Lecture 4: OCL Semantics

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1 Software Design, Modelling and Analysis in UML Lecture 4: OCL Semantics Prof. Dr. Andreas Podelski, Dr. Bernd Westphal Albert-Ludwigs-Universität Freiburg, Germany main

2 Content The Object Constraint Language (OCL): Semantics Overview OCL Types Arithmetic / Logical Operators OCL Expressions Iterate A Complete Example Scontent 2/29

3 Recall Soclsyn OCL Syntax 1/4: Expressions expr ::= w : (w) expr 1 = expr 2 : Bool oclisunde ned (expr 1 ) : Bool {expr 1,...,expr n } : Set( ) isempty(expr 1 ) : Set( ) Bool size(expr 1 ) : Set( ) Int allinstances C : Set( C) Where, given S = (T, C,V,atr), w W {self C : C C C} is a set of typed logical variables, w has type (w) is any type from T T B T C {Set( 0) 0 T T B T C} T B is a set of (OCL) basic types, in the following we use T B = {Bool,Int,String} T C = { C C C} is the set of object types, Set( 0) denotes the set-of- 0 type for 0 T B T C (sufficient because of flattening (cf. standard)). v(expr 1 ) : C where v : atr(c), T, r 1(expr 1 ) : C D where r 1 : D 0,1 atr(c),c,d C, r 2(expr 1 ) : C Set( D) where r 2 : D atr(c),c,d C. 7/ Soclsyn OCL Syntax 2/4: Constants & Arithmetics For example: expr ::=... true, false : Bool expr 1 {and,or,implies}expr 2 : Bool Bool Bool notexpr 1 : Bool Bool 0, 1,1, 2,2,... : Int expr 1 {+,,...}expr 2 : Int Int Int expr 1 {<,,...}expr 2 : Int Int Bool OclUnde ned : Generalised notation: with {+,,...} expr ::= (expr 1,...,expr n ) : 1 n 10/24 OCL Syntax 3/4: Iterate OCL Syntax 4/4: Context or, with a little renaming, expr ::= expr 1 ->iterate(w 1 :T 1 ;w 2 :T 2 =expr 2 expr 3 ) Syntax: (Assuming signature S = (T, C,V,atr).) context ::= contextw 1 :T 1,...,w n :T n inv :expr wheret i C andw i : Ti W for all1 i n,n Sthetask Soclsyn where expr ::= expr 1 ->iterate(iter :T 1 ;result :T 2 =expr 2 expr 3 ) expr 1 is of a collection type (here: a setset( 0) for some 0), iter W is called iterator, of the type denoted byt 1 (ift 1 is omitted, 0 is assumed as type ofiter) result W is called result variable, gets type 2 denoted byt 2, expr 2 in an expression of type 2 giving the initial value forresult, (OclUnde ned 2, if omitted) expr 3 is an expression of type 2, in particulariter andresult may appear inexpr 3. 12/ Soclsyn Semantics: contextw 1 : C 1,...,w n : C n inv :expr is (just) an abbreviation for allinstances C1 ->forall(w 1 : C1... allinstances Cn ->forall(w n : Cn expr )... ) 17/24 3/29

4 OCL Semantics: The Task Given an OCL expression (over signature S ), e.g. expr 1 = context CP inv : wen implies dd.wis > 0 and a system state σ 1 = {7 VM {dd {1 DD },cp {3 DD,5 DD }}, 1 DD {wis 13}, 3 CP {dd {1 DD },wen true}, 5 CP {dd {1 DD },wen false}} Σ D S and a valuation of the logical variables β 1 : W I(T T B T C ), compute the value I expr 1 (σ 1,β 1 ) {true, false, Bool } ofexpr 1 inσ 1 underβ 1. More general: Define the interpretationi expr (σ,β) ofexpr inσ underβ: I (, ) : OCLExpressions(S) Σ D S (W I(T T B T C )) I(Bool) Sthetask 4/29

5 OCL Semantics OMG (2006) 5/ main

6 Basically business as usual... (i) Equip each OCL (!) type with a reasonable domain, i.e. define function I with dom(i) = T T B T C (ii) Equip each set typeset(τ 0 ) with reasonable domain, i.e. define function I with dom(i) = {Set(τ 0 ) τ 0 T T B T C } (iii) Equip each arithmetical operation with a reasonable interpretation (that is, with a function operating on the corresponding domains), i.e. define function I with dom(i) = {+,,,...}, e.g.,i(+) I(Int) I(Int) I(Int) (iv) Same game for set operations: define function I with dom(i) = {isempty,...} (v) Equip each expression with a reasonable interpretation, i.e. define function I : Expr Σ D S (W I(T T B T C )) I(Bool) Soclsem...except for OCL being a three-valued logic, and the iterate expression. 6/29

7 (i) Domains of OCL and (!) Model Basic Types Recall: OCL basic types T B = {Bool,Int,String} We set: I(Bool) := {true,false, Bool } I(Int) := Z { Int } I(String) :=... { String } We may omit indexτ of τ if it is clear from context. Given signature S with model basic types T and domain D, set for each model basic typet T. I(T) := D(T) { T } Soclsemtypes 7/29

8 OCL and Model Types?! An Example. S = ({Bool,Nat},{VM,CP,DD}, {cp : CP,dd : DD 0,1,wen : Bool,wis : Nat}, {VM {cp,dd},cp {wen,dd},dd {wis}) Soclsemtypes 8/29

9 (i) Domains of Object and (ii) Set Types Letτ C be an (OCL) object type for a classc C. We set I(τ C ) := D(C) { τc } Letτ be a type from T T B T C. We set I(Set(τ)) := 2 I(τ) { Set(τ) } Note: in the OCL standard, only finite subsets of I(τ). Infinity doesn t scare us, so we simply allow it Soclsemtypes 9/29

10 (iii) Interpretation of Arithmetic Operations Literals map to fixed values: I(true) := true, I(false) := false, I(0) := 0, I(1) := 1,... I(OclUndefined τ ) := τ Boolean operations (defined point-wise forx 1,x 2 I(τ)): true, ifx 1 τ x 2 andx 1 = x 2 I(= τ )(x 1,x 2 ) := false, ifx 1 τ x 2 andx 1 x 2, otherwise Bool Logical connectives, e.g. I(and)(, ) : {true, false, } {true, false, } {true, false, } is defined by the following truth table: Soclsemarith x 1 true true true false false false x 2 true false true false true false I(and)(x 1,x 2 ) true false false false false false We assume common logical connectives not, or,... with the canonical 3-valued interpretation. 10/29

11 (iii) Interpretation of OclIsUndefined The is-undefined predicate (defined point-wise for x I(τ)): { true, ifx = τ I(oclIsUndefined τ )(x) := false, otherwise Note: I(oclIsUndefined τ ) is definite, i.e., it never yields. Integer operations (defined point-wise forx 1,x 2 I(Int)): { x 1 +x 2, ifx 1 x 2 I(+)(x 1,x 2 ) :=, otherwise Note: There is a common principle. The interpretation of an operation (symbol) Soclsemarith ω : τ 1...τ n τ, n 0 is a function I(ω) : I(τ 1 ) I(τ n ) I(τ) on corresponding semantical domain(s) of OCL (!) types. 11/29

12 (iv) Interpretation of Set Operations Basically the same principle as with arithmetic operations... Letτ T T B T C. Set comprehension (x 1,...,x n I(τ)): I({} τ n)(x 1,...,x n ) := {x 1,...,x n } for alln N 0 Empty-ness check (x I(Set(τ))): I(isEmpty τ )(x) := true Bool false, ifx =, ifx = Set(τ), otherwise Soclsemarith Counting (x I(Set(τ))): I(size τ )(x) := { x Int, ifx Set(τ), otherwise 12/29

13 (v) Interpretation of OCL Expressions Soclsyn OCL Syntax 1/4: Expressions expr ::= w : (w) expr 1 = expr 2 : Bool oclisunde ned (expr 1 ) : Bool {expr 1,...,expr n } : Set( ) isempty(expr 1 ) : Set( ) Bool size(expr 1 ) : Set( ) Int allinstances C : Set( C) Where, given S = (T, C,V,atr), w W {self C : C C C} is a set of typed logical variables, w has type (w) is any type from T T B T C {Set( 0) 0 T T B T C} T B is a set of (OCL) basic types, in the following we use T B = {Bool,Int,String} T C = { C C C} is the set of object types, Set( 0) denotes the set-of- 0 type for 0 T B T C (sufficient because of flattening (cf. standard)). v(expr 1 ) : C where v : atr(c), T, r 1(expr 1 ) : C D where r 1 : D 0,1 atr(c),c,d C, r 2(expr 1 ) : C Set( D) where r 2 : D atr(c),c,d C. 7/ Soclsyn OCL Syntax 2/4: Constants & Arithmetics For example: expr ::=... true, false : Bool expr 1 {and,or,implies}expr 2 : Bool Bool Bool notexpr 1 : Bool Bool 0, 1,1, 2,2,... : Int expr 1 {+,,...}expr 2 : Int Int Int expr 1 {<,,...}expr 2 : Int Int Bool OclUnde ned : Generalised notation: with {+,,...} expr ::= (expr 1,...,expr n ) : 1 n 10/24 OCL Syntax 3/4: Iterate OCL Syntax 4/4: Context or, with a little renaming, expr ::= expr 1 ->iterate(w 1 :T 1 ;w 2 :T 2 =expr 2 expr 3 ) Syntax: (Assuming signature S = (T, C,V,atr).) context ::= contextw 1 :T 1,...,w n :T n inv :expr wheret i C andw i : Ti W for all1 i n,n Soclsemexpr Soclsyn where expr ::= expr 1 ->iterate(iter :T 1 ;result :T 2 =expr 2 expr 3 ) expr 1 is of a collection type (here: a setset( 0) for some 0), iter W is called iterator, of the type denoted byt 1 (ift 1 is omitted, 0 is assumed as type ofiter) result W is called result variable, gets type 2 denoted byt 2, expr 2 in an expression of type 2 giving the initial value forresult, (OclUnde ned 2, if omitted) expr 3 is an expression of type 2, in particulariter andresult may appear inexpr 3. 12/ Soclsyn Semantics: contextw 1 : C 1,...,w n : C n inv :expr is (just) an abbreviation for allinstances C1 ->forall(w 1 : C1... allinstances Cn ->forall(w n : Cn expr )... ) 17/24 13/29

14 Valuations of Logical Variables Recall: we have typed logical variables(w )W,τ(w) is the type ofw. Byβ, we denote a valuation of the logical variables, i.e. for eachw W, β(w) I(τ(w)) Soclsemexpr 14/29

15 (v) Interpretation of OCL Expressions expr ::= w ω(expr 1,...,expr n ) allinstances C v(expr 1 ) r 1 (expr 1 ) r 2 (expr 1 ) expr 1 ->iterate(v 1 :τ 1 ;v 2 :τ 2 =expr 2 expr 3 ) I w (σ,β) := β(w) I ω(expr 1,...,expr n ) (σ,β) := I(ω)(I expr 1 (σ,β),...i expr n (σ,β)) I allinstances C (σ,β) := dom(σ) D(C) Note: in the OCL standard, dom(σ) is assumed to be finite. Again: doesn t scare us Soclsemexpr 15/29

16 Example S = ({Bool,Nat},{VM,CP,DD}, {cp : CP,dd : DD 0,1,wen : Bool,wis : Nat}, {VM {cp,dd},cp {wen,dd},dd {wis}) σ 1 = {7 VM {dd {1 DD },cp {3 DD,5 DD }}, 1 DD {wis 13}, 3 CP {dd {1 DD },wen true}, 5 CP {dd {1 DD },wen false}} I w (σ,β) := β(w) I allinstances C (σ,β) := dom(σ) D(C) I ω(expr 1,...,expr n ) (σ,β) := I(ω)(I expr 1 (σ,β),...i expr n (σ,β)) I allinstances CP (σ 1,β) I allinstances CP ->size (σ 1,β) Soclsemexpr β 1 := {3 CP }, I self (σ 1,β 1 ) 16/29

17 (v) Interpretation of OCL Expressions expr ::= w ω(expr 1,...,expr n ) allinstances C v(expr 1 ) r 1 (expr 1 ) r 2 (expr 1 ) expr 1 ->iterate(v 1 :τ 1 ;v 2 :τ 2 =expr 2 expr 3 ) Assumeexpr 1 : τ C for somec C. Setu 1 := I expr 1 (σ,β) D(τ C ). I v(expr 1 ) (σ,β) := { σ(u 1 )(v), ifu 1 dom(σ), otherwise Soclsemexpr 17/29

18 Example S = ({Bool,Nat},{VM,CP,DD}, {cp : CP,dd : DD 0,1,wen : Bool,wis : Nat}, {VM {cp,dd},cp {wen,dd},dd {wis}) σ 1 = {7 VM {dd {1 DD },cp {3 DD,5 DD }}, 1 DD {wis 13}, 3 CP {dd {1 DD },wen true}, 5 CP {dd {1 DD },wen false}} Assumeexpr 1 : τ C for somec C. Setu 1 := I expr 1 (σ,β) D(τ C ). { σ(u 1 )(v), ifu 1 dom(σ) I v(expr 1 ) (σ,β) :=, otherwise β 1 := {3 CP }, I wen(self) (σ 1,β 1 ) Soclsemexpr 18/29

19 (v) Interpretation of OCL Expressions expr ::= w ω(expr 1,...,expr n ) allinstances C v(expr 1 ) r 1 (expr 1 ) r 2 (expr 1 ) expr 1 ->iterate(v 1 :τ 1 ;v 2 :τ 2 =expr 2 expr 3 ) Assumeexpr 1 : τ C for somec C. Setu 1 := I expr 1 (σ,β) D(τ C ). I v(expr 1 ) (σ,β) := { σ(u 1 )(v), ifu 1 dom(σ), otherwise I r 1 (expr 1 ) (σ,β) := { u, ifu 1 dom(σ) andσ(u 1 )(r 1 ) = {u}, otherwise I r 2 (expr 1 ) (σ,β) := { σ(u 1 )(r 2 ), ifu 1 dom(σ), otherwise Soclsemexpr Recall: σ evaluatesr 2 of typec to a set. 19/29

20 Iterate: Intuitive Semantics expr ::= expr 1 ->iterate(iter :T 1 ; result :T 2 =expr 2 expr 3 ) Set(τ 0 )hlp := expr 1 ; τ 1 iter; τ 2 result := expr 2 ; while (!hlp.empty()) do iter := hlp.pop(); result := expr 3 ; od; return result; Soclsemiter 20/29

21 Iterate: Intuitive Semantics expr ::= expr 1 ->iterate(iter :T 1 ; result :T 2 =expr 2 expr 3 ) Set(τ 0 )hlp := expr 1 ; τ 1 iter; τ 2 result := expr 2 ; while (!hlp.empty()) do iter := hlp.pop(); result := expr 3 ; od; return result; Soclsemiter Recall: In our (simplified) setting, we always haveexpr 1 : Set(τ 0 ) andτ 1 = τ 0. In the type hierarchy of full OCL with inheritance andoclany,τ 0 andτ 1 may be different and still type consistent. 20/29

22 (v) Interpretation of OCL Expressions expr ::= w ω(expr 1,...,expr n ) allinstances C v(expr 1 ) r 1 (expr 1 ) r 2 (expr 1 ) expr 1 ->iterate(v 1 :τ 1 ;v 2 :τ 2 =expr 2 expr 3 ) I expr 1 ->iterate(v 1 :τ 1 ;v 2 :τ 2 =expr 2 expr 3 ) (σ,β) { I expr := 2 (σ,β), ifi expr 1 (σ,β) = iterate(hlp,v 1,v 2,expr 3,σ,β ), otherwise whereβ = β[hlp I expr 1 (σ,β), v 2 I expr 2 (σ,β)] and iterate(hlp,v 1,v 2,expr 3,σ,β ) { I expr := 3 (σ,β [v 1 x]), ifβ (hlp) = {x} I expr 3 (σ,β ), ifβ (hlp) = X {x} andx whereβ = β [v 1 x, v 2 iterate(hlp,v 1,v 2,expr 3,σ,β [hlp X])] Soclsemiter Quiz: Is (our) I a function? 21/29

23 Example S = ({Bool,Nat},{VM,CP,DD}, {cp : CP,dd : DD 0,1,wen : Bool,wis : Nat}, {VM {cp,dd},cp {wen,dd},dd {wis}) σ 1 = {7 VM {dd {1 DD },cp {3 DD,5 DD }}, 1 DD {wis 13}, 3 CP {dd {1 DD },wen true}, 5 CP {dd {1 DD },wen false}} context CP inv : wen implies dd.wis > Soclsemexa 22/29

24 Example S = ({Bool,Nat},{VM,CP,DD}, {cp : CP,dd : DD 0,1,wen : Bool,wis : Nat}, {VM {cp,dd},cp {wen,dd},dd {wis}) σ 1 = {7 VM {dd {1 DD },cp {3 DD,5 DD }}, 1 DD {wis 13}, 3 CP {dd {1 DD },wen true}, 5 CP {dd {1 DD },wen false}} context CP inv : wen implies dd.wis > 0 allinstances CP ->forall(self self.wen implies self.dd.wis > 0) Soclsemexa 22/29

25 Example S = ({Bool,Nat},{VM,CP,DD}, {cp : CP,dd : DD 0,1,wen : Bool,wis : Nat}, {VM {cp,dd},cp {wen,dd},dd {wis}) σ 1 = {7 VM {dd {1 DD },cp {3 DD,5 DD }}, 1 DD {wis 13}, 3 CP {dd {1 DD },wen true}, 5 CP {dd {1 DD },wen false}} context CP inv : wen implies dd.wis > 0 allinstances CP ->iterate(self;r : Bool = true and(r,implies(wen(self),>(wis(dd(self)),0)))) Soclsemexa 22/29

26 Tell Them What You ve Told Them... Given an OCL expression expr, and a system stateσ, and a valuation β of the logical variables we can compute the value I expr (σ,β) {true, false, Bool } ofexpr inσ underβ using the interpretation function I (, ) : OCLExpressions(S) Σ D S (W I(T T B T C )) I(Bool) Sttwytt 23/29

27 User s Guide Example: Approach: Task: Given a square with side length a = What is the length of the longest The lectures straight isline supposed fully inside to work the square? as a lecture: spoken word+slides+discussion It is not our goal to make any of the three work in isolation. Submission A: Submission B: Interaction: 27 The length of the longest straight line fully inside the square with side lengtha = 19.1 is (rounded). The longest straight line inside the square is the diagonal. By Pythagoras, its length is a2 + a 2. Insertinga = 19.1 yields Absence often moaned but it takes two: please ask/comment immediately. (rounded). Exercise submissions: Each task is a tiny little scientific work: (i) Briefly rephrase the task in your own words. (ii) State your claimed solution main Sformalia (iii) Convince your reader that your proposal is a solution (proofs are very convincing). 31/34 24/29

28 User s Guide Example: Approach: Task: Given a square with side length a = What is the length of the longest The lectures straight isline supposed fully inside to work the square? as a lecture: spoken word+slides+discussion It is not our goal to make any of the three work in isolation. Submission A: Submission B: Interaction: 27 Absence often moaned but it takes two: please ask/comment immediately. The length of the longest straight line fully inside the square with side lengtha = 19.1 is (rounded). The longest straight line inside the square is the diagonal. By Pythagoras, its length is a2 + a 2. Insertinga = 19.1 yields (rounded). Exercise submissions: Each task is a tiny little scientific work: (i) Briefly rephrase the task in your own words. (ii) State your claimed solution main Sformalia (iii) Convince your reader that your proposal is a solution (proofs are very convincing). 31/34 25/29

29 Formalia: Exercises and Tutorials You should work in groups of approx. 3, clearly give names on submission. Please submit via ILIAS (cf. homepage); paper submissions are tolerated. Schedule: Week N, Thursday, 8 10 Lecture A1 (exercise sheet A online) Week N + 1, Tuesday 8 10 Lecture A2 Thursday 8 10 Lecture A3 Week N + 2, Monday, 12:00 (exercises A early submission) Tuesday, 8:00 (exercises A late submission) 8 10 Tutorial A Thursday 8 10 Lecture B1 (exercise sheet B online)... Rating system: most complicated rating system ever Admission points (good-will rating, upper bound) ( reasonable proposal given student s knowledge before tutorial ) Exam-like points (evil rating, lower bound) ( reasonable proposal given student s knowledge after tutorial ) 10% bonus for early submission main Sformalia Tutorial: Plenary, not recorded. Together develop one good solution based on selection of early submissions (anonymous) there is no Musterlösung for modelling tasks. 29/34 26/29

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31 References 28/ main

32 References OMG (2006). Object Constraint Language, version 2.0. Technical Report formal/ OMG (2007a). Unified modeling language: Infrastructure, version Technical Report formal/ OMG (2007b). Unified modeling language: Superstructure, version Technical Report formal/ main 29/29

Lecture 04: OCL Semantics

Software Design, Modelling and Analysis in UML Lecture 04: OCL Semantics 2014-10-30 Prof. Dr. Andreas Podelski, Dr. Bernd Westphal 04 2014-10-30 main Albert-Ludwigs-Universität Freiburg, Germany Contents