Pipelined MIPS Processor - Computer Architecture - Lecture Slides, Slides of Computer Science

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Arithmetic I

CPSC 350

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.

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

Pipelined MIPS Processor

We concentrate first on the arithmetic-logic unit

Computer Arithmetic

Unsigned Numbers

• 32 bits are available
• Range 0..2 32 -
• 11012 = 2^3 +2 2 +2 0 = 13 10
• Upper bound 2 32 –1 = 4,294,967,

Number representations

What signed integer number

representations do you know?

Signed Numbers

• Sign-magnitude representation
  • MSB represents sign, 31bits for magnitude
• One’s complement
  • Use 0..2 31 -1 for non-negative range
  • Invert all bits for negative numbers
• Two’s complement
  • Same as one’s complement except
  • negative numbers are obtained by inverting all bits and adding 1

Two’s Complement

Suppose we want to express -30 as an 8bit integer in two’s complement representation.

30 = 0001 1110 (^2)

Invert the bits to obtain the negative number:

1110 0001 (^2)

Add one:

-30 = 1110 0010 (^2)

Advantages and Disadvantages

• sign-magnitude representation
• one’s complement representation
• two’s complement representation

Two’s complement

• The unsigned sum of an n-bit number and its

negative yields?

• Example with 3 bits:
  • (^011 )
  • (^101 )
  • 1000 2 = 2 n^ => negate(x) = 2n-x
• Explain one’s complement

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

Conversions

• Suppose that you have 3bit two’s complement

number

• Convert into a 6bit two’s complement number
  • 111101 2 = -
• Replicate most significant bit!

Comparisons

What can go wrong if you accidentally

compare unsigned with signed numbers?