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Integers Today ! Numeric Encodings ! Programming Implications ! Basic operations ! Programming Implications Next time ! Floats Tuesday, September 27, 2011 Checkpoint Tuesday, September 27, 2011 Encoding integers in binary Positive integers,

  1. Integers Today ! Numeric Encodings ! Programming Implications ! Basic operations ! Programming Implications Next time ! Floats Tuesday, September 27, 2011

  2. Checkpoint Tuesday, September 27, 2011

  3. Encoding integers in binary Positive integers, easy binary to unsigned What about negative integers? 3 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  4. Encoding integers in binary Idea #1: sign bit – use 1 in the most significant (leftmost) bit like a minus sign • 3 = 0011, -3 = 1011 – intuitive, but simple arithmetic is complicated • 5 + -3 = 0101 + 1011 = a miracle occurs = 0010 Idea #2: ones' complement – flip all bits for negatives • 3 = 0011, -3 = 1100 – addition not too bad (just add and then add carry bit if any) • 5 + -3 = 0101 + 1100 = 0001 + 1 (carry) = 0010 4 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  5. Encoding integers Both ideas lead to two representations of zero, positive and negative: – sign bit: 0000 and 1000 – ones' complement: 0000 1111 – 5 + -5 = 0101 + 1010 = 1111 = -0 5 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  6. Encoding integers Idea #3: Two’s complement – Informal encoding view: – To encode –N, encode N, flip all bits, add 1 • 5 = 0101, • -5 = 1010 + 1 = 1011 – More formally, given w bits [x w-1 , x w-2 , …, x 1 , x 0 ], • N = -(2 w -1 )* x w-1 + ! 2 i * x i for i from 0 to w -2 • 1011 = -2 3 + 3 = -8 + 3 = -5 Addition is now simple: always add, ignore overflow – 5 + -5 = 0101 + 1011 = 0000 Only one zero (why?) Significant bit still serves as sign bit 6 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  7. Encoding integers Unsigned Two ʼ s Complement Sign Bit C short 2 bytes long short int x = 15213; short int y = -15213; 7 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  8. Encoding example x = 15213: 00111011 01101101 y = -15213: 11000100 10010011 8 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  9. Numeric ranges Unsigned Values Two’s Complement Values – Tmin = –2 w–1 – Umin = 0 • 100…0 • 000…0 – TMax = 2 w–1 – 1 – UMax = 2 w -1 • 011…1 • 111…1 Values for W = 16 9 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  10. Values for other word sizes Observations C constants – #include <limits.h> – | TMin | = | TMax | + 1 • Asymmetric range – Declares – UMax = 2 * TMax + 1 • ULONG_MAX • INT_MAX, INT_MIN • LONG_MAX, LONG_MIN – Values platform-specific 10 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  11. Unsigned & signed numeric values Equivalence X B2U( X ) B2T( X ) – Same encodings for 0000 0 0 nonnegative values 0001 1 1 0010 2 2 Uniqueness (bijections) 0011 3 3 – Every bit pattern represents 0100 4 4 unique integer value 0101 5 5 – Each representable integer has 0110 6 6 unique bit encoding 0111 7 7 ! Can invert mappings 1000 8 –8 1001 9 –7 – U2B( x ) = B2U -1 ( x ) 1010 10 –6 • Bit pattern for unsigned integer 1011 11 –5 – T2B( x ) = B2T -1 ( x ) 1100 12 –4 • Bit pattern for two’s comp 1101 13 –3 integer 1110 14 –2 1111 15 –1 11 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  12. Casting signed to unsigned C allows conversions from signed to unsigned short int x = 15213; unsigned short int ux = (unsigned short) x; short int y = -15213; unsigned short int uy = (unsigned short) y; Resulting value – No change in bit representation – Non-negative values unchanged • ux = 15213 – Negative values change into (large) positive values • uy = 50323 12 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  13. Relation between signed & unsigned Casting from signed to unsigned Unsigned Two ʼ s Complement T2U T2B B2U x ux X Maintain same bit pattern Consider B2U and B2T equations and a bit pattern X; compute B2U(X) – B2T(X) weighted sum of for bits from 0 to w – 2 cancel each other If we let 13 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  14. Relation between signed & unsigned ux = x + 2 16 = -15213 + 65536 14 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  15. Conversion - graphically 2’s Comp. " Unsigned – Ordering inversion UMax 1111 – Negative " Big positive UMax – 1 TMax + 1 Unsigned 0111 TMax TMax Range 2 ʼ s Comp. 0000 0 0 Range –1 –2 1000 TMin 15 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  16. Signed and unsigned in C Constants – By default are considered to be signed integers – Unsigned if have “U” as suffix 0U, 4294967259U Casting – Explicit casting bet/ signed & unsigned same as U2T and T2U int tx, ty; unsigned ux, uy; tx = (int) ux; uy = (unsigned) ty; – Implicit casting tx = ux; uy = ty; – Mixed expressions – cast to unsigned first tx + ux; uy < ty; 17 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  17. Sign extension Task: – Given w-bit signed integer x – Convert it to w+k-bit integer with same value Rule: – Make k copies of sign bit: – X’= x w–1 ,…, x w–1 , x w–1 , x w–2 ,…, x 0 w k copies of MSB • • • X • • • • • • • • • X # w k 19 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  18. Sign extension example Converting from smaller to larger integer data type C automatically performs sign extension short int x = 15213; int ix = (int) x; short int y = -15213; int iy = (int) y; Decimal Hex Binary 15213 x 3B 6D 00111011 01101101 15213 00 00 3B 6D ix 00000000 00000000 00111011 01101101 -15213 y C4 93 11000100 10010011 -15213 FF FF C4 93 iy 11111111 11111111 11000100 10010011 20 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  19. Justification for sign extension Prove correctness by induction on k – Induction Step: extending by single bit maintains value w - • • • X X # - + • • • w +1 – Key observation: –2 w +2 w–1 = –2 w–1 = – Look at weight of upper bits: • X –2 w–1 x w–1 • X’ –2 w x w–1 + 2 w–1 x w–1 = –2 w–1 x w–1 21 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  20. Why should I use unsigned? Don’t use just because number nonzero – C compilers on some machines generate less efficient code – Easy to make mistakes (e.g., casting) – Few languages other than C supports unsigned integers Do use when need extra bit’s worth of range – Working right up to limit of word size 22 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  21. Checkpoint Tuesday, September 27, 2011

  22. Negating with complement & increment Claim: Following holds for 2’s complement – ~x + 1 == -x Complement – Observation: ~x + x == 1111…11 2 == -1 x 1 0 0 1 1 1 0 1 + ~x 0 1 1 0 0 0 1 0 -1 1 1 1 1 1 1 1 1 Increment – ~x + x + (-x + 1) == -1 + (-x + 1) – ~x + 1 == -x 23 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  23. Comp. & incr. examples x = 15213 0 24 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  24. Unsigned addition Standard addition function – Ignores carry output Implements modular arithmetic – s = UAdd w (u , v) = u + v mod 2 w • • • u Operands: w bits + • • • v True Sum: w +1 bits • • • u + v • • • Discard Carry: w bits UAdd w ( u , v ) 25 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  25. Visualizing integer addition Integer addition – 4-bit integers u, v – Compute true sum Add4(u , v) – Values increase linearly with u and v – Forms planar surface Integer Addition Add 4 ( u , v ) 32 24 16 8 0 1 2 3 4 5 6 7 v 8 u 9 10 11 12 13 14 15 0 26 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  26. Visualizing unsigned addition Wraps around Overflow – If true sum " 2 w – At most once UAdd 4 ( u , v ) 15 True Sum 11 2 w +1 Overflow 8 2 w 4 v 0 1 2 3 4 5 6 7 8 0 Modular Sum 9 10 11 12 13 14 15 0 u 27 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

  27. Two’s complement addition TAdd and UAdd have identical Bit-level behavior – Signed vs. unsigned addition in C: – int s, t, u, v; – s = (int) ((unsigned) u + (unsigned) v); – t = u + v – Will give s == t • • • u Operands: w bits + • • • v True Sum: w +1 bits • • • u + v • • • Discard Carry: w bits TAdd w ( u , v ) 28 EECS 213 Introduction to Computer Systems Northwestern University Tuesday, September 27, 2011

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