2,147,483,648ten
• 1000 0000 0000 0000 0000 0000 0000 0001two
  2,147,483,647ten
• 1000 0000 0000 0000 0000 0000 0000 0010two
  2,147,483,646ten
• ...
• 1111 1111 1111 1111 1111 1111 1111 1101two
  3ten
• 1111 1111 1111 1111 1111 1111 1111 1110two
  2ten
• 1111 1111 1111 1111 1111 1111 1111 1111two
  1ten

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Conversions

• How do you convert an n-bit number
• into a 2n-bit number?
• (Assume twos complement representation)

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Conversions

• Suppose that you have 3bit twos complement
  number
• 1012 -3
• Convert into a 6bit twos complement number
• 1111012 -3
• Replicate most significant bit!

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Comparisons

• What can go wrong if you accidentally
• compare unsigned with signed numbers?

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Comparisons for unsigned

• Register s0
• 1111 1111 1111 1111 1111 1111 1111 1111
• Register s1
• 0000 0000 0000 0000 0000 0000 0000 0001
• Compare registers (set less than)
• slt t0, s0, s1 true, since 1 lt 1
• sltu t1, s0, s1 false, since 232-1gt1

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Addition Subtraction

• Just like in grade school (carry/borrow 1s)
  0111 0111 0110  0110 - 0110 - 0101
• 1101 0001 0001
• Two's complement operations are simple
• subtraction using addition of negative numbers
  0111 7  1010 -6 0001

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MIPS instructions

• lb loads a byte and stores the sign-extended
  version in a word.
• lbu loads a byte and stores it in a word
• Which of these two is typically used to process
  characters?

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Overflow

• Overflow means that the result is too large for
• a finite computer word. For instance, adding two
  n-bit numbers does not yield an n-bit number.
• Suppose we add two 3-bit numbers 111  
  001
• 1000

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Detecting Overflow

• No overflow when adding a positive and a negative
  number
• No overflow when signs are the same for
  subtraction
• Overflow occurs when the value affects the sign
• overflow when adding two positives yields a
  negative
• or, adding two negatives gives a positive
• or, subtract a negative from a positive and get a
  negative
• or, subtract a positive from a negative and get a
  positive

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Detecting Overflow
Operation Operand A Operand B Overflow if result
AB gt0 gt0 lt0
AB lt0 lt0 gt0
A-B gt0 lt0 lt0
A-B lt0 gt0 gt0
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Effects of Overflow

• An exception (interrupt) occurs
• Control jumps to predefined address for exception
• Interrupted address is saved for possible
  resumption
• Don't always want to detect overflow
• MIPS instructions addu, addiu, subu note
  addiu still sign-extends!

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Building an Arithmetic Logic Unit
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Logic Gates AND
a
c
a b c 0 0 0 0 1 0 1 0 0 1 1 1
b
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Logic Gates OR
a
c
b
a b c 0 0 0 0 1 1 1 0 1 1 1 1
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An ALU (arithmetic logic unit)

• Let's build an ALU to support the andi and ori
  instructions
• Selection of operation 0 and, 1 or
• we'll just build a 1 bit ALU, and use 32 of
  them
• Possible Implementation (sum-of-products)

a
b
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The Multiplexor

• Selects one of the inputs to be the output,
  based on a control input
• Build (and/or) ALU using a MUX

note it is called a 2-input mux even
though it has 3 inputs!
0
1
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Different Implementations

• Not easy to decide the best way to build
  something
• Don't want too many inputs to a single gate
• for our purposes, ease of comprehension is
  important
• Dont want to have to go through too many gates
• Let's look at a 1-bit ALU for addition

cout a b a cin b cin sum a xor b xor cin
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Different Implementations

• How could we build a 1-bit ALU for add, and, and
  or?
• How could we build a 32-bit ALU?

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Building a 32 bit ALU
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What about subtraction (a b) ?

• Two's complement approach just negate b and
  add.
• How do we negate?
• A solution