Quiz 1

Statistics:

mean     21.750 (261.000/12)
stddev   7.037
median   24.000
midrange 16.000-27.500

#   avg
1   11.50 / 15
2    5.92 / 10
3    4.33 /  5

Question 1-1

Prove the following using our axiomatic scheme.

{ i * k = n AND k mod 2 = 0 }
k := k / 2;
i := i * 2;
{i * k = n }

Question 1-2

Consider the following Ada program.

with Text_IO; use Text_IO;
procedure Test is
    X : Integer;

    procedure A is begin
        X := 3;
    end A;

    procedure B is
        X : Integer;
    begin
        X := 4;
        A;
        Put_Line(Integer'Image(X));
    end B;
begin
    X := 2;
    B;
    Put_Line(Integer'Image(X));
end Test;

• a. If Ada used statically scoped variables, what would this program print?

• b. If it used dynamically scoped variables, what would it print?

• c. Which does Ada use?

Question 1-3

Consider the following grammar.

S -> a S b S | x

Solutions

Solution 1-1

 
1.  i * k = n AND k mod 2 = 0 implies (i * 2) * (k / 2) = n (Math fact)

2.  { i * k = n AND k mod 2 = 0 }                           (Conseq: 1)
    { (i * 2) * (k / 2) = n }

3.  { (i * 2) * (k / 2) = n }                               (Assignment)
    k := k / 2;
    { (i * 2) * k = n }

4.  { i * k = n AND k mod 2 = 0 }                           (Seq: 2, 3)
    k := k / 2;
    { (i * 2) * k = n }

5.  { (i * 2) * k = n }                                     (Assignment)
    i := i * 2;
    {i * k = n }

6.  { i * k = n AND k mod 2 = 0 }                           (Seq: 4, 5)
    k := k / 2;
    i := i * 2;
    {i * k = n }

Solution 1-2

• a. 4, 3
• b. 3, 2
• c. static scoping

Solution 1-3