ITI Introduction to Computing II

Transcription

1 ITI Intoduction to Computing II Macel Tucotte School of Electical Engineeing and Compute Science Abstact data type: Stack Stack-based algoithms Vesion of Febuay 2, 2013 Abstact These lectue notes ae meant to be looked at on a compute sceen. Do not pint them unless it is necessay.

2 Evaluating aithmetic expessions Stack-based algoithms ae used fo syntactical analysis (pasing).

3 Evaluating aithmetic expessions Stack-based algoithms ae used fo syntactical analysis (pasing). Fo example to evaluate the following expession:

4 Evaluating aithmetic expessions Stack-based algoithms ae used fo syntactical analysis (pasing). Fo example to evaluate the following expession: Compiles use simila algoithms to check the syntax of you pogams and geneate machine instuctions (executable). To veify that paentheses ae balanced: ([]) is ok, but not ([)] o )((())(.

5 The fist two steps of the analysis of a souce pogam by a compile ae the lexical analysis and the syntactical analysis.

6 The fist two steps of the analysis of a souce pogam by a compile ae the lexical analysis and the syntactical analysis. Duing the lexical analysis (scanning) the souce code is ead fom left to ight and the chaactes ae egouped into tokens, which ae successive chaactes that constitute numbes o identifies. One of the tasks of the lexical analyse is to emove spaces fom the input. E.g.: whee epesent blank spaces, is tansfomed into the following list of tokens: [10,+,2,+,300]

7 The next step is the syntactical analysis (pasing) and consists in egouping the tokens into gammatical units, fo example the sub-expessions of RPN expessions (seen in class this week). In the next slides, we look at simple examples of lexical and syntactical analysis.

8 public class Test { public static void scan( Sting expession ) { Reade eade = new Reade( expession ); } while ( eade.hasmoetokens() ) { System.out.pintln( eade.nexttoken() ); } public static void main( Sting[] ags ) { scan( " * 567 " ); } } // > java Test // INTEGER: 3 // SYMBOL: + // INTEGER: 4 // SYMBOL: * // INTEGER: 567

9 public class Token { pivate static final int INTEGER = 1; pivate static final int SYMBOL = 2; pivate int ivalue; pivate Sting svalue; pivate int type; public Token( int ivalue ) { this.ivalue = ivalue; type = INTEGER; } public Token( Sting svalue ) { this.svalue = svalue; type = SYMBOL; } public int ivalue() {... } public Sting svalue() {... } } public boolean isintege() { etun type == INTEGER; } public boolean issymbol() { etun type == SYMBOL; }

10 LR Scan public static int execute( Sting expession ) { Token op = null; int l = 0, = 0; Reade eade = new Reade( expession ); l = eade.nexttoken().ivalue(); } while ( eade.hasmoetokens() ) { op = eade.nexttoken(); = eade.nexttoken().ivalue(); l = eval( op, l, ); } etun l;

11 eval( Token op, int l, int ) public static int eval( Token op, int l, int ) { int esult = 0; if ( op.svalue().equals( "+" ) ) esult = l + ; else if ( op.svalue().equals( "-" ) ) esult = l - ; else if ( op.svalue().equals( "*" ) ) esult = l * ; else if ( op.svalue().equals( "/" ) ) esult = l / ; else System.e.pintln( "not a valid symbol" ); } etun esult;

12 Evaluating an aithmetic expession: LR Scan Left-to-ight algoithm: Declae L, R and OP Read L While not end-of-expession do: Read OP Read R Evaluate L OP R Stoe esult in L At the end of the loop the esult can be found in L.

13 3 * 8-10

14 ^ L = 3 OP = R = > Read L While not end-of-expession do: Read OP Read R Evaluate L OP R Stoe esult in L

15 ^ L = 3 OP = + R = Read L While not end-of-expession do: > Read OP Read R Evaluate L OP R Stoe esult in L

16 ^ L = 3 OP = + R = 4 Read L While not end-of-expession do: Read OP > Read R Evaluate L OP R Stoe esult in L

17 ^ L = 3 OP = + R = 4 Read L While not end-of-expession do: Read OP Read R > Evaluate L OP R (7) Stoe esult in L

18 ^ L = 7 OP = + R = 4 Read L While not end-of-expession do: Read OP Read R Evaluate L OP R > Stoe esult in L

19 ^ L = 7 OP = - R = 4 Read L While not end-of-expession do: > Read OP Read R Evaluate L OP R Stoe esult in L

20 ^ L = 7 OP = - R = 5 Read L While not end-of-expession do: Read OP > Read R Evaluate L OP R Stoe esult in L

21 ^ L = 7 OP = - R = 5 Read L While not end-of-expession do: Read OP Read R > Evaluate L OP R (2) Stoe esult in L

22 ^ L = 2 OP = - R = 5 Read L While not end-of-expession do: Read OP Read R Evaluate L OP R > Stoe esult in L

23 ^ L = 2 OP = - R = 5 > Read L While not end-of-expession do: Read OP Read R Evaluate L OP R Stoe esult in L end of expession, exit the loop, L contains the esult.

24 What do you think?

25 What do you think? Without paentheses the following expession cannot be evaluated coectly: 7 - (3-2)

26 What do you think? Without paentheses the following expession cannot be evaluated coectly: 7 - (3-2) Because the esult of the left-to-ight analysis coesponds to: (7-3) - 2

27 What do you think? Without paentheses the following expession cannot be evaluated coectly: 7 - (3-2) Because the esult of the left-to-ight analysis coesponds to: (7-3) - 2 Similaly the following expession cannot be evaluated by ou simple algoithm: 7-3 * 2

28 What do you think? Without paentheses the following expession cannot be evaluated coectly: 7 - (3-2) Because the esult of the left-to-ight analysis coesponds to: (7-3) - 2 Similaly the following expession cannot be evaluated by ou simple algoithm: 7-3 * 2 Since the left-to-ight analysis coesponds to: (7-3) * 2

29 What do you think? Without paentheses the following expession cannot be evaluated coectly: 7 - (3-2) Because the esult of the left-to-ight analysis coesponds to: (7-3) - 2 Similaly the following expession cannot be evaluated by ou simple algoithm: 7-3 * 2 Since the left-to-ight analysis coesponds to: (7-3) * 2 But accoding to the opeato pecedences, the evaluation should have poceeded as follows: 7 - (3 * 2)

30 Remaks The left-to-ight algoithm: Does not handle paentheses; No pecedence.

31 Remaks The left-to-ight algoithm: Does not handle paentheses; No pecedence. Solutions:

32 Remaks The left-to-ight algoithm: Does not handle paentheses; No pecedence. Solutions: 1. Use a diffeent notation; 2. Develop moe complex algoithms.

33 Remaks The left-to-ight algoithm: Does not handle paentheses; No pecedence. Solutions: 1. Use a diffeent notation; 2. Develop moe complex algoithms. Both solutions involve stacks!

34 Notations Thee ae 3 ways to epesent the following expession: L OP R.

35 Notations Thee ae 3 ways to epesent the following expession: L OP R. infix: this is the usual notation, the opeato is sandwiched in between its opeands: L OP R;

36 Notations Thee ae 3 ways to epesent the following expession: L OP R. infix: this is the usual notation, the opeato is sandwiched in between its opeands: L OP R; postfix: in postfix notation, the opeands ae placed befoe the opeato, L R OP. This notation is also called Revese Polish Notation o RPN, it s the notation used by cetain scientific calculatos (such as the HP-35 fom Hewlett-Packad o the Texas Instuments TI-89 using the RPN Inteface by Las Fedeiksen 1 ) o PostScipt pogamming language. 7 - (3-2) = (7-3) - 2 =

37 Notations Thee ae 3 ways to epesent the following expession: L OP R. infix: this is the usual notation, the opeato is sandwiched in between its opeands: L OP R; postfix: in postfix notation, the opeands ae placed befoe the opeato, L R OP. This notation is also called Revese Polish Notation o RPN, it s the notation used by cetain scientific calculatos (such as the HP-35 fom Hewlett-Packad o the Texas Instuments TI-89 using the RPN Inteface by Las Fedeiksen 1 ) o PostScipt pogamming language. 7 - (3-2) = (7-3) - 2 = pefix: the thid notation consists in placing the opeato befoe the opeands, OP L R. The pogamming language Lisp uses a combination of paentheses and pefix notation: (- 7 (* 3 2)).

38 Notations Thee ae 3 ways to epesent the following expession: L OP R. infix: this is the usual notation, the opeato is sandwiched in between its opeands: L OP R; postfix: in postfix notation, the opeands ae placed befoe the opeato, L R OP. This notation is also called Revese Polish Notation o RPN, it s the notation used by cetain scientific calculatos (such as the HP-35 fom Hewlett-Packad o the Texas Instuments TI-89 using the RPN Inteface by Las Fedeiksen 1 ) o PostScipt pogamming language. 7 - (3-2) = (7-3) - 2 = pefix: the thid notation consists in placing the opeato befoe the opeands, OP L R. The pogamming language Lisp uses a combination of paentheses and pefix notation: (- 7 (* 3 2)). 1

39 Infix postfix (mentally) Successively tansfom, one by one, all the sub-expessions following the same ode of evaluation that you would nomally follow to evaluate an infix expession.

40 Infix postfix (mentally) Successively tansfom, one by one, all the sub-expessions following the same ode of evaluation that you would nomally follow to evaluate an infix expession. An infix expession l becomes l, whee l and ae sub-expessions and is an opeato.

41 9 / ( )

42 9 / ( ) 9 / ( 2 }{{} l }{{} 4 }{{} 5 )

43 9 / ( ) 9 / ( 2 }{{} l 9 / ( l {}}{ 2 }{{} {}}{ 4 4 }{{} 5 ) {}}{ 5 )

44 9 / ( ) 9 / ( 2 }{{} l 9 / ( l {}}{ 2 }{{} {}}{ 4 4 }{{} 5 ) {}}{ 5 ) 9 / ( [ 2 4 ] 5 )

45 9 / ( ) 9 / ( 2 }{{} l 9 / ( l {}}{ 2 }{{} {}}{ 4 4 }{{} 5 ) {}}{ 5 ) 9 / ( [ 2 4 ] 5 ) 9 / ( [ 2 4 ] }{{} l }{{} 5 }{{} )

46 9 / ( ) 9 / ( 2 }{{} l 9 / ( l {}}{ 2 }{{} {}}{ 4 4 }{{} 5 ) {}}{ 5 ) 9 / ( [ 2 4 ] 5 ) 9 / ( [ 2 4 ] }{{} l l {}}{ 9 / [ [ 2 4 ] }{{} {}}{ 5 5 }{{} ) {}}{ ]

47 9 / ( ) 9 / ( 2 }{{} l 9 / ( l {}}{ 2 }{{} {}}{ 4 4 }{{} 5 ) {}}{ 5 ) 9 / ( [ 2 4 ] 5 ) 9 / ( [ 2 4 ] }{{} l l {}}{ 9 / [ [ 2 4 ] }{{} {}}{ 5 9 / [ ] 5 }{{} ) {}}{ ]

48 9 / ( ) 9 / ( 2 }{{} l 9 / ( l {}}{ 2 }{{} {}}{ 4 4 }{{} 5 ) {}}{ 5 ) 9 / ( [ 2 4 ] 5 ) 9 / ( [ 2 4 ] }{{} l l {}}{ 9 / [ [ 2 4 ] 9 }{{} l }{{} {}}{ 5 9 / [ ] / }{{} 5 }{{} ) {}}{ ] [ ] }{{}

49 9 / ( ) 9 / ( 2 }{{} l 9 / ( l {}}{ 2 }{{} {}}{ 4 4 }{{} 5 ) {}}{ 5 ) 9 / ( [ 2 4 ] 5 ) 9 / ( [ 2 4 ] }{{} l l {}}{ 9 / [ [ 2 4 ] 9 }{{} l l {}}{ 9 }{{} {}}{ 5 9 / [ ] / }{{} 5 }{{} ) {}}{ ] [ ] }{{} {}}{ [ ] {}}{ /

50 9 / ( ) 9 / ( 2 }{{} l 9 / ( l {}}{ 2 }{{} {}}{ 4 4 }{{} 5 ) {}}{ 5 ) 9 / ( [ 2 4 ] 5 ) 9 / ( [ 2 4 ] }{{} l l {}}{ 9 / [ [ 2 4 ] 9 }{{} l l {}}{ 9 }{{} {}}{ 5 9 / [ ] / }{{} 5 }{{} ) {}}{ ] [ ] }{{} {}}{ [ ] / {}}{ /

51 Evaluating a postfix expession (mentally) Scan the expession fom left to ight. When the cuent element is an opeato, apply the opeato to its opeands, i.e. eplace l by the esult of the evaluation of l /

52 Evaluating a postfix expession (mentally) Scan the expession fom left to ight. When the cuent element is an opeato, apply the opeato to its opeands, i.e. eplace l by the esult of the evaluation of l / 9 2 }{{} l 4 }{{} }{{} 5 /

53 Evaluating a postfix expession (mentally) Scan the expession fom left to ight. When the cuent element is an opeato, apply the opeato to its opeands, i.e. eplace l by the esult of the evaluation of l / 9 2 }{{} l 9 4 }{{} }{{} l {}}{ 8 5 / 5 /

54 Evaluating a postfix expession (mentally) Scan the expession fom left to ight. When the cuent element is an opeato, apply the opeato to its opeands, i.e. eplace l by the esult of the evaluation of l. 9 2 }{{} l / 9 9 }{{} 8 l }{{} 4 l {}}{ }{{} 8 5 / }{{} 5 }{{} 5 / /

55 Evaluating a postfix expession (mentally) Scan the expession fom left to ight. When the cuent element is an opeato, apply the opeato to its opeands, i.e. eplace l by the esult of the evaluation of l. 9 2 }{{} l / 9 9 }{{} 8 l }{{} 4 l {}}{ 9 }{{} 8 5 / }{{} 5 }{{} l {}}{ 3 / 5 / /

56 Evaluating a postfix expession (mentally) Scan the expession fom left to ight. When the cuent element is an opeato, apply the opeato to its opeands, i.e. eplace l by the esult of the evaluation of l. 9 2 }{{} l / 9 9 }{{} 8 l 9 }{{} 9 l }{{} 4 l {}}{ }{{} 8 5 / }{{} 5 }{{} l {}}{ 3 / 3 }{{} / }{{} 5 / /

57 Evaluating a postfix expession (mentally) Scan the expession fom left to ight. When the cuent element is an opeato, apply the opeato to its opeands, i.e. eplace l by the esult of the evaluation of l. 9 2 }{{} l / 9 9 }{{} 8 l 9 }{{} 9 l }{{} 4 l {}}{ }{{} 8 5 / }{{} 5 }{{} l {}}{ 3 / 3 }{{} l {}}{ 3 / }{{} 5 / /

58 Evaluating a postfix expession Until the end of the expession has been eached: 1. Fom left to ight until the fist opeato; 2. Apply the opeato to the two peceding opeands; 3. Replace the opeato and its opeands by the esult. At the end we have esult.

59 9 3 / * - +

60 9 2 4 * 5 - /

61 Remaks: infix vs postfix The ode of the opeands is the same fo both notations, howeve opeatos ae inseted at diffeent places: 2 + (3 * 4) = * + (2 + 3) * 4 = *

62 Remaks: infix vs postfix The ode of the opeands is the same fo both notations, howeve opeatos ae inseted at diffeent places: 2 + (3 * 4) = * + (2 + 3) * 4 = * Evaluating an infix expession involves handling opeatos pecedence and paenthesis in the case of the postfix notation, those two concepts ae embedded in the expession, i.e. the ode of the opeands and opeatos.

63 What ole will the stack be playing? Algoithm: Eval Infix

64 What ole will the stack be playing? opeands = new stack; Algoithm: Eval Infix while ( "has moe tokens" ) { t = next token; if ( "t is an opeand" ) { opeands.push( "the intege value of t" ); } else { // this is an opeato op = "opeato value of t"; = opeands.pop(); l = opeands.pop(); opeands.push( "eval( l, op, )" ); } } etun opeands.pop();

65 Evaluating a postfix expession The algoithm equies a stack (Numbes), a vaiable that contains the last element that was ead (X) and two moe vaiables, L and R, whose pupose is the same as befoe. Numbes = [ While not end-of-expession do: Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes (ight befoe left?!) L = POP Numbes Evaluate L X R; PUSH esult onto Numbes To obtain the final esult: POP Numbes.

66 9 3-2 /

67 9 3 / * - +

68 9 / ((2 * 4) - 5) = * 5 - / ^ > Numbes = [ X = L = R = While not end-of-expession do: Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes Evaluate L X R; PUSH esult onto Numbes Ceate a new stack

69 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [ X = 9 L = R = While not end-of-expession do: > Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes Evaluate L X R; PUSH esult onto Numbes Read X

70 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 X = 9 L = R = While not end-of-expession do: Read X > If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes Evaluate L X R; PUSH esult onto Numbes Push X

71 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 X = 2 L = R = While not end-of-expession do: > Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes Evaluate L X R; PUSH esult onto Numbes Read X

72 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 2 X = 2 L = R = While not end-of-expession do: Read X > If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes Evaluate L X R; PUSH esult onto Numbes Push X

73 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 2 X = 4 L = R = While not end-of-expession do: > Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes Evaluate L X R; PUSH esult onto Numbes Read X

74 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 2 4 X = 4 L = R = While not end-of-expession do: Read X > If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes Evaluate L X R; PUSH esult onto Numbes Push X

75 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 2 4 X = * L = R = While not end-of-expession do: > Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes Evaluate L X R; PUSH esult onto Numbes Read X

76 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 2 X = * L = R = 4 While not end-of-expession do: Read X If X isnumbe, PUSH X onto Numbes If X isopeato, > R = POP Numbes L = POP Numbes Evaluate L X R; PUSH esult onto Numbes Ah! X is an opeato, pop the top element save into R

77 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 X = * L = 2 R = 4 While not end-of-expession do: Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes > L = POP Numbes Evaluate L X R; PUSH esult onto Numbes Top element is emoved and saved into L

78 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 8 X = * L = 2 R = 4 While not end-of-expession do: Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes > Evaluate L X R; PUSH esult onto Numbes Push the esult of L X R, 2 4 = 8, onto the stack

79 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 8 X = 5 L = 2 R = 4 While not end-of-expession do: > Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes Evaluate L X R; PUSH esult onto Numbes Read X

80 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 8 5 X = 5 L = 2 R = 4 While not end-of-expession do: Read X > If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes Evaluate L X R; PUSH esult onto Numbes Push X

81 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 8 5 X = - L = 2 R = 4 While not end-of-expession do: > Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes Evaluate L X R; PUSH esult onto Numbes Read X

82 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 8 X = - L = 2 R = 5 While not end-of-expession do: Read X If X isnumbe, PUSH X onto Numbes If X isopeato, > R = POP Numbes L = POP Numbes Evaluate L X R; PUSH esult onto Numbes Remove the top element and save it into R

83 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 X = - L = 8 R = 5 While not end-of-expession do: Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes > L = POP Numbes Evaluate L X R; PUSH esult onto Numbes Remove the top element and save it into L

84 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 3 X = - L = 8 R = 5 While not end-of-expession do: Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes > Evaluate L X R; PUSH esult onto Numbes Push the esult of L X R, 8 5 = 3, onto the stack

85 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 3 X = / L = 8 R = 5 While not end-of-expession do: > Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes Evaluate L X R; PUSH esult onto Numbes Read X

86 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [9 X = / L = 8 R = 3 While not end-of-expession do: Read X If X isnumbe, PUSH X onto Numbes If X isopeato, > R = POP Numbes L = POP Numbes Evaluate L X R; PUSH esult onto Numbes R = POP Numbes.

87 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [ X = / L = 9 R = 3 While not end-of-expession do: Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes > L = POP Numbes Evaluate L X R; PUSH esult onto Numbes L = POP Numbes.

88 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [3 X = / L = 9 R = 3 While not end-of-expession do: Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes > Evaluate L X R; PUSH esult onto Numbes Push L X R, 9 3 = 3, onto the stack su la pile.

89 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [3 X = / L = 9 R = 3 While not end-of-expession do: Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes > Evaluate L X R; PUSH esult onto Numbes End-of-expession

90 9 / ((2 * 4) - 5) = * 5 - / ^ Numbes = [ X = / L = 9 R = 3 > While not end-of-expession do: Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes Evaluate L X R; PUSH esult onto Numbes The esult is POP Numbes = 3 ; the stack is now empty

91 Poblem Rathe than evaluating an RPN expession, we would like to convet an RPN expession to infix (usual notation).

92 Poblem Rathe than evaluating an RPN expession, we would like to convet an RPN expession to infix (usual notation). Hum?

93 Poblem Rathe than evaluating an RPN expession, we would like to convet an RPN expession to infix (usual notation). Hum? Do we need a new algoithm?

94 Poblem Rathe than evaluating an RPN expession, we would like to convet an RPN expession to infix (usual notation). Hum? Do we need a new algoithm? No, a simple modification will do, eplace Evaluate L OP R by Concatenate (L OP R). Note: paentheses ae essential (not all of them but some ae). This time the stack does not contain numbes but chaacte stings that epesent sub-expessions.

95 / - /

96 Postfix infix Sting pntoinfix(sting[] tokens) Numbes = [ X = L = R = While not end-of-expession do: Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes Concatenate ( L X R ); PUSH esult onto Numbes

97 Postfix? While not end-of-expession do: Read X If X isnumbe, PUSH X onto Numbes If X isopeato, R = POP Numbes L = POP Numbes Pocess L X R; PUSH esult onto Numbes We ve seen an example whee Pocess == Evaluate, then one whee Pocess == Concatenate, but Pocess could also poduce assembly code (i.e. machine instuctions). This shows how pogams ae compiled o tanslated.

98 Memoy management heap (instances) fee basept Stack fo method calls Vaiables static pogam (byte code) Java Vitual Machine Method n activation ecod... Method 2 activation ecod Method 1 (main) activation ecod Pogam counte Local vaiables Paametes Retun addess Pevious basept Retun value Schematic and simplified epesentation of the memoy duing the execution of a Java pogam.

99 The Java Vitual Machine (JVM) must: Method call

100 The Java Vitual Machine (JVM) must: Method call 1. Ceate a new activation ecod/block (which contains space fo the local vaiables and the paametes among othe things);

101 The Java Vitual Machine (JVM) must: Method call 1. Ceate a new activation ecod/block (which contains space fo the local vaiables and the paametes among othe things); 2. Save the cuent value of basept in the activation ecod and set the basept to the addess of the cuent ecod;

102 The Java Vitual Machine (JVM) must: Method call 1. Ceate a new activation ecod/block (which contains space fo the local vaiables and the paametes among othe things); 2. Save the cuent value of basept in the activation ecod and set the basept to the addess of the cuent ecod; 3. Save the value of the pogam counte in the designated space of the activation ecod, set the pogam counte to the fist instuction of the cuent method;

103 The Java Vitual Machine (JVM) must: Method call 1. Ceate a new activation ecod/block (which contains space fo the local vaiables and the paametes among othe things); 2. Save the cuent value of basept in the activation ecod and set the basept to the addess of the cuent ecod; 3. Save the value of the pogam counte in the designated space of the activation ecod, set the pogam counte to the fist instuction of the cuent method; 4. Copy the values of the effective paametes into the designated aea of the cuent activation ecod;

104 The Java Vitual Machine (JVM) must: Method call 1. Ceate a new activation ecod/block (which contains space fo the local vaiables and the paametes among othe things); 2. Save the cuent value of basept in the activation ecod and set the basept to the addess of the cuent ecod; 3. Save the value of the pogam counte in the designated space of the activation ecod, set the pogam counte to the fist instuction of the cuent method; 4. Copy the values of the effective paametes into the designated aea of the cuent activation ecod; 5. Initial the local vaiables;

105 The Java Vitual Machine (JVM) must: Method call 1. Ceate a new activation ecod/block (which contains space fo the local vaiables and the paametes among othe things); 2. Save the cuent value of basept in the activation ecod and set the basept to the addess of the cuent ecod; 3. Save the value of the pogam counte in the designated space of the activation ecod, set the pogam counte to the fist instuction of the cuent method; 4. Copy the values of the effective paametes into the designated aea of the cuent activation ecod; 5. Initial the local vaiables; 6. Stat executing the instuction designated by the pogam counte.

106 The Java Vitual Machine (JVM) must: Method call 1. Ceate a new activation ecod/block (which contains space fo the local vaiables and the paametes among othe things); 2. Save the cuent value of basept in the activation ecod and set the basept to the addess of the cuent ecod; 3. Save the value of the pogam counte in the designated space of the activation ecod, set the pogam counte to the fist instuction of the cuent method; 4. Copy the values of the effective paametes into the designated aea of the cuent activation ecod; 5. Initial the local vaiables; 6. Stat executing the instuction designated by the pogam counte. activation block = stack fame, call fame o activation ecod.

107 The JVM must: When the method ends

108 When the method ends The JVM must: 1. Save the etun value (at the designated space)

109 When the method ends The JVM must: 1. Save the etun value (at the designated space) 2. Retun the contol to the calling method, i.e. set the pogam counte and basept back to thei pevious value;

110 When the method ends The JVM must: 1. Save the etun value (at the designated space) 2. Retun the contol to the calling method, i.e. set the pogam counte and basept back to thei pevious value; 3. Remove the cuent block;

111 When the method ends The JVM must: 1. Save the etun value (at the designated space) 2. Retun the contol to the calling method, i.e. set the pogam counte and basept back to thei pevious value; 3. Remove the cuent block; 4. Execute instuction designated by the cuent value of the pogam counte.

112 Example 1 (simplified) public class Calls { public static int c( int v ) { int n; n = v + 1; etun n; } public static int b( int v ) { int m,n; m = v + 1; n = c( m ); etun n; } public static int a( int v ) { int m,n; m = v + 1; n = b( m ); etun n; }

113 } public static void main( Sting[] ags ) { int m,n; m = 1; n = a( m ); System.out.pintln( n ); }

114 Example 1 (simplified) a: main: v m n ags m n 1 1 a(1)

115 Example 1 (simplified) b: a: main: v m n v m n ags m n b(2) a(1)

116 Example 1 (simplified) c: v n 3 c(3) b: v m 2 3 n b(2) a: v 1 m n 2 a(1) main: ags m 1 n

117 Example 1 (simplified) 4 c(3) b: a: main: v m n v m n ags m n b(2) a(1)

118 Example 1 (simplified) 4 b(2) a: v m n a(1) main: ags m n 1

119 Example 1 (simplified) 4 a(1) main: ags m n 1 4

120 Example 1: summay c: 4 v n 3 4 c(3) b: 4 v m n b(2) a: 4 v m n a(1) main: ags m n 1 4

121 Example 2 (with a pogam counte) 01 public class Fact { 02 public static int fact( int n ) { 03 // pe-condition: n >= 0 04 int a, ; 05 if ( n == 0 n == 1 ) { 06 a = 1; 07 } else { 08 = fact( n-1 ); 09 a = n * ; 10 } 11 etun a; 12 } 13 public static void main( Sting[] ags ) { 14 int a, n; 15 n = 3; 16 a = fact( n ); 17 } 18}

122 main ags n a pogam counte 15

123 main ags n 3 a pogam counte 16

124 fact n a etun addess main ags n 3 a fact( 3 ) pogam counte 16

125 fact n 3 a etun addess main ags n 3 a fact( 3 ) pogam counte 16

126 fact n 3 a etun addess 16 main ags n 3 a fact( 3 ) pogam counte 16

127 fact n 3 a etun addess 16 main ags n 3 a fact( 3 ) pogam counte 5

128 fact n 3 a etun addess 16 main ags n 3 a fact( 3 ) pogam counte 8

129 fact n a etun addess fact n 3 a etun addess 16 main ags n 3 a fact( 2 ) fact( 3 ) pogam counte 8

130 fact n 2 a etun addess fact n 3 a etun addess 16 main ags n 3 a fact( 2 ) fact( 3 ) pogam counte 8

131 fact n 2 a etun addess 8 fact n 3 a etun addess 16 main ags n 3 a fact( 2 ) fact( 3 ) pogam counte 8

132 fact n 2 a etun addess 8 fact n 3 a etun addess 16 main ags n 3 a fact( 2 ) fact( 3 ) pogam counte 5

133 fact n 2 a etun addess 8 fact n 3 a etun addess 16 main ags n 3 a fact( 2 ) fact( 3 ) pogam counte 8

134 fact n a etun addess fact n 2 a etun addess 8 fact n 3 a etun addess 16 main ags n 3 a fact( 1 ) fact( 2 ) fact( 3 ) pogam counte 8

135 fact n 1 a etun addess fact n 2 a etun addess 8 fact n 3 a etun addess 16 main ags n 3 a fact( 1 ) fact( 2 ) fact( 3 ) pogam counte 8

136 fact n 1 a etun addess 8 fact n 2 a etun addess 8 fact n 3 a etun addess 16 main ags n 3 a fact( 1 ) fact( 2 ) fact( 3 ) pogam counte 8

137 fact n 1 a etun addess 8 fact n 2 a etun addess 8 fact n 3 a etun addess 16 main ags n 3 a fact( 1 ) fact( 2 ) fact( 3 ) pogam counte 5

138 fact n 1 a etun addess 8 fact n 2 a etun addess 8 fact n 3 a etun addess 16 main ags n 3 a fact( 1 ) fact( 2 ) fact( 3 ) pogam counte 6

139 fact n 1 a 1 etun addess 8 fact n 2 a etun addess 8 fact n 3 a etun addess 16 main ags n 3 a fact( 1 ) fact( 2 ) fact( 3 ) pogam counte 11

140 fact n 1 a 1 etun addess 8 1 fact n 2 a etun addess 8 fact n 3 a etun addess 16 main ags n 3 a fact( 1 ) fact( 2 ) fact( 3 ) pogam counte 11

141 fact n 1 a 1 etun addess 8 1 fact n 2 a etun addess 8 fact n 3 a etun addess 16 main ags n 3 a fact( 1 ) fact( 2 ) fact( 3 ) pogam counte 8

142 1 fact n 2 a etun addess 8 fact n 3 a etun addess 16 main ags n 3 a fact( 2 ) fact( 3 ) pogam counte 8

143 1 fact n 2 a 1 etun addess 8 fact n 3 a etun addess 16 main ags n 3 a fact( 2 ) fact( 3 ) pogam counte 9

144 1 fact n 2 a 2 1 etun addess 8 fact n 3 a etun addess 16 main ags n 3 a fact( 2 ) fact( 3 ) pogam counte 11

145 1 fact n 2 a 2 1 etun addess 8 2 fact n 3 a etun addess 16 main ags n 3 a fact( 2 ) fact( 3 ) pogam counte 11

146 1 fact n 2 a 2 1 etun addess 8 2 fact n 3 a etun addess 16 main ags n 3 a fact( 2 ) fact( 3 ) pogam counte 8

147 2 fact n 3 a etun addess 16 main ags n 3 a fact( 3 ) pogam counte 8

148 2 fact n a 3 2 etun addess 16 main ags n 3 a fact( 3 ) pogam counte 9

149 2 fact n 3 a 6 2 etun addess 16 main ags n 3 a fact( 3 ) pogam counte 11

150 2 fact n 3 a 6 2 etun addess 16 6 main ags n 3 a fact( 3 ) pogam counte 11

151 2 fact n 3 a 6 2 etun addess 16 6 main ags n 3 a fact( 3 ) pogam counte 16

152 main 6 ags n 3 a pogam counte 16

153 main 6 ags n 3 a 6 pogam counte 17