Understanding Binary Numbers and Number Representations in Computer Systems

1. Chapter Three

2. Numbers • Computer words are composed of bits, thus words can be represented as binary numbers • Binary numbers (base 2) 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001... decimal: 0...2n-1 • Of course it gets more complicated: numbers are finite (overflow) fractions and real numbers negative numbers e.g., no MIPS subi instruction; addi can add a negative number • How do we represent negative numbers? i.e., which bit patterns will represent which numbers?

3. Possible Representations • Sign Magnitude: One's Complement Two's Complement 000 = +0 000 = +0 000 = +0 001 = +1 001 = +1 001 = +1 010 = +2 010 = +2 010 = +2 011 = +3 011 = +3 011 = +3 100 = -0 100 = -3 100 = -4 101 = -1 101 = -2 101 = -3 110 = -2 110 = -1 110 = -2 111 = -3 111 = -0 111 = -1 • Issues: balance, number of zeros, ease of operations • Which one is best? Why?

4. The MIPS word is 32 bits long, so we can represent 232 different 32-bit patterns. It is natural to let these combinations represent the numbers from 0 to 232 -1. • 0000 0000 0000 0000 0000 0000 0000 0000two = 0ten0000 0000 0000 0000 0000 0000 0000 0001two = + 1ten0000 0000 0000 0000 0000 0000 0000 0010two = + 2ten...1111 1111 1111 1111 1111 1111 1111 1101two = 4,294,967,293ten1111 1111 1111 1111 1111 1111 1111 1110two = 4,294,967,294ten1111 1111 1111 1111 1111 1111 1111 1111two = 4,294,967,295ten

5. MIPS • 32 bit signed numbers:0000 0000 0000 0000 0000 0000 0000 0000two = 0ten0000 0000 0000 0000 0000 0000 0000 0001two = + 1ten0000 0000 0000 0000 0000 0000 0000 0010two = + 2ten...0111 1111 1111 1111 1111 1111 1111 1110two = + 2,147,483,646ten0111 1111 1111 1111 1111 1111 1111 1111two = + 2,147,483,647ten1000 0000 0000 0000 0000 0000 0000 0000two = – 2,147,483,648ten1000 0000 0000 0000 0000 0000 0000 0001two = – 2,147,483,647ten1000 0000 0000 0000 0000 0000 0000 0010two = – 2,147,483,646ten...1111 1111 1111 1111 1111 1111 1111 1101two = – 3ten1111 1111 1111 1111 1111 1111 1111 1110two = – 2ten1111 1111 1111 1111 1111 1111 1111 1111two = – 1ten

6. Two's Complement Operations • Two’s complement representation has the advantage that all negative numbers have a 1 in the most significant bit. Consequently hardware needs to test only this bit to see if a number is positive or negative (with 0 considered positive). This bit is often called the sign bit. • What is the decimal value of this 32 bit two’s complement number? 1111 1111 1111 1111 1111 1111 1111 1100 Ans : (1X -231) + (1 X 230) +………………..+ 22 + 0 + 0 = -4ten • Negating a two's complement number: invert all bits and add 1 • remember: “negate” and “invert” are quite different! • Converting n bit numbers into numbers represented with more than n bits: • MIPS 16 bit immediate gets converted to 32 bits for arithmetic • copy the most significant bit (the sign bit) into the other bits 0010 -> 0000 0010 1010 -> 1111 1010 • "sign extension" (lbu vs. lb), (lhu vs lh), (slt vs sltu), (slti vs sltiu)

7. The function of a signed load is to copy the sign repeatedly to fill the rest of the register. Unsigned loads simply fill with 0s to the left of the data • Suppose $s0 has the binary number 1111 1111 1111 1111 1111 1111 1111 1111two and register $s1 has the binary number 0000 0000 0000 0000 0000 0000 0000 0001two What are the values of register $t0 and $t1 after these two instructions? slt $t0, $s0, $s1 sltu $t1, $s0, $s1

8. Addition & Subtraction • Just like in grade school (carry/borrow 1s) 0111 0111 0110+ 0110 - 0110 - 0101 • Two's complement operations easy • subtraction using addition of negative numbers 0111 + 1010 • Overflow (result too large for finite computer word): i.e the result of an operation cannot be represented with the available hardware. • e.g., adding two n-bit numbers does not yield an n-bit number

9. Detecting Overflow • In 2’s complement, an addition or subtraction overflow occurs when the result overflows in to the sign bit. • 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

10. Effects of Overflow • An exception (interrupt) occurs • Control jumps to predefined address for exception • Interrupted address is saved for possible resumption • Details based on software system / language • example: flight control vs. homework assignment • Don't always want to detect overflow — new MIPS instructions: addu, addiu, subu note: addiu still sign-extends! note: sltu, sltiu for unsigned comparisons

11. Multiplication • More complicated than addition • accomplished via shifting and addition • More time and more area • 1101 (multiplicand)x1011 (multiplier)

12. Multiplication: Implementation Datapath Control

13. Final Version • Multiplier starts in right half of product What goes here?

14. MIPS provides a separate pair of 32 bit registers to contain the 64 bit product, called Hi and Lo. • To produce a properly signed or unsigned product, MIPS has two instructions • Mult (multiply) and multu (multiply unsigned) • To fetch the integer 32 bit product, the programmer uses mflo (move from lo ) and mfhi (move from hi)

15. Faster multiplications are possible by essentially providing one 32 bit adder for each bit of the multiplier: one input is the multiplicand ANDed with a multiplier bit and the other is the output of a prior adder.