Programming Languages Meeting 3 September 9/10, 2014

Homework Results Look for sequences. Every RHS is one. ::= 0|1|2|3|4|5|6|7|8|9 ::= | ::= ::= ( ) | λ ::= ( ) | | ::= |,

Homework (2) Spelling: it is grammar There are four different grammars cited in the slides from the first lecture. I expected four parse trees. You cannot include ( ) in the parse tree if they don’t appear in the input string.

From Last Time Example 2 (Scope): How many identifiers used? How many defining points? What is the program output?

Plan for Tonight A survey and acting on the results Assignment statements – Simple semantics – Making them more complex and testing the limits of our definitions – Reference values – Multiple assignment statements – Syntax and semantics of complex lhs expressions

Survey On the small piece of paper, write down all the programming languages you have installed on your computer because you have: – An IDE for the language – An interpreter for the language – A compiler for the language Hint: Check the Program Files folder or the Applications folder

On to Assignment Statements In addition to Env and Loc, which have been defined before, define Value, the set of storable values for a particular type V* = Value + {undefined} Store = Loc  V*

Assignment (2) Investigating the semantics of an assignment statement, which stores a value in a named location: Abstract syntax shows we need – An l-expression: the target of the assignment – A value: to store in the target

Assignment (3) Possible l-expressions are – a simple identifier – an array component – a field in a record – a pointer image – and others

Assignment (4) At declaration time each identifier is associated with an attribute record, which contains – A type descriptor – An addressing function (the identity function for nonstructured types) – Storage requirements (# of words and layout)

Assignment (5) At instantiation, which occurs at procedure invocation, each identifier is mapped to a base location, which becomes an argument to the addressing function. Let M = Ide  Loc + {unbound}  V* with the understanding that {unbound}  {undefined}

Assignment (6) In the simplest cases we can work with M = Ide  V* bypassing consideration of locations. Ready for the formal semantics of an assignment statement:

Assignment (7) The meaning of the assignment statement A := E is a function that maps M to M, sometimes denoted by [A:=E] defined so that M goes to M’ where M is the original memory state: a function assigning identifiers to values M’ is the new memory state, defined by

Assignment (8) M(x) if x ≠ A M’(x) = v if x = A Note that this says: The previous value of A is irrelevant No other values are affected

Your Turn Exercise 1: Assume Ide = { a,b,c }, Value = { 0,1,2,3 } Find [a := 2] Caution: There are several sets of functions involved. Keep them straight.

Your Turn (2) Exercise 2: (Refer to the handout: Syntax Definition for REF) 1.Determine the four components of the context- free grammar specification for REF – start symbol, nonterminals, etc. 2.Write a small (no more than 5 lines) syntactically correct program in REF. Put your final version on the special sheet of paper.

Exercise 2 (2) 3.What is the output of your program? 4.How many different types does the language support? 5.Create the parse tree for the program you have been given.

Exercise 3 For the program fragment on the following slide, give the elements of Env, Loc, Store, and M that represent the program state after the fragment loads. If you need memory addresses, make them up. Hint: Figure out what the elements of each of the semantic domains look like.

Exercise 3 (2) program declare ONE : integer constant = 1; declare TWO : integer constant = 2; declare A,B : integer; declare RC,RD : ref integer; declare RRE, RRF : ref ref integer;

Semantics of REF Semantics of a REF assignment statement: First the LHS: It is an identifier with a type. – Type is integer constant: invalid assignment—type mismatch – Type is [ref…] integer: Assume j occurrences of ref, j = 0,1,2,…

Semantics (2) Now the RHS: three kinds of expressions a)+ operation included b)literal integer c)single identifier

Semantics (3) Generating a RHS value: Case a) RHS involves + – Find location bound to variable, perform operation as indicated in RHS dereferencing identifiers as necessary, obtaining value v Case b) RHS is an integer – The value v is the value of the integer

Semantics (4) Case c) RHS is a single identifier, with type – Integer constant: Value v is the value of the constant – [ref]… integer: There are 3 subcases, depending on the reference level k of the identifier Case 1: j < k+1 Case 2: j = k+1 Case 3: J > k+1

Semantics (5) Taking each of these subcases in turn j < k+1 : Dereference RHS identifier to level j. Set v to the value of the dereferenced identifier. j = k+1 : Set v to the location of the RHS identifier j > k+1 : Invalid statement—type mismatch

Semantics (6) Store v in location determined by LHS. – If v is an integer, the reference level of the LHS, j, must be 0. Otherwise, there is a type mismatch. – If v is a reference value, it is of the same level as the LHS, so is storable.

Exercise 4 Using the declaration fragment from before, execute the following statements and determine the memory function M : Ide -> Loc -> V* when the fragment completes.

Exercise 4 (2) A := ONE; B := A + ONE; RC := A; RD := B; RRE := RC; RRF := RD;

Exercise 5 Finally, assume that you have executed the 6 lines of code in Exercise 4. Make each of the statements below the 7 th line of code and determine how M changes. 1. A := RRF + B; 2. B := RRE; 3. RC := RRF + RD; 4. RRF : = RRE; 5. RRE : = RD;

Installing Systems Make sure that you have installed on your computer: ChucK ( ) -- chuck.cs.princeton.edu Python (3.4.1) -- python.org Perl (5.20.0) – perl.org

For Next Time Install the systems noted on the previous slide. Execute the five statements in REF in Exercise 5. Post the results on piazza.com