Homework Assignment #2: Control, Subprograms, Scoping, ...
(due on 06/05 at the beginning of class)



Problem #1: Iterative Control Structures

Consider the following ALGOL 60-style for statement:
for i := j + 1 step i * j until 3*j do j := j + 1
The semantics should be clear, but you can refer to page 327 of the book for all details. Assume that the initial value of j is 1. List the sequence of values taken by variable i, assuming the following semantics:
a. (10 points) All expressions are evaluated once at the loop entry

i = 2
i = 4

b. (10 points) All expressions are evaluated before each iteration

i = 2
i = 4
i = 12

c. (10 points) step expressions are evaluated once at loop entry, and until expressions are evaluated before each iteration

i = 2
i = 4
i = 6
i = 8
...

d. (10 points) until expressions are evaluated once at loop entry, and step expressions are evaluated before each iteration, just after the loop counter is incremented

i = 2
i = 4

In all cases, when more than one expression is evaluated at the same time, they are evaluated in left-to-right order. Also, the assignment is always done as soon as its RHS (Right-Hand-Side) is evaluated.

Problem #2: Parameter passing

Consider the following program written in C syntax:
void main() {
    int value = 2, list[5] = {1,3,4,7,9};

    swap(value,list[0]);
    printf("%d and [%d,%d,%d,%d,%d]\n",
                value,list[0],list[1],list[2],list[3],list[4]);

    swap(list[0],list[1]);
    printf("%d and [%d,%d,%d,%d,%d]\n",
                value,list[0],list[1],list[2],list[3],list[4]);

    swap(value,list[value]);
    printf("%d and [%d,%d,%d,%d,%d]\n",
                value,list[0],list[1],list[2],list[3],list[4]);

return 0;
}

void swap(int a, int b) {
    int temp;

    temp = a;
    a = b;
    b = temp;
}
 For each of the following parameter-passing methods, what is the output of the program:
a. (10 pts)    Passed by value.

2 and [1,3,4,7,9]
2 and [1,3,4,7,9]
2 and [1,3,4,7,9]

b. (10 pts)    Passed by reference.

1 and [2,3,4,7,9]
1 and [3,2,4,7,9]
2 and [3,1,4,7,9]

c. (10 pts)    Passed by name.

1 [2,3,4,7,9]
1 [3,2,4,7,9]
2 [3,2,1,7,9]

d. (10 pts)    Passed by value-result.

1 and [2,3,4,7,9]
1 and [3,2,4,7,9]
2 and [3,1,4,7,9]

Problem #3: Scoping

Consider the following program written in C-like syntax:
int x;

void set_x(int n) {
    x = n;
}

void print_x() {
   printf("%d\n",x);
}

void first() {
   set_x(1);
   print_x();
}

void second() {
   int x;
   set_x(2);
   print_x();
}

int main() {
   set_x(0);
   first();
   print_x();
   second();
   print_x();
}
 a. (20 points) What does this program print if the language uses static scoping?

1
1
2
2

b. (20 points) What does this program print if the language uses dynamic scoping?

1
1
2
1

Problem #4: Functions as Parameters

Consider the following program written in C-like syntax:
int x;

void set_x(int n) {
    x = n;
}

void print_x() {
   printf("%d\n",x);
}

void foo(void(S*)(int m), void(P*)(void), int n) {
   int x = 0;
   if ((n == 1) || (n == 3))
      set_x(n);
   else
      S(n);

   if ((n == 1) || (n == 2))
      print_x;
   else
      P();
}

int main() {
   set_x(0); foo(set_x,print_x,1); print_x();
   set_x(0); foo(set_x,print_x,2); print_x();
   set_x(0); foo(set_x,print_x,3); print_x();
   set_x(0); foo(set_x,print_x,4); print_x();
}
 Assume that the language uses dynamic scoping.
a. (20 points) What does the program print if the language uses shallow binding?

Under shallow binding all called to print_x and set_x use dynamic scoping

1
0
2
0
3
0
4
0

b. (20 points) What does the program print if the language uses deep binding?

Under deep binding the called to print_x and set_x as P() and S() use static scoping.

1
0
0
2
0
0
4
4