programming languages meeting 4 september 16/17, 2014

Programming Languages

Meeting 4September 16/17, 2014

Planning Ahead

• Short exam next time– Construct a grammar given a description of strings

in the language– Parse a sentence in a grammar– Describe a language defined by a grammar

Short Exam (2)

– Inferences from the grammar• Precedence of operators• Associativity of operators• Ambiguity of sentences

– Construct one of the various semantic domains in a specific situation

REF Results

Declarations followed by 1. A := RRF + B;

Ans: 42. B := RRE;

Ans: 13. RC := RRF + RD; Ans: Type mismatch (see slide Semantics(6) from last week).

REF Results (2)

4. RRF : = RRE;Ans: Location of RC

5. RRE : = RD;Ans: Location of RD

Multiple Assignments

Let’s now investigate statements of the form

ref1,ref2,…,refm := exp1,exp2,…,expn

What conditions should we put on m and n?

MA (2)

m = 1, n = 1The ordinary assignment statement

A := 0;

(ref1 is A and the location it is mapped to; exp1 has value 0)

MA (3)

m > 1, n = 1A,B,C := 0;

should represent the sequenceA := 0;B := 0;C := 0;

or any of the other 5 arrangements of it.(Why 5?)

MA (4)

m = 2, n = 2

A,B = 0,1;A,B = X,Y;A,B = B,A;

If initially A:=9; B:=23; X:=7; Y:=12; what is the state after each of the statements above, accumulating the changes?

MA (5)

More complexity:Suppose A has been declared to be an array of

size 10 of integers, all initialized to 0. Also j has been declared to be an integer and currently has value 4. What is the result of executing

A[j-1], j, A[j+1] := j+1;

Work with your partner to find several plausible answers.

MA(6)

Case 1:Evaluate exp to obtain a value vfor i=1 to m step 1

evaluate refi to obtain loci

assign v to loci

In our case j becomes 5 and A is [0,0,5,0,0,5,0,0,0,0]

MA (7)

Case 2:Evaluate exp to obtain a value vfor i=m to 1 step −1

evaluate refi to obtain loci

assign v to loci

In our case j becomes 5 and A is [0,0,0,5,5,0,0,0,0,0]

MA (8)

Case 3:for i=1 to m step 1

evaluate refi to obtain loci

evaluate exp to obtain value v

assign v to loci

In our case j becomes 5 and A is [0,0,5,0,0,6,0,0,0,0]

MA (9)

Case 4:for i=m to 1 step −1

evaluate refi to obtain loci

evaluate exp to obtain value v

assign v to loci

In our case j becomes 5 and A is [0,0,0,6,5,0,0,0,0,0]

MA (10)

Case 5:for i=1 to m step 1

evaluate refi to obtain loci

evaluate exp to obtain value vfor i=1 to m step 1

assign v to loci

In our case j becomes 5 and A is [0,0,5,0,5,0,0,0,0,0]

Control Structures

Start with the three fundamental structures and assume the existence of an empty statement (NOP)• Sequence S1;S2

• If-then-else if P then S1 else S2

• While-do while P do S

CS Semantics

Use program functions, defined as followsLet P be a program (or program fragment)Let a be an identifier declared in P

representing valuesLet Va be the set of legal values for a as

determined by its type. Let Va* = Va + {undefined}

CS Semantics (2)

Question: What are the possible outcomes after executing a program fragment?1. 2. 3. 4. 5.

CS Semantics (3)

The program function for P is

[P]: Va* × Vb* × … −> Va* × Vb* × … + {error,

nonterminating}

defined to show the state after executing P.

Example 1

P = X := X+3; Y:=Y−1 where X,Y are type integer

Example 2

P = while Y > 0 do begin X := X−1; Y := Y−1 endwhere X,Y are type 0..maxint

Example 3

P = if X <= Y then X := Y – X else X := 0 where X,Y are type integer

Constructing a PF

Given a program P1. Find the set of identifiers: X, Y, Z, …2. Find the set of values VX* , … associated with

each identifier, which includes {undefined}3. Construct the function [P] with

domain: VX* × VY* × … codomain: VX* × VY* × …

+ {error, nonterminating}

Constructing (2)

Use the following notational conventions:1. If an identifier in P does not appear as the

target of an assignment statement, it can be suppressed in the domain and codomain

2. [P] (error) = error [P] (nonterminating) = nonterminating

so are not written

Constructing (3)

3. undefined is denoted by 4. The identifier and its value are denoted by

the same letter5. Account for all possible values of the domain,

perhaps by partitioning with conditions, e.g. x<0, x≥0, x =

Example 4Program Function Exercise II.1

P = while x <= b do x := x + a end

If-Then Semantics

Let P = if C then S1 else S2

Define [P](v) = [S1](v) if [C](v) = true [S2](v) if [C](v) =

false error if [C](v) =

error

If-Then-Else (2)

Theorem:

[if C then S] = [if C then S else Φ] where [Φ] = id

While-Do Semantics

Let P = while C do S

Define [P](v) = [S]k(v) where nonterminating error

Sequence Semantics

Let P = S1 ; S2

Define [P](v) = [S2] o [S1](v)

Böhm – Jacopini Theorem

The simple assignment statement and the three control structures sequence, if-then-else, while-do are enough• to write any block structured program• to compute any computable function• to convert any flowchart specified program to

code

Additional Control Structures

Useful for human understanding of programsAnother example of “syntactic sugar”

Definition:[repeat S until C] = [S ; while (not C) do S]

Additional (2)

Let P = case F of F1:S1; F2:S2; … ; FN:SN endDefinition[P] = [if F=F1 then S1 else if F=F2 then S2 else if … else if F=FN then SN]Issues: Two selector values with same statement What if F does not equal any FI? Can some SI be empty?

Additional (3)

Let P = switch(E) {case E1:S1;case E2:S2; … ; case EN:SN; default: SD}Definition[P] = [ if E=E1 then S1;S2;…;SN;SD else if E=E2 then S2;…;SN;SD

else if E=E3 then S3;…;SN;SD

else SD]

Assignment

• Due at beginning of next class meeting. Work by yourself. Ask questions only of the instructor.

1. Program Function Exercise I.52. Program Function Exercise II.5and under the assumptions of Section III3. Program Function Exercise I.44. Program Function Exercise II.4