我创造了一台自动售货机，工作正常。一旦交易完成，我想从项目数量中减去1。我在我的代码中提供了意见以供理解。忽略pred巧克力中的一些评论。不知怎的，我试图减去，但它不会。我不知道这似乎是什么问题。任何帮助表示赞赏。
sig button {
qty:Int} // buttons on vending machine for selecting chocolates
//sig coin{}
sig choco{
value:Int, // Each chocolate has some cost as an attribute aka value.
choice :one button // Selecting option
}
fact {
// all c:choco | all c1:choco -c | c1.choice != c.choice
}
sig machine{
cust : one customer, // Customer entity
a,b,c,d,nullb ,ip: one button, // buttons on vending machine ,ip is the input selected by user
//oners,twors,fivers ,tenrs,null1: set coin,
ipp,opc2 : one Coin, // ipp = input rs , opc = balance rs
customeripb: cust one -> one ip, // customer presses a button
customeripc: cust one -> one ipp, // customer enters coins
customeropc: opc2 one -> one cust, //customer receives balance of coins
op: one choco , // output of chocolate from machine
customerop: op one -> one cust, // customer receives a chocolate
cadbury, kitkat, eclairs , gum,null: lone choco // types of chocolate
}
{
//#(fivers+tenrs+null+twors+oners) = 5
#(a+b+c+d) = 4 // no of buttons of a b c and d are 4 on machine
# (cadbury+kitkat+ eclairs +gum) =4 // no of options to choose = 4
cadbury=choice.a // cadbury corresponds to button a
cadbury.value= 10 // cadbury costs 10rs
kitkat=choice.b // kitkat corresponds to button b
kitkat.value=5 // kitkat costs 5rs
null.value=0 // null costs 0 rs
null=choice.nullb
// as such null doesnt exist it is just to specify no i/p no o/p and nulb is an imaginary button
eclairs=choice.c // eclairs corresponds to button c
eclairs.value=1 // eclairs costs 1 rs
gum=choice.d // gum corresponds to button d
gum.value=2 // gum costs 1 rs
a.qty>=10 and a.qty<=40
b.qty>=11 and b.qty<=40
c.qty>=12 and c.qty<=40
d.qty>=13 and d.qty<=40
nullb.qty=0
//ip=nullb //input button selection is never nullb(which is imaginary button)
ipp.value!=0 // input of coins is never = 0rs
/* all m:machine|all o:opc2
|all opp: op| all i:ip|all ii:ipp| all c:m.cust
|c -> i in m.customeripb and c->ii in m.customeripc and o->c in m.customerop and opp->c in m.customerop
*/
//button=!=none
}
sig customer //user of machine
{
}
abstract sig Coin { //each coin has a valueof rs
value: Int
}
sig Nullrs extends Coin {} { value = 0 } // void rs
sig Oners extends Coin {} { value = 1 } // one rs
sig Twors extends Coin {} { value = 2 } // twors
sig Fivers extends Coin {}{ value = 5 } // five rs
sig Tenrs extends Coin {} { value = 10 } // ten rs
sig Threers extends Coin {} { value = 3 } // this is only used in o/p to specify 3rs will come out
sig Fourrs extends Coin {} { value = 4 }// this is only used in o/p to specify 4rs will come out
sig Sixrs extends Coin {} { value = 6 }// this is only used in o/p to specify 6rs will come out
sig Sevenrs extends Coin {}{ value = 7 }// this is only used in o/p to specify 7rs will come out
sig Eightrs extends Coin {} { value = 8 } // this is only used in o/p to specify 8rs will come out
sig Niners extends Coin {} { value = 9} //// this is only used in o/p to specify 9rs will come out
pred show{} // show
pred chocolate [before,after:machine ] // machine has two states one before o/p and one after
{
before.cadbury=after.cadbury
before.kitkat=after.kitkat
before.eclairs=after.eclairs
before.gum=after.gum
//all chocolates will not change and are fixed
before.ipp.value=after.ipp.value
// input value of rs remains same i.e i/p is inside machine once inputed so it cant change
before.opc2.value=0 // before state o/p value of balance coins =0
before.op=before.null // beforestate o/p = no chocolate
before.ip!=before.nullb // input button pressed never equals nullb
after.ip!=after.nullb // input button pressed never equals nullb
//before.ip=after.ip // input button pressed remains same
after.op=after.kitkat or after.op=after.eclairs
before.null=after.null // imaginary null chocolate remains in same state
before.opc2!=none and after.opc2 !=none
// balance of coins is never empty in case of empty I have defined nullrs
(after.op.value=before.ipp.value=>after.opc2.value=0)
//
(after.op=after.null=>after.opc2.value=before.ipp.value)
(before.ipp.value > after.op.value=>after.opc2.value=before.ipp.value-after.op.value)
//(before.ipp.value=after.op.value=>after.opc2.value=0)
//opc2.value!=ipp.value
before.ip=before.a or before.ip=before.b or before.ip=before.c or before.ip=before.d
(after.op=after.cadbury ) => ( ( after.ip=after.a and after.a.qty=minus[before.a.qty,1])) else
(after.op=after.kitkat ) => ( (after.ip=after.b and after.b.qty=minus[before.b.qty, 1])) else
(after.op=after.eclairs ) =>( (after.ip=after.c and after.c.qty=minus[before.c.qty,1])) else
(after.op=after.gum ) =>((after.ip=after.d and after.d.qty=minus[before.d.qty,1])) else
(after.ip=before.ip and after.ip.qty=minus[before.ip.qty,0] )
after.op!=before.null => after.op.choice=before.ip
(after.op=before.gum=>before.ipp.value>=Twors.value)
after.op=before.cadbury=>before.ipp.value>=Tenrs.value
after.op=before.eclairs=>before.ipp.value>=Oners.value
after.op=before.kitkat=>before.ipp.value>=Fivers.value
(before.ipp=Oners or before.ipp=Twors or before.ipp=Fivers or before.ipp=Tenrs or before.ipp=Nullrs) and
before.ipp!=Threers and before.ipp!=Fourrs and before.ipp !=Sixrs and before.ipp!=Sevenrs and before.ipp!=Eightrs and before.ipp!=Niners
(before.ip=before.b and before.ipp.value < 5) => (after.op!=before.kitkat or after.op!=before.eclairs or after.op!=before.cadbury or after.op!=before.gum)and after.op=before.null
(before.ip=before.d and before.ipp.value < 2) => (after.op!=before.kitkat or after.op!=before.eclairs or after.op!=before.cadbury or after.op!=before.gum)and after.op=before.null
(before.ip=before.a and before.ipp.value < 10 )=> (after.op!=before.kitkat or after.op!=before.eclairs or after.op!=before.cadbury or after.op!=before.gum) and after.op=before.null
(before.ip=before.c and before.ipp.value >= 1) => (after.op!=before.kitkat or after.op!=before.null or after.op!=before.cadbury or after.op!=before.gum) and after.op=before.eclairs
(before.ip=before.c and before.ipp.value = 0) => (after.op!=before.kitkat or after.op!=before.null or after.op!=before.cadbury or after.op!=before.gum) and after.op=before.null
(before.ip=before.a and before.ipp.value =10) => (after.op!=before.kitkat or after.op!=before.null or after.op!=before.eclairs or after.op!=before.gum) and after.op= before.cadbury
(before.ip=before.d and before.ipp.value >= 2) => (after.op!=before.kitkat or after.op!=before.null or after.op!=before.cadbury or after.op!=before.eclairs) and after.op=before.gum
(before.ip=before.b and before.ipp.value >= 5) => (after.op!=before.eclairs or after.op!=before.null or after.op!=before.cadbury or after.op!=before.gum) and after.op=before.kitkat
}
run chocolate for exactly 2 machine, 8 button, 5 choco,9 Int,5 Coin,1 customer
答案 0 :(得分：1)
一般来说，比“它不起作用”更具体是有意义的。
我假设你所说的“它不起作用”的意思是在after机器中你期望所选巧克力的数量减少1，但相反，它保持不变。原因是你的(if-then-else) or (if-then-else) or ...表达式，这在逻辑上是有缺陷的。您可能想要表达的是强制执行至少一个then分支（因为您知道如果满足条件，则只有一个分支），但这不是满足整个分离的必要条件。
更具体地说，在
中((after.op=after.cadbury)
=> (... and after.a.qty=minus[before.a.qty,1] and ...)
else (... and after.a.qty=before.a.qty and ...)
)
or
((after.op=after.kitkat)
=> (... and after.b.qty=minus[before.b.qty,1] and ...)
else (... and after.b.qty=before.b.qty and ...)
)
即使after.op等于after.cadbury，也不会强制该子句的then分支为真，因为为了满足整个表达式，它足以满足该分支的下一条款，规定所有数量应保持不变。
你想要的是if-then-elsif-...-else构造的一些柔和，例如，
(after.op = after.cadbury) => {
...
} else (after.op = after.kitkat) => {
...
} else {
...
}
如果你这样做，你的机器仍然无法工作，也就是说，你的模型将变得不可满足：你的约束强制要求after和before机器共享相同的按钮 1 < / sup>和数量与按钮相关联（qty字段位于button sig），这意味着after和before台机器中的数量必须相同。我真的没有理由把qty放在sig button中。
[1]：在before.cadbury=after.cadbury and ...谓词中加chocolate，在附加的sig cadbury=choice.a and ...事实中加machine