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CS 230: Computer CS 230: Computer Organization and Organization and

Assembly LanguageAssembly LanguageAviral

ShrivastavaDepartment of Computer Science and

EngineeringSchool of Computing and Informatics

Arizona State University

Slides courtesy: Prof. Yann Hang Lee, ASU, Prof. Mary Jane Irwin, PSU, Ande Carle, UCB

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AnnouncementsAnnouncements

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CSE 230 Road mapCSE 230 Road map

• So far– Write any program in MIPS assembly

language– Convert into binary instructions

• Today– How are numbers represented– How are they added, subtracted,

multiplied divided

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NumbersNumbers• Bits are just bits (no inherent meaning)

– conventions define relationship between bits and numbers

• Unsigned Numbers– 0 - 231

• Large Numbers– Avogadro's number

• 6.0221415 × 1023 ~ 64 bits

• Signed Numbers– How to represent –ve numbers

• Fractions– How to represent 14159265.3

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Unsigned NumbersUnsigned Numbers• Representation

(d31 d30 … d2 d1 d0)2 = d31*231 + d30*230 + … d2*22 + d1 * 21 + d0*20

0000 0000 0000 0000 0000 0000 0000 0000 = 010

0000 0000 0000 0000 0000 0000 0000 0001 = 110

0000 0000 0000 0000 0000 0000 0000 0010 = 210

. . . . . . . . . .1111 1111 1111 1111 1111 1111 1111 1111 =

4,294,967,29510

• Minimum Number= 0 = (0000….0000)2

• Maximum Number= (1111….1111)2 = (1 0000….0000)2 – 1 = 232-132

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Addition of Unsigned Addition of Unsigned NumbersNumbers

• Addition 0 0 1 1 3 + 0 0 1 0 +2--------------------- 0 1 0 1 5

• What happens when– Addition results in a number that does

not fit in 32-bits• 232-1 + 232-1 = 2*232 – 2 = 233 -2 -- needs

33 bits

– Overflow

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Subtraction of Unsigned Subtraction of Unsigned NumbersNumbers

• Subtraction 0 0 1 1 3 - 0 0 1 0 -2--------------------- 0 0 0 1 1

• What happens when– You subtract greater number from

smaller number• Need –ve numbers

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Signed NumbersSigned Numbers• One simple strategy

– Sign Magnitude Representation• Leftmost bit is sign bit• Rest 31-bits are unsigned

– Representation• (d31 d30 … d2 d1 d0)2 = (-1)*d31 + d30*230 + … d2*22 + d1 * 21 +

d0*20

– Number Range = -(231-1), …,-1,-0, +0,+1, …, +(231-1)• 2 zero’s

– How to find –ve of a number• Just change the sign-bit

– Addition & Subtraction• Add/Sub the 31-bits, and change the sign bit logically• Need a seamless way to perform these very frequent

operations

31-bit magnitude1-bit sign

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2s Complement 2s Complement RepresentationRepresentation

• Representation• (d31 d30 … d2 d1 d0)2 = d31*(-2)31 + d30*230 + … d2*22 + d1 * 21 + d0*20

• Examples

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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Negate 2's Complement Negate 2's Complement NumberNumber

• Negating a two's complement number: invert all bits and

add 1

– remember: “negate” and “invert” are quite different!• 0000 0000 0000 0000 0000 0000 0000 0010two = + 2ten

• 1111 1111 1111 1111 1111 1111 1111 1110two = – 2ten

• Converting n bit numbers into numbers with more than n

bits:

– MIPS 16 bit immediate gets converted to 32 bits for arithmetic

• "sign extension"

– copy the most significant bit (the sign bit) into the other bits

0010 -> 0000 0010

1010 -> 1111 1010

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Add 2’s Complement Add 2’s Complement NumbersNumbers

• Just like unsigned numbers7 + 6 = 13

-4 + -5 = -9

0 1 1 10 1 1 0+1 1 0 1

11

1 1 0 01 0 1 1+0 1 1 1

1

00

0

11

1

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Subtract 2’s Complement Subtract 2’s Complement NumbersNumbers

• A – B = A + (-B) = A + (!B + 1)• Just like unsigned numbers

6 – 7 = 6 + (~7 + 1) = -1

-3 – 5 = -3 + (~(5)+1) = -81 1 0 10 1 0 1-1 0 0 0

0 1 1 00 1 1 1-1 1 1 1

001

0 1 1 01 0 0 1+1 1 1 1

011

101

1 1 0 11 0 1 1+1 0 0 0

111

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OverflowOverflow• When result of operation too large to fit in 32-bits

7 – (-13) = 20

• Detect Overflows– Positive + Positive -> Negative– Negative + Negative -> Positive– Positive – Negative -> Negative– Negative – Positive -> Positive

• Consider the operations A + B, and A – B– Can overflow occur if B is 0 ?– Can overflow occur if A is 0 ?

0 1 1 10 0 1 1-

01

0 1 1 11 1 0 1+0 1 0 0

001

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Effects of OverflowEffects 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

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Yoda says…Yoda says…

• Luke: I can’t believe it. • Yoda: That is why you fail