C Puzzles The course that gives CMU its Zip! Integer Arithmetic Operations Jan. 25, Unsigned Addition. Visualizing Integer Addition

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1 1513 The corse that gies CMU its Zip! Integer Arithmetic Operations Jan. 5, 1 Topics Basic operations Addition, negation, mltiplication Programming Implications Conseqences of oerflow Using shifts to perform powerof mltiply/diide C Pzzles Taen from Exam #, CS 37, Spring 97 Assme machine with 3 bit word size, two s complement integers For each of the following C expressions, either: Arge that is tre for all argment ales Gie example where not tre Initialization int x = foo(); int y = bar(); nsigned x = x; nsigned y = y; x < ((x*) < ) x >= x & 7 == 7 (x<<3) < x > 1 x > y x < y x * x >= x > && y > x y > x >= x <= x <= x >= Operands: w bits Tre Sm: w1 bits Unsigned Addition Standard Addition Fnction Ignores carry otpt Implements Modlar Arithmetic s = (, ) = mod w Discard Carry: w bits (, ) (,) = w < w w Visalizing Integer Addition Integer Addition bit integers and Compte tre sm Add (, ) Vales increase linearly with and Forms planar srface Add (, ) Page 1

2 Visalizing Unsigned Addition Wraps Arond If tre sm At most once Tre Sm w1 Oerflow w Modlar Sm w Oerflow UAdd (, ) Mathematical Properties Modlar Addition Forms an Abelian Grop Closed nder addition (, ) w 1 Commtatie (, ) = (, ) Associatie (t, (, )) = ( (t, ), ) is additie identity (, ) = Eery element has additie inerse Let UComp w ( ) = w (, UComp w ( )) = 5 Detecting Unsigned Oerflow Tas Gien s = (, ) Determine if s = Application nsigned s,, ; s = ; Did addition oerflow? Claim Oerflow iff s < of = (s < ) Or symmetrically iff s < Proof Know that < w No Oerflow No oerflow s = = Oerflow s = w < = s w Oerflow w s Operands: w bits Two s Complement Addition Tre Sm: w1 bits TAdd and UAdd hae Identical BitLeel Behaior Signed s. nsigned addition in C: int s, t,, ; s = (int) ((nsigned) (nsigned) ); t = Will gie s == t Discard Carry: w bits TAdd w (, ) 7 Page

3 Fnctionality Tre sm reqires w1 bits Drop off MSB Treat remaining bits as s comp. integer TAdd(, ) > < NegOer < > Characterizing TAdd PosOer TAdd w (,) = w 1 w w 1 1 w w 1 w 1 Tre Sm PosOer NegOer TAdd Reslt < TMin (NegOer) w TMin w TMax w TMax w < (PosOer) Vales Visalizing s Comp. Addition bit two s comp. Range from to 7 Wraps Arond If sm w 1 Becomes negatie At most once If sm < w 1 Becomes positie At most once NegOer TAdd (, ) PosOer 9 1 Detecting s Comp. Oerflow Tas Gien s = TAdd w (, ) Determine if s = Add w (, ) Example int s,, ; s = ; Claim Oerflow iff either: w 1 w 1, <, s (NegOer),, s < (PosOer) of = (< == <) && (<!= s<); Proof PosOer NegOer Mathematical Properties of TAdd Isomorphic Algebra to UAdd TAdd w (, ) = UT( (TU( ), TU())) Since both hae identical bit patterns Two s Complement Under TAdd Forms a Grop Closed, Commtatie, Associatie, is additie identity Eery element has additie inerse Let TComp w ( ) = UT(UComp w (TU( )) TAdd w (, TComp w ( )) = TComp w () = TMin w TMin w = TMin w Easy to see that if and <, then TMin w TMax w Symmetrically if < and Other cases from analysis of TAdd 11 1 Page 3

4 Two s Complement Negation Mostly lie Integer Negation TComp() = TMin is Special Case TComp(TMin) = TMin Negation in C is Actally TComp mx = x mx = TComp(x) Comptes additie inerse for TAdd x x == Tcomp( ) w w 1 w w 1 Negating with Complement & Increment In C ~x 1 == x Complement Obseration: ~x x == == 1 x ~x Increment ~x x (x 1) == 1 (x 1) ~x 1 == x Warning: Be catios treating int s as integers OK here: We are sing grop properties of TAdd and TComp 1 x = 1513 Comp. & Incr. Examples Decimal Hex Binary x B D ~x 151 C ~x C y 1513 C TMin Decimal Hex Binary TMin 37 1 ~TMin 377 7F FF ~TMin Decimal Hex Binary ~ 1 FF FF ~1 Comparing Two s Complement Nmbers Tas Gien signed nmbers, Determine whether or not > Retrn 1 for nmbers in shaded region below Bad Approach Test ( ) > > < < < > TSb(,) = TAdd(, TComp()) == > Problem: Thrown off by either Negatie or Positie Oerflow 15 1 Page

5 Will Get Wrong Reslts NegOer: <, > Comparing with TSb bt > PosOer: >, < bt < NegOer > < < > PosOer NegOer 17 TSb (, ) PosOer Woring Arond Oerflow Problems Partition into Three Regions <, < < < <, same sign does not oerflow Can safely se test ( ) > < < < < 1, < > < < < < < Mltiplication Compting Exact Prodct of wbit nmbers x, y Either signed or nsigned Ranges Unsigned: x * y Up to w bits Two s complement min: x * y Up to w 1 bits Two s complement max: x * y Up to w bits, bt only for TMin w Maintaining Exact Reslts w 1) = w w1 1 w 1 )*( w 1 1) = w w 1 w 1 ) = w Wold need to eep expanding word size with each prodct compted Done in software by arbitrary precision arithmetic pacages Also implemented in Lisp, ML, and other adanced langages Operands: w bits Tre Prodct: *w bits Unsigned Mltiplication in C Discard w bits: w bits UMlt w (, ) Standard Mltiplication Fnction Ignores high order w bits Implements Modlar Arithmetic UMlt w (, ) = mod w * 19 Page 5

6 Unsigned s. Signed Mltiplication Unsigned Mltiplication nsigned x = (nsigned) x; nsigned y = (nsigned) y; nsigned p = x * y Trncates prodct to wbit nmber p = UMlt w (x, y) Simply modlar arithmetic p = x y mod w Two s Complement Mltiplication int x, y; int p = x * y; Compte exact prodct of two wbit nmbers x, y Trncate reslt towbit nmber p = TMlt w (x, y) Relation Signed mltiplication gies same bitleel reslt as nsigned p == (nsigned) p 1 x = 1513: txx = : xx = 7753: xx = 13: Mltiplication Examples short int x = 1513; int txx = ((int) x) * x; int xx = (int) (x * x); int xx = (int) ( * x * x); Obserations Casting order important If either operand int, will perform int mltiplication If both operands short int, will perform short int mltiplication Really is modlar arithmetic Comptes for xx: 1513 mod 553 = 7753 Comptes for xx: (int) 555U = 13 Note that xx == (xx << 1) Powerof Mltiply with Shift Operation << gies * Both signed and nsigned Operands: w bits Tre Prodct: w bits Discard bits: w bits UMlt w (, ) Examples << 3 == * << 5 << 3 == * 1 Most machines shift and add mch faster than mltiply Compiler will generate this code atomatically 3 * TMlt w (, ) Unsigned Powerof Diide with Shift Qotient of Unsigned by Power of >> gies / Uses logical shift Operands: Diision: Qotient: / / / 1. Binary Point Diision Compted Hex Binary x B D x >> D B x >> B x >> B Page

7 s Comp Powerof Diide with Shift Qotient of Signed by Power of >> gies / Uses arithmetic shift Ronds wrong direction when < Operands: Diision: Reslt: / / RondDown( / ) 1. Binary Point Diision Compted Hex Binary y C y >> E y >> FC y >> FF C Correct Powerof Diide Qotient of Negatie Nmber by Power of Want / (Rond Toward ) Compte as ( 1)/ In C: ( (1<<)1) >> Biases diidend toward Case 1: No ronding Diidend: Diisor: / / Biasing has no effect Binary Point Correct Powerof Diide (Cont.) Case : Ronding Diidend: Correct Powerof Diide Examples Diisor: / / 1 Incremented by 1 Binary Point Incremented by 1 y/ Compted Hex Binary y C y1 151 C (y1)>> E A y C y C A (y15)>> FC A y C y C (y55)>> FF C Biasing adds 1 to final reslt 7 Page 7

8 Properties of Unsigned Arithmetic Unsigned Mltiplication with Addition Forms Commtatie Ring Addition is commtatie grop Closed nder mltiplication UMlt w (, ) w 1 Mltiplication Commtatie UMlt w (, ) = UMlt w (, ) Mltiplication is Associatie UMlt w (t, UMlt w (, )) = UMlt w (UMlt w (t, ), ) 1 is mltiplicatie identity UMlt w (, 1) = Mltiplication distribtes oer addtion UMlt w (t, (, )) = (UMlt w (t, ), UMlt w (t, )) Properties of Two s Comp. Arithmetic Isomorphic Algebras Unsigned mltiplication and addition Trncating to w bits Two s complement mltiplication and addition Trncating to w bits Both Form Rings Isomorphic to ring of integers mod w Comparison to Integer Arithmetic Both are rings Integers obey ordering properties, e.g., > > >, > > These properties are not obeyed by two s complement arithmetic TMax 1 == TMin 1513 * 3 == 13 (1bit words) 9 3 C Pzzle Answers Assme machine with 3 bit word size, two s complement integers TMin maes a good conterexample in many cases x < ((x*) < ) False: TMin x >= Tre: = UMin x & 7 == 7 (x<<3) < Tre: x 1 = 1 x > 1 False: x > y x < y False: 1, TMin x * x >= False: 5535 (x*x = 13171) x > && y > x y > False: TMax, TMax x >= x <= Tre: TMax < x <= x >= False: TMin 31 Page