Lösungsvorschlag Software Architecture FS08: Unterschied zwischen den Versionen

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<math> balance(charge(c_n, a)) = balance(c_n) + a \geq 0 </math> using A6, induction assumption for <math> balance(c_n) \geq 0 </math>, and P2 for <math> a > 0 </math>
 
<math> balance(charge(c_n, a)) = balance(c_n) + a \geq 0 </math> using A6, induction assumption for <math> balance(c_n) \geq 0 </math>, and P2 for <math> a > 0 </math>
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'''Prove that <math> balance(c) \leq limit(c) </math>'''
 
'''Prove that <math> balance(c) \leq limit(c) </math>'''
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<math> limit(c_0) = l </math> using A1
 
<math> limit(c_0) = l </math> using A1
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Hence: <math> balance(c_0) = 0 \leq limit(c_0) = l > 0 </math> using P1 for the last inequality
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Induction assumption:
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<math> c_n </math> is a correct expression of type CREDIT_CARD and length <math> n </math>, that is:
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<math> balance(c_n) \leq limit(c_n) </math> and <math> limit(c_n) > 0 </math>
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Induction step (reuse inequalities found for balance above):
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<math> balance(settle(c_n)) = 0</math> (see above) and <math> limit(settle(c_n)) = limit(c_n) > 0 </math> using A3 and induction assumption, hence
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<math> balance(settle(c_n)) \leq limit(settle(c_n)) </math>
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<math> balance(charge(c_n, a)) = balance(c_n) + a \leq limit(c_n) \iff balance(c_n) \leq limit(c_n) - a </math> using P2
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<math> limit(charge(c_n,a)) = limit(c_n) </math> using A5
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<math> balance(charge(c_n, a)) \leq limit(charge(c_n,a)) </math> since above we could show that a stronger statement was true

Version vom 21. Mai 2009, 22:12 Uhr

Modularity, ADT, Designy by Contract and Concurrency (8 points)

b

d

f: true, since the visitor needs to know implementation details of the objects it visits

b: true because class invariants are 'and'-ed, hence stronger assertions can be introduced

c

Design by Contract (12 points)

<eiffel> class

 BANK_ACCOUNT

feature withdraw(v: INTEGER) is

 require TRUE

do end

feature balance: INTEGER

end

class STUDENT_ACCOUNT inherit BANK_ACCOUNT redefine withdraw end

feature withdraw(v: INTEGER) is require else balance >= v do

 balance := balance - v

ensure

 balance_altered: balance = balance - v

end

end

class B_A_NORMAL inherit BANK_ACCOUNT redefine withdraw end

feature withdraw(v: INTEGER) is require else balance - fee >= v do

 balance := balance -v

ensure

 balance_altered: balance = balance - v

end

feature

 fee: INTEGER

end

class B_A_BUSINESS inherit BANK_ACCOUNT redefine withdraw end

feature withdraw(v: INTEGER) is require else balance + credit >= v do

 balance := balance -v

ensure

 balance_altered: balance = balance - v

end

feature

 credit: INTEGER

end </eiffel>

Abstract Data Types (16 points)

Writing and ADT for CREDIT_CARD (7 points)

TYPES

CREDIT_CARD

FUNCTIONS

Creators:

new_card: INTEGER x INTEGER --|--> CREDIT_CARD

Queries:

limit: CREDIT_CARD ----> INTEGER

balance: CREDIT_CARD ----> INTEGER

Commands:

settle: CREDIT_CARD ----> CREDIT_CARD

charge: CREDIT_CARD x INTEGER --|--> CREDIT_CARD

PRECONDITIONS

P1 new_card (limit: INTEGER) require limit > 0

P2 charge (c: CREDIT_CARD; amount: INTEGER) require amount + balance(c) <= limit(c) and amount > 0

AXIOMS

A1 limit(new_card(l: INTEGER)) = l

A2 balance(new_card(l: INTEGER)) = 0

A3 limit(settle(c: CREDIT_CARD)) = limit(c)

A4 balance(settle(c: CREDIT_CARD)) = 0

A5 limit(charge(c: CREDIT_CARD; a: INTEGER)) = limit(c: CREDIT_CARD)

A6 balance(charge(c: CREDIT_CARD; a: INTEGER)) = balance(c: CREDIT_CARD) + a

Proof of balance properties (6 points)

Prove that balance(c) >= 0:

Base case: using A2, new_card(l)

Induction assumption: is a correct expression of type CREDIT_CARD of length , i.e.

Induction step:

using A4

using A6, induction assumption for , and P2 for


Prove that

Base case:

new_card(l: INTEGER)

using A2

using A1

Hence: using P1 for the last inequality

Induction assumption:

is a correct expression of type CREDIT_CARD and length , that is:

and

Induction step (reuse inequalities found for balance above):

(see above) and using A3 and induction assumption, hence


using P2

using A5

since above we could show that a stronger statement was true