List of lectures TDDA69 Data and Program Structure
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Listoflectures
IntroductionandFunctionalProgramming
ImperativeProgrammingandDataStructures
3
Environment
4
Evaluation
5
ObjectOrientedProgramming
6
Macrosand
7
VirtualMachinesandBytecode
8
GarbageCollectionandNativeCode
9
DistributedComputing
10
Logic
11
Guestlecture
12
Summary
1
2
TDDA69DataandProgramStructure
Evaluation
CyrilleBerger
2/57
Lecturegoal
Lecturecontent
Learnhowcomputerprogramsare
evaluatedandinterpreted
AbstractSyntax
Parsingaprogram
ContextFreeGrammar
AbstractSyntaxTree
Evaluation
SubstitutionModel
Environmentmodel
Writinganevaluator
3/57
4/57
Howisaprograminterpreted?
Sourcecode
Parser
Parser
AbstractSyntaxTree
Treevisitor
Generator
Sourcecode
Bytecode
VirtualMachine
Assembler
Assembly
AbstractSyntaxTree
...
OperatingSystem
CPU
5/57
LexerandParser
Parsingaprogram
Atokenisawordoranatomic
element,ieif,1,+...
Alexerisconvertingasequence
ofcharactersintoasequenceof
tokens
Aparsercombinestokensintoa
datastructure
8
ReversePolishNotationCalculator(1/2)
ReversePolishNotationCalculator(2/2)
Evaluate:'145*23+/
Lexer:['1','4','5','*','2','3',
'+','/','+']
Parser:
145*23+/+
Currenttoken:'<eof>'
Currentstack: 5
Iftokenisanumber,pushitonthestack
Iftokenisanoperator,removethefirsttwo
elementsofthestackandpushtheresult
Whatistheresultof
evaluation:www.govote.at46
3415
9
10
Calculator
1+4*5/(2+
Lexer:['1''+''4''*''5''/''(''2'
'+''3'')']
Howtoparse?Andhowto
evaluate?
ContextFreeGrammar
11
Recursivestructure
ContextFreeGrammar
Acontext-freegrammarisanatural
notationforarecusivestructure
Acontext-freegrammarisdefinedby
G=(V,Σ,R,S)
Programminglanguageshaverecursive
structures
STATEMENTcan
ifEXPRESSIONthenSTATEMENTelseSTATEMENT
whileEXPRESSIONdoSTATEMENTend
STATEMENT;STATEMENT;....
EXPRESSION
...
Σisasetofterminals
keywords,
Visasetofnon-terminals
EXPRESSIONcan
STATEMENT,
S∊Visthestartsymbol
Risthesetofproductionrules
EXPRESSION+EXPRESSION
EXPRESSION*EXPRESSION
EXPRESSION(EXPRESSION)
...
X→Y1,...,
X∊VYi∊Σ⋃V⋃{Ɛ
13
14
Derivation
ContextFreeGrammarforaCalculator
Aderivationisa
sequenceof
productions
TerminalsΣ={'+','*','/','-','(',')','0',...
'9'}
Non-terminalsEXPR,LITERAL
Rules
S→...→...→...→...
EXPR→EXPR'+'EXPR
Aderivationcanbe
drawnasatree
|EXPR'-'EXPR
|EXPR'/'EXPR
|EXPR'*'EXPR
|'('EXPR')'
|
Startsymbolisthetree'sroot
ForaproductionX→Y1...Yn,
Y1...YnarethechildrenofX
LITERAL→LITERAL
...
LITERAL→LITERAL
15
X
Y1
...
Yn
16
Parsetree
Derivationofcalculator
Grammar:
E→E{'+'|'-'|'*'|
'/'}E
|'('E
|[0-9]+
Derive:
E
1
1+4*5/(2+
E
+
E
4
E
*
E
5
Aparse
E
/
(
E
Terminalsattheleaves
Non-terminalsatthe
interiornodes
E
E
+
2
)
E
3
17
Anin-orderofthe
leavesisthe
originalinput
Theparse
treeshows
the
associationof
operations
ParsetreeUnicity
Grammar:
E→E{'+'|'-'|'*'|'/'}E
|'('E')
|[0-9]+
Derive:
1+4*5/(2+3)
Whyisthat
notunique?
www.govote.at
571354
E
1
E
+
E
4
E
*
E
5
E
1
E
+
E
4
E
*
E
5
E
/
(
E
2
E
E
+
)
E
3
18
Ambiguity
Thisstringhastwoparsetrees:2*3+1
E
/
(
E
2
E
E
+
E
)
2
E
E
*
E
3
E
+
E
1
E
2
E
*
E
+
E
3
E
1
3
19
20
WhatisanAbstractSyntaxTree?
Aparsertracesthederivationof
asequenceoftokens
Weneedastructural
representationofaprogram,such
asaparsetree
AnAbstractSyntaxTreeisa
parsetreewithlessdetail
AbstractSyntaxTree
22
AbstractSyntaxTreevsParseTree(1/3)
E
1
E
+
E
4
E
*
E
5
E
/
(
E
2
E
E
+
)
E
3
Dowereally
needall
thosenodes?
Canwe
removethe'E'
and
parenthesis?
AbstractSyntaxTreevsParseTree(2/3)
E
1
E
+
E
4
+
E
*
E
5
E
/
(
E
2
23
1
E
E
+
)
E
4
*
5
/
2
+
3
3
24
AbstractSyntaxTreevsParseTree(3/3)
Theybothcapturesthenesting
structure
Theyare
ASTaremorecompactand
easiertouse
Lisp(1/2)
Representinga
treeasan
Firstelementis
array
thevalue
Secondelementisa
listofchildren
['+'[1['*'[4['/'[5
['+'[23]]]]]]]]
InLisp:
+
1
4
*
5
/
2
+
3
(+1(*4(/5(+2
25
26
Lisp(2/2)
ThesyntaxofLispis
EXPR→'('operatorEXPR...')'
Operators
Evaluation
defineanewfunction
(define(<name><formalparameters>)<body>)
condanewfunction
(cond(exprresult)(exprresult)...(Tdefault_result))
27
SubstitutionModel
Toevaluateacombinationwitha
compoundprocedureasoperator:
SubstitutionModel
evaluatetheelementsofthecombination
applytheprocedure(i.e.valueftheoperator)
tothearguments(i.e.valuesoftheoperands):
evaluatethebodyoftheprocedurewhere
eachformalparameterisreplacedbythe
correspondingargument
30
SubstitutionModel-Example(1/)
SubstitutionModel-Example(1/2)
(define(squarex)(*xx))
(define(sum-of-squaresxy)
(+(squarex)(squarey)))
(define(fa)
(sum-of-squares(+a1)(*a2)))
Howtoevaluate(f5)?
Toevaluate:
(f5)
Fetchthevalueoff
(sum-of-squares(+a1)(*a2)))
Replaceaby5
(sum-of-squares(+51)(*52))
Evaluatethe
(sum-of-squares610)
Fetchthevalueofsum-of-squaresandreplacethe
arguments:
(+(square6)(square10))
Andso
(+(*66)(*1010))
(+36100)
136
Thisisthesubstitutionmodelintheapplicativeorder
31
32
ApplicativevsNormal
ApplicativevsNormal-Example(1/2)
Recursivelyevaluateallthe
subexpressions,butinwhatorder?
Applicativeorder:evaluatetheargumentsand
thenapply.Evaluatetheprocedurebodywith
everyformalparameterreplacedwiththe
correspondingargumentvalue
Normalorder:fullyexpandandthenreduce.
Recursivelyexpandeveryprocedureuntilonly
primitiveoperatorsareleft.Evaluate(reduce)
theresult.
(define(squarex)(*xx))
Applicative
order
(square(*42))
(square(*8))
(*88)
64
Normalorder
(square(*42))
(*(*42)(*42))
(*88)
64
Forprocedureapplicationsthatcanbemodeledwiththe
substitutionmodelthatterminatewithlegitimatevalues,
applicativeandnormalorderproducethesamevalue.
33
Normalorderandlazyevaluation
ApplicativevsNormal-Example(2/2)
if(objectandobject.count()
andobject.sum()/
object.count()>0.90):
print('Highsuccessrate!')
(define(fooxy
(if(=x0)y(*2x))))
Applicative
order
(foo10(/200))
error:divisionby
zero
34
Normalorder
(foo10(/200))
(if(=100)(/200)
(*210))
(*210)
20
Whatistheresult?www.govote.at366333
35
36
Assignmentandthesubstitutionmodel(1/2)
(define(make-withdrawbalance)
(lambda(amount)
(set!balance(-balanceamount))
balance))
(defineW(make-withdraw25))
(W20)
5
(W10)
-5
Assignmentandthesubstitutionmodel(2/2)
(define(make-withdrawbalance)
(lambda(amount)
(set!balance(-balanceamount))
balance))
(defineW(make-withdraw25))
(W
((make-withdraw25)20)
((lambda(amount)(set!balance(-25amount))25)
20)
((set!balance(-2520))25)
25
37
38
Whydoesn’tthesubstitutionmodelwork?
Substitutionisbasedonthenotion
thatsymbolsarenamesforvalues
Usingset!changessymbolstoplaces
wherevaluescanbestored
Thevalueinsuchaplacecanchange
usingassignmentwithset!
Anewmodelofevaluationisneeded:
theenvironmentmodelof
evaluation.
Environmentmodel
39
Environment
Aframeisaset
ofbindings
betweenvalue
andname
Anenvironmentis
asetofframe
nil:...
pi:3.14...
...
Evaluationrules
(global)I
II
a:6
f:[34]
...
a:2
m:4
...
Afunctionisevaluated,bycreatinganew
frameandbindingtheargumentsintheframe
Afunctionisstoredasalambdarepresenting
thefunctionbodyandapointertothe
environmentwherethefunctioniscreated
VI
41
Evaluationrules-Example(1/5)
42
Evaluationrules-Example(2/5)
(define(make-withdrawbalance)...)
(defineW1(make-withdraw100))
43
44
Evaluationrules-Example(3/5)
Evaluationrules-Example(4/5)
(W150)
Afterthecall:
45
46
Evaluationrules-Example(5/5)
(defineW2(make-withdraw100))
Writinganevaluator
47
Usingthevisitorpatternforevaluation
Howdoes
interpeterwork
inpractice?
Wecanuse
thevisitor
pattern.
+
1
4
*
5
/
Visitorinterpreter
['+'[1
['*'[4
['/'[5
['+'[23
]
]]]]]]]
defevaluate(node):
ifnotisinstance(node,list):
returnnode
elifnode[0]=='+':
returnevaluate(node[1]
[0])
+evaluate(node[1]
[1])
elifnode[0]=='-':
returnevaluate(node[1]
[0])
-evaluate(node[1][1])
elifnode[0]=='*':
returnevaluate(node[1]
[0])
*evaluate(node[1][1])
elifarray[0]=='/':
returnevaluate(node[1]
[0])
/evaluate(node[1][1])
+
2
3
+
1
4
*
5
/
+
2
49
Condstatement
3
50
Whilestatement
['while'[<'x'10]
['set'['x'['+'['x'
1]]]
defevaluate(node):
...
elifnode[0]=='while':
whileevaluate(node[1]
[0]):
evaluate(node[1][1])
['cond'[[[<'x'0]['-'x]]],
[[='x'0]0],
['T''x']]]]
defevaluate(node):
...
elifarray[0]=='cond':
foreltinnode[1]:
ifelt[0]=='T'
orevaluate(elt[0]):
returnevaluate(elt[2])
51
52
Definingvalue
Setvalue
['define'['x'
defevaluate(node):
...
elifnode[0]=='define':
currentFrame.define(
node[1][0],
evaluate(node[1]
[1]))
['set'['x'
defevaluate(node):
...
elifnode[0]=='set':
currentFrame.set(
node[1][0],
evaluate(node[1]
[1]))
53
DefiningaLambda
54
CallingaLambda
['lambda'[['x']['+''x'
defevaluate(node):
...
elifnode[0]==
'lambda':
return(array[1][0],
array[1][1],
currentFrame)
['increment'
defevaluate(node):
...
else:
(args,body,frame)=
currentFrame.get(node[0])
previousFrame=currentFrame
currentFrame=createFrame(frame)
foriinrange(0,len(args)):
name=args[i]
value=evaluate(node[1][i])
currentFrame.define(name,value)
ret=evaluate(body)
currentFrame=previousFrame
returnret
55
56
Summary
ParserandAbstractSyntax
EvaluationModels
Howtobuildasimpleinterpreter
57/57
