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Home , Positional notation

1/21/2019

CS 362: Design Positional Lecture 3: System Review • The meaning of a digit depends on its position in a number.

• A number, written as the sequence of digits

dndn‐1…d2d1d0 in base b represents the value d * bn + d * bn‐1 + ... + d * b2 + d * b1 + d * b0 Cynthia Taylor n n‐1 2 1 0 University of Illinois at Chicago • For a base b, digits will range from 0 to b‐1 September 5th, 2017

Consider 101 1015 = ?

• In base 10, it represents the number 101 (one A. 26 hundred one) = B. 51

• In base 2, 1012 = C. 126

• In base 8, 1018 = D. 130

101‐3=? Binary: Base 2

A. ‐10 • Used by

B. 8 • A number, written as the sequence of digits

dndn‐1…d2d1d0 where d is in {0,1}, represents C. 10 the value n n‐1 2 1 0 dn * 2 + dn‐1 * 2 + ... + d2 * 2 + d1 * 2 + d0 * 2 D. ‐30

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Binary to Decimal to Binary

• Use expansion • Repeatedly divide by 2, recording the remainders. • The remainders form the binary digits of the number. 101102 = • Converting 25 to binary

3410=?2 : Base 16

A. 010001 • Like binary, but shorter! • Each digit is a “nibble”, or half a byte • Indicated by prefacing number with 0x B. 010010

• A number, written as the sequence of digits dndn‐ C. 100010 1…d2d1d0 where d is in {0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F}, represents the value D. 111110 n n‐1 2 1 0 dn * 16 + dn‐1 * 16 + ... + d2 * 16 + d1 * 16 + d0 * 16

Hexadecimal to binary 23C16 = ?2

• Each hex digit corresponds directly to four A. 10001 binary digits B. 10111110

• 35AE16 = C. 1000111100

D. 1000011011000

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If every hex digit corresponds to 4 binary digits, how many binary digits Negative : Two’s Complement does an digit correspond to? • Flip all the bits and add 1 A. 2

• For n bits, in unsigned version ‐x = 2n –x B. 3

C. 4

D. 5

‐6 in Two’s Complement Two’s Complement: 11012 = ?10

A. 11110110 A. ‐2

B. 11111001 B. ‐3

C. 11111010 C. ‐4

D. 11111110 D. ‐5

Negating Two’s Compliment The negation of 111100012 is _____2

• To negate any number, flip all the bits and add A. 00001110 1 B. 00001111

C. 00011110

D. 01110001

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In n bits, the sum of a number and its Addition negation will be

A. 0n…00 • Use the same place‐by‐place algorithm that you use for decimal numbers, but do the in the appropriate base B. 1n0n‐1…00

• C. 1 …1 Works in a set number of bits – carry out from n 0 the most significant bit is thrown away D. It will vary

Addition and Subtraction 2A5C16 + 38BE16 = ?

• Positive and negative numbers are handled in A. 586A the same way. B. 631A • The carry out from the most significant bit is ignored. C. 6986

• To perform the subtraction A ‐ B, compute A + (two's complement of B)

111101102 + 000011002 = ?2 011110012 + 000001112 = ?

A. 00000010 A. 01000000

B. 00001100 B. 01111110

C. 11110010 C. 10000000

D. 11111110 D. 11111111

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Overflow Overflow can occur when adding

• Overflow occurs when an addition or A. Two positive numbers subtraction results in a value which cannot be represented using the number of bits B. Two negative numbers available. C. A positive and a

• In that case, the algorithms we have been D. A and B using produce incorrect results. E. All of the above

To detect overflow, we should How To Detect Overflow

A. Check if the most significant bit is a 1 • On an addition, an overflow occurs if and only if the carry into the bit differs from the B. Check if the carry‐out bit is a 1 carry out from the sign bit.

• C. Check if the carry‐out bit is different from the Overflow occurs if adding two negative most significant bit numbers produces a positive result or if adding two positive numbers produces a negative result.

Reading

• Next lecture: Electricity & Transistors – Reading: 3.1‐3.4

• First lab due: Wednesday, January 30th

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