COP 5021 Lecture -*- Outline -*-

* Denotational Semantics (2.2)

*** denotational example
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   SYNTAX OF WHILE LANGUAGE EXPRESSIONS

Abstract Syntax (1.2):

  x,y \in Var
  n \in NumericLiteral
  a \in AExp
  b \in BExp
  op_a \in Op_a
  op_b \in Op_b
  op_r \in Op_r

  a ::= x | n | a1 op_a a2
  b ::= true | false | not b | b1 opb b2
        | a1 op_r a2

Lexical syntax:

  n \in NumericLiteral
  dgs \in Digits
  d \in Digit
  
  n ::= d | dgs d
  dgs ::= d | dgs d
  d ::= 0 | 1 | 2 | 3 | 4 | 5 | 6
      | 7 |8 | 9
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   DENOTATIONAL SEMANTICS OF EXPRESSIONS
  
Domains:
        Z = { ..., -1, 0, 1, ... }
        T = { true, false }
  s \in State = Var -> Z

Meaning Functions:

  A : Aexp -> State -> Z

 A[[x]]s = s(x)
 A[[n]]s = N(n)
 A[[a1 op_a a2]]s
    = (O_a[[op_a]]) (A[[a1]]s) (A[[a2]]s))

where

  N : NumericLiteral -> Z
  D : Digit -> Z
  D[[0]] = 0,  D[[1]] = 1,  D[[2]] = 2,
  D[[3]] = 3,  D[[4]] = 4,  D[[5]] = 5,
  D[[6]] = 6,  D[[7]] = 7,  D[[8]] = 8,
  D[[9]] = 9,
  N[[n]] = Nval[[n]](1)
  Nval: Digits -> Z -> Z
  Nval[[d]](p) = p*D[[d]]
  Nval[[dgs d]](p)
     = Nval[[dgs]](10*p) + Nval[[d]](p)

  O_a : Op_a -> (Z x Z -> Z)

  O_a[[+]] = \n1 n2 . n1 + n2
  O_a[[-]] = \n1 n2 . n1 - n2
  ...

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     In the semantic function Nval, the p argument gives the power of 10
        that denotes the significance of the digit

     Other languages may have different definitions of N
        E.g., in C or C++, N[[015]] = 13, since 015 is in octal.

     Q:  Why is the state a parameter to the meaning of arithmetic
         expressions?
       It's needed to evaluate variable references

     Q: How would you define B : Bexp -> (State -> T)?

     Recall that the synatax is:
     b ::= true | false | not b | b1 opb b2
        | a1 op_r a2


     O_r : Op_r -> (Z x Z -> T)
     O_b : Op_b -> (T x T -> T)

     B[[true]](s) = true
     B[[false]](s) = false
     B[[not b]](s)
       = if B[[b]](s) then false else true
     B[[b1 op_b b2]](s)
       = O_b[[op_b]](B[[b1]](s), B[[b2]](s))
     B[[a1 op_r a2]](s)
       = O_r[[op_r]](A[[a1]](s), A[[a2]](s))