In the following Pascal program segment, what is the value of X after the execution of the program segment?
X:=-10; Y:=20;
If X > Y then if X < 0 then X:=abs(X) else X:=2*X;

Which of the following strings can definitely be said to be tokens without looking at the next input character while compiling a Pascal program?

 I. begin           II. program           III. <>   

Assume that X and Y are non-zero positive integers. What does the following Pascal program segment do?

 while X <>Y do
 if X > Y then
    X := X – Y
 else
    Y := Y – X;
 write(X); 

What is the value of X printed by the following program?

 program COMPUTE (input, output);
    var
                  X:integer;
    procedure FIND (X:real);
                  begin
                  X:=sqrt(X);
                  end;
   begin
                  X:=2
                  Find(X)
                  Writeln(X)
   end 

A variant record in Pascal is defined by

                    type varirec      =   record
                                          number : integer;
                                          case (var1,var2) of
                                          var1: (x,y : integer);
                                          var2: (p.q.: real)
                        end
                    end   

Suppose an array of 100 records was declared on a machine which uses 4 bytes for an integer and 8 bytes for a real. How much space would the compiler have to reserve for the array?

Consider the following high level program segment. Give the contents of the memory locations for variables W, X, Y and Z after the execution of the program segment. The values of the variables A and B are 5 CH and 92H, respectively. Also indicate error conditions if any.

 var
     A, B, W, X, Y   :unsigned byte;
     Z               :unsigned integer, (each integer is represented by two bytes)
 begin
     X               :=A+B
     Y               :=abs(bA-b);
     W               :=A-B
     Z               :=A*B
 End;  

(a) Consider the following Pascal function where A and B are non-zero positive integers. What is the value of GET(3,2)?

 function GET(A,B:integer);integer;
 begin
          if B = 0 then
    GET:=1
 else if A < B then
    GET:=0
 else
    GET:=GET(A-1,B)+GET(A-1,B-1)
 end ; 

(b) The Pascal procedure given for computing the transpose of an N × N (N>1) matrix A of integers has an error. Find the error and correct it.
Assume that the following declaration are made in the main program

 const
     MAXSIZE=20;
 type
     INTARR=array [1.,MAXSIZE,1..MAXSIZE] of integer;
 Procedure TRANSPOSE (var A: INTARR; N : integer);
 var
     I, J, TMP, integer;
 begin
     for I:=1 to NO – 1 do
     for J:=1 to N do
     begin
          TMP: = A[I,J];
          A[I,J]:=A[J,I];
          A(J,I):=TMP
     end
 end; 

(a) Using the scope rules of Pascal determine the declaration that apply to each occurrence of the names A and B in the following program segment.

 procedure T(U, V, X, Y: integer);
 var
    A: record
        A, B : integer
    end; 
    B: record
        B, A : integer
    end;
 begin
    with A do
        begin
          A:=4;
          B:=V
 end;
 with B do
    begin
         A:=X;
         B:=Y
    end
 end; 

(b) Find the lexical errors in the following Pascal statement:

     if A > 1, then B = 2.5A else read (C);  

The following is an incomplete Pascal function to convert a given decimal integer (in the range -8 to +7) into a binary integer in 2’s complement representation. Determine the expression A, B, C that complete program.

 function TWOSCOMP (N:integer):integer;
 var
 RAM, EXPONENT:integer;
 BINARY :integer;
 begin
 if(N>=-8) and (N<=+7) then
     begin
 if N<0 then
     N : = A;
 BINARY:=0;
 EXPONENT:=1;
 while N<>0 do
     begin
       REM:=N mod 2;
       BINARY:=BINARY + B*EXPONENT;
       EXPONENT:=EXPONENT*10;
       N := C
     end
 TWOSCOMP:=BINARY
 end
 end; 

Consider the following C functions.

  int fun1 (int n)  {                              int fun2 (int n)  {
    static int i = 0;                                static int i = 0;  
    if (n > 0)  {                                    if (n > 0)  { 
       ++i;                                              i = i + fun1 (n);   
       fun1 (n-1);                                       fun2 (n-1);
    }                                                 }
    return (i);                                           return (i);
  }                                                 } 

The return value of fun2 (5) is _______.

Given the following Pascal-like program segment

Procedure A;
  x,y: integer;
   Procedure B;
    x,z: real
    S1
   end B;
   Procedure C;
      i: integer;
       S2
   end C;
end A; 

The variables accessible in S1 and S2 are

Consider the following C program.

#include <stdio.h>
struct Outnode { 
    char x, y, z ; 
};

int main () { 
    struct Outnode p = {'1', '0', 'a'+2} ; 
    struct Outnode *q = &p ; 
    printf ("%c, %c", *((char*)q+1), *((char*)q+2)) ; 
    return 0 ; 
} 

The output of this program is

Consider the following C program:

#include<stdio.h> 

void fun1(char *s1, char *s2)  { 
     char *tmp;  
     tmp = s1; 
     s1 = s2; 
     s2 = tmp; 
} 

void fun2(char **s2, char **s2)  { 
     char *tmp; 
     tmp = *s1; 
     *s1 = *s2; 
     *s2 = tmp; 
}

int main ()  { 
     char *str1 = "Hi", *str2 = "Bye"; 
     fun1(str1, str2);     printf("%s %s", str1, str2); 
     fun2(&str1, &str2);   printf("%s %s", str1, str2); 
     return 0; 
} 

The output of the program above is

Consider the following C code. Assume that unsigned long int type length is 64 bits.

        unsigned long int fun(unsigned long int n)  { 
            unsigned long int i, j, j=0, sum = 0; 
            for (i = n; i > 1; i = i/2) j++; 
            for ( ; j > 1; j = j/2) sum++; 
            return sum; 
        } 

The value returned when we call fun with the input 2⁴⁰ is

Consider the following program written in pseudo-code. Assume that x and y are integers.

Count (x, y)  {
    if (y != 1)  {
        if (x != 1)  {
            print("*") ;
            Count (x/2, y) ;
        }
        else  {
             y = y - 1 ; 
             Count (1024, y) ; 
        }
    }
}

The number of times that the print statement is executed by the call Count(1024, 1024) is ______.

What value would the following function return for the input x=95?

                 Function fun (x:integer):integer;
      Begin
                 If x > 100 then fun = x - 10
                 Else fun = fun(fun(x + 11))
      End;  

What is the result of the following program?

program side-effect (input, output);
    var x, result: integer;
    function f (var x:integer):integer;
    begin
        x:x+1;f:=x;
    end;
begin
    x:=5;
    result:=f(x)*f(x);
    writeln(result);
end; 

Given the programming constructs (i) assignment (ii) for loops where the loop parameter cannot be changed within the loop (iii) if-then-else (iv) forward go to (v) arbitrary go to (vi) non-recursive procedure call (vii) recursive procedure/function call (viii) repeat loop, which constructs will you not include in a programming language such that it should be possible to program the terminates (i.e., halting) function in the same programming language.

Consider the following C function definition.

int Trial (int a, int b, int c)
{
    if ((a >= b) && (c < b) return b;
    else if (a >= b) return Trial(a, c, b);
    else return Trial(b, a, c);
} 

The function Trial:

Suppose we have a function HALTS which when applied to any arbitrary function f and its arguments will say TRUE if function f terminates for those arguments and FALSE otherwise. Example, Given the following function definition.
FACTORIAL (N) = IF(N=0) THEN 1 ELSE N*FACTORIAL (N-1)
Then HALTS(FACTORIAL 4) = TRUE and HATS(FACTORIAL - 5) = FALSE

Let us define the function FUNNY(f) = IF HALTS(ff) THEN not(ff) ELSE TRUE
(a) Show that FUNNY terminates for all functions f.
(b) Use (a) to prove (by contradiction) that it is not possible to have a function like HALTS which for arbitrary functions and inputs says whether it will terminate on that input or not.

What will be the output of the following program assuming that parameter passing is

(i) call by value
(ii) call by reference
(iii) call by copy restore

      procedure P{x, y, z};
                        begin y:y+1; z: x+x end; 
      begin
                        a:= b:= 3;
                        P(a+b, a, a);
                        Print(a) 
      end. 

Consider the following pascal program skeleton:

                        var a,x,...;
                        procedure readarray;
                              var i,....;
                              begin
                                             ...:=a...
                              end;
      procedure exchange(...);
                       begin
                                             ...:=a...
                                             ...:=x...
                       end;
      procedure qsort(...);
                       var k,v,...;
                       function partition (...)...;
                              var i,j,...;
                              begin
                              ...:=a...
                              ...:=v...
                              end;
                       begin
                              .
                              .
                       end;
      begin
                       .
                       .
      end; 

Assume that at a given point in time during program execution, following procedures are active: sort, qsort(1,9), qsort(1.3), partition(1,3), exchange(1,3).

Show snapshots of the runtime stack with access links after each of the activations.

The following C declarations

struct node
{
   int i;
   float j;
};
struct node *s[10]; 

define s to be

The most appropriate matching for the following pairs

    X: m=malloc(5); m= NULL;        1: using dangling pointers
    Y: free(n); n->value=5;         2: using uninitialized pointers
    Z: char *p; *p = ’a’;           3. lost memory  

is:

Aliasing in the context of programming languages refers to

Consider the following C declaration

struct {
    short s [5]
    union {
    float y;
    long z;
    } u;
} t; 

Assume that objects of the type short, float and long occupy 2 bytes, 4 bytes and 8 bytes, respectively. The memory requirement for variable t, ignoring alignment considerations, is

The value of j at the end of the execution of the following C program.

int incr(int i)
{
   static int count = 0;
   count = count + i;
   return (count);
}
main()
{
   int i,j;
   for (i = 0; i <= 4; i++)
      j = incr(i);
} 

is

Consider the following program is pseudo-Pascal syntax.

    program main;
       var x: integer;
       procedure Q [z:integer);
       begin
          z: z + x;
          writeln(z)
       end;
    procedure P (y:integer);
          var x: integer;
       begin
          x: y + 2;
          Q(x);
          writeln(x)
       end;
       begin 
          x:=5;
          P(x);
          Q(x);
          writeln(x)
       end. 

What is the output of the program, when
(a) The parameter passing mechanism is call-by-value and the scope rule is static scooping?

(b) The parameter passing mechanism is call-by-reference and the scope rule is dynamic scooping?

What is printed by the print statements in the program P1 assuming call by reference parameter passing?

Program P1()
{
   x = 10;
   y = 3;
   func1(y,x,x);
   print x;
   print y;
}
func1(x,y,z)
{
   y = y+4;
   z = x+y+z;
} 

Consider the following program

                Program P2 
                   var n: int: 
                   procedure W(var x: int) 
                   begin 
                          x=x+1; 
                          print x;   
                   end 

                procedure D 
                begin  
                          var n: int; 
                          n=3; 
                          W(n);  
                End 
            begin               //beginP2 
                n=10; 
                D; 
            end  

If the language has dynamic scoping and parameters are passed by reference, what will be printed by the program?

Consider the following C program:

           void abc(char*s)
           {
              if(s[0]==’\0’)return;
              abc(s+1);
              abc(s+1);
              printf(“%c”,s[0]);
           }
           main()
           {  abc(“123”)
           }  

(a) What will be the output of the program?
(b) If abc(s) is called with a null-terminated string s of length n characters (not counting the null (‘\0’) character), how many characters will be printed by abc(s)?

In the C language

Consider the following logic program P

A(x) <- B(x, y), C(y) 
     <- B(x,x) 

Which of the following first order sentences is equivalent to P?

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

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:

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:

Px = new Q();
Qy = new Q();
Pz = 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

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

Consider the C program shown below.

#include 
#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

The goal of structured programming is to:

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.

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

Consider the following program fragment for reversing the digits in a given integer to obtain a new integer. Let n = d₁d₂…d_m.

     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:

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

It is desired to design an object-oriented employee record system for a company. Each employee has a name, unique id and salary. Employees belong to different categories and their salary is determined by their category. The functions to get Name, getld and compute salary are required. Given the class hierarchy below, possible locations for these functions are:

(i) getld is implemented in the superclass
(ii) getld is implemented in the subclass
(iii) getName is an abstract function in the superclass
(iv) getName is implemented in the superclass
(v) getName is implemented in the subclass
(vi) getSalary is an abstract function in the superclass
(vii) getSalary is implemented in the superclass
(viii) getSalary is implemented in the subclass

Choose the best design

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