Questions

Runtime Stack
```
  program main();        Assume parameters are passed by value
    var A,B: integer;
    procedure one(X,Y: integer);
      procedure two(A: integer);
      begin
        B := B + A;
      end;
    begin
      if (X = 1) then two(Y);
      else begin B := B + 1; one(X-1,Y); end;
     L1: end;
  begin
     A := 3; B := A+1;
     one(A,B); 
 L0: write(B);
  end.
```
- Show what the runtime stack would look like for the executing program when the procedure two is ready to start executing (i.e. immediately after the activation record is placed on the stack with its initial values set). Be sure to clearly mark the fields in each activation record.
  
  First activation is for main. Last activation is for two. The three in the middle are for one.
- What gets printed out at the end?
```
10 get printed out
```

Parameter Passing
Given the following program:

program main();
var A,B: integer;
  procedure new(B,C : integer);
  begin
    B := B + C; C := 5;
    write(A,B,C);
  end;
begin
  A := 3; B := 4; 
  new(A,B); 
  write(A,B);
end.

What five values get printed out if the parameters are passed

by value?
```
3 7 5 3 4
```
by reference?
```
7 7 5 7 5
```
by value-result?
```
3 7 5 7 5
```

Arrays
Suppose we have a 3 dimensional array int A[1..5][0..4][2..8].

Assuming an integer is size 2, what is the size of the entire array?

# elements = (5-1+1)*(4-0+1)*(8-2+1) = 5 * 5 * 7 = 175
total size = #elements * size = 175*2 = 350

Assuming that the array is stored as row-major, integers are of size 2, and the array is stored starting at location 100 (i.e. A[1][0][2] is stored at location 100), where is element A[3][2][5] stored?

Compute the strides:
  d3 = 2 (base element size)
  d2 = # elements in dimension 3 * d3 = 7 * 2 = 14
  d1 = # elements in dimension 2 * d2 = 5 * 14 = 70


address = 100 + (3 - 1)*70 + (2 - 0)*14 + (5 - 2)*2
        = 100 + 140 + 28 + 6 = 274

With the same assumptions as part (b), what element is stored at location 130?

Since the base address is 100 and the element size is 2, this
must be the 15th element of the array.  You can count this
out or find values (in range for the given dimension)
that fit the equation:

130 = 100 + (i - 1)*70 + (j-0)*14 + (k-1)*2
30 = (i-1)*70 + (j-0)*14 + (k-1)*2
  i = 1, j = 2, and k = 2 work here.
A[1][2][2]