Q. The following program fragment is written in a programming language that allows variables and does not allow nested declarations of functions.

global int i = 100, j = 5;
void P(x)
{
int i = 10;
print(x + 10);
i = 200;
j = 20;
print(x);
}
main()
{
P(i + j);
}

If the programming language uses static scoping and call by need parameter passing mechanism, the values printed by the above program are

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Q. The following program fragment is written in a programming language that allows variables and does not allow nested declarations of functions.

global int i = 100, j = 5;
void P(x)
{
int i = 10;
print(x + 10);
i = 200;
j = 20;
print(x);
}
main()
{
P(i + j);
}

If the programming language uses dynamic scoping and call by name parameter passing mechanism, the values printed by the above program are :

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Q. Consider the following class definitions in a hypothetical Object Oriented language that supports inheritance and uses dynamic binding. The language should not be assumed to be either Java or C++, though the syntax is similar.

Class P
{
void f(int i)
{
print(i);
}
}

Class Q subclass of P
{
void f(int i)
{
print(2*i);
}
}

Now consider the following program fragment:

P x = new Q();
Q y = new Q();
P z = new Q();
x.f(1); ((P)y).f(1); z.f(1);

Here ((P)y) denotes a typecast of y to P. The output produced by executing the above program fragment will be

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Q. In the following C program fragment, j, k n and TwoLog_n are interger variables, and A is an array of integers. The variable n is initialized to an integer ≥ 3, and TwoLog_n is initialized to the value of 2*⌈log2(n)⌉

for (k = 3; k < = n; k++)
A[k] = 0;
for (k = 2; k < = TwoLog_n; k++)
for (j = k + 1; j < = n; j++)
A[j] = A[j] || (j % k);
for (j = 3; j < = n; j++)
if (!A[j]) printf("%d", j);

The set of numbers printed by this program fragment is

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Q. Consider the C program shown below.

#define print(x) printf("%d ", x)
int x;
void Q(int z)
{
z += x;
print(z);
}
void P(int *y)
{
int x = *y + 2;
Q(x);
*y = x - 1;
print(x);
}
main(void)
{
x = 5;
P(&x);
print(x);
}

The output of this program is

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Q. The goal of structured programming is to

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Q. Consider the following C function

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

In order to exchange the values of two variables x and y.

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Q. The following numbers are inserted into an empty binary search tree in the given order: 10, 1, 3, 5, 15, 12, 16. What is the height of the binary search tree (the height is the maximum distance of a leaf node from the root)?

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Q. The best data structure to check whether an arithmetic expression has balanced parentheses is a

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Q. Consider the following C function:

int f(int n)
{
static int i = 1;
if (n >= 5)
return n;
n = n+i;
i++;
return f(n);
}

The value returned by f(1) is

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Q. Consider the following program fragment for reversing the digits in a given integer to obtain a new integer. Let n = D1D2…Dm

int n, rev;
rev = 0;
while (n > 0)
{
rev = rev*10 + n%10;
n = n/10;
}

The loop invariant condition at the end of the ith iteration is:

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Q. Consider the following C program segment:

char p[20];
char *s = "string";
int length = strlen(s);
int i;
for (i = 0; i < length; i++)
p[i] = s[length — i];
printf("%s",p);

The output of the program is

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Q. Consider the label sequences obtained by the following pairs of traversals on a labeled binary tree. Which of these pairs identify a tree uniquely ?

(i) preorder and postorder
(ii) inorder and postorder
(iii) preorder and inorder
(iv) level order and postorder

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Q. Two matrices M1 and M2 are to be stored in arrays A and B respectively. Each array can be stored either in row-major or column-major order in contiguous memory locations. The time complexity of an algorithm to compute M1 × M2 will be

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Q. Consider the following C program

main()
{
int x, y, m, n;
scanf ("%d %d", &x, &y);
/* Assume x > 0 and y > 0 */
m = x;
n = y;
while (m! = n)
{
if (m > n)
m = m - n;
else
n = n - m;
}
print f ("% d", n);
}

The program computes

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Q. What does the following algorithm approximate?

x = m;
y = 1;
while (x - y > e)
{
x = (x + y)/2;
y = m/x;
}
print(x);

(Assume m > 1, e > 0).

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Q. Consider the following C program segment

struct CellNode
{
struct CelINode *leftchild;
int element;
struct CelINode *rightChild;
}

int Dosomething(struct CelINode *ptr)
{
int value = 0;
if (ptr != NULL)
{
if (ptr->leftChild != NULL)
value = 1 + DoSomething(ptr->leftChild);
if (ptr->rightChild != NULL)
value = max(value, 1 + DoSomething(ptr->rightChild));
}
return (value);
}

The value returned by the function DoSomething when a pointer to the root of a non-empty tree is passed as argument is

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Q. The order of an internal node in a B+ tree index is the maximum number of children it can have. Suppose that a child pointer takes 6 bytes, the search field value takes 14 bytes, and the block size is 512 bytes. What is the order of the internal node?

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Q. Let A[1, ..., n] be an array storing a bit (1 or 0) at each location, and f(m) is a unction whose time complexity is θ(m). Consider the following program fragment written in a C like language:

counter = 0;
for (i = 1; i < = n; i++)
{
if (A[i] == 1)
counter++;
else {
f(counter);
counter = 0;
}
}

The complexity of this program fragment is

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Q. The time complexity of the following C function is (assume n > 0)

int recursive (int n) {
if (n == 1)
return (1);
else
return (recursive (n - 1) + recursive (n - 1));
}

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