Programming
Question 1 
int SimpleFunction (int Y[], int n, int x)
{
int total = Y[0], loopIndex;
for (loopIndex = 1; loopIndex <= n  1; loopIndex++)
total = x * total + Y[loopIndex]
return total;
}
Let Z be an array of 10 elements with Z[i]=1, for all i such that 0 ≤ i ≤ 9. The value returned by SimpleFunction(Z, 10, 2) is _______
A  1023 
n=10,x=2
Initial total value is 1 => total=1.
For loop will execute 9 times.
loopindex=1, 1<=9 condition is true then
total = x * total + Y[loopIndex]= 2*1+Y[1]=2+1=3
loopindex=2, 2<=9 condition is true then
total=2*3+Y[2]=6+1=7
loopindex=3, 3<=9 condition is true then
total=2*7+Y[3]=14+1 =15
loopindex=4, 4<=9 condition is true then
total= 2*15+Y[4]=30+1=31
loopindex=5, 5<=9 condition is true then
total=2*31+Y[5]=62+1=63
loopindex=6, 6<=9 condition is true then
total=2*63+Y[6]=126+1=127
loopindex=7, 7<=9 condition is true then
Total =2*127+Y[7]=254+1=255
loopindex=8, 8<=9 condition is true then
total=2*255+Y[8]=510+1=511
loopindex=9, 9<=9 condition is true then
total=2*511+Y[9]=1022+1=1023
loopindex=10, 10<=9 condition is false then
Total value is returned which is 1023.
You can also write generalized formulae 2101=1023
Question 2 
# include <stdio.h>
int main()
{
int i, j, count;
count = 0;
i = 0;
for (j = 3; j <= 3; j++)
{
if ((j >= 0) && (i++))
count = count + j;
}
count = count + i;
printf(“%d”, count);
return 0;
}
Which one of the following options is correct?
A  The program will compile successfully and output 13 when executed. 
B  The program will compile successfully and output 10 when executed. 
C  The program will compile successfully and output 8 when executed. 
D  The program will not compile successfully. 
Input: count=0 , i=0 and j=3
For(j = 3; j <= 3; j++) → Condition TRUE then enters into the loop.
if((j >= 0) && (i++)) → Condition fails because they are given logical AND.
So, we are not entering the “if” condition and come out of the loop. The count and “i” values remain “0”.
For(j = 2; j <= 3; j++) → Condition TRUE then enters into the loop.
if((j >= 0) && (i++)) → Condition fails because they are given logical AND.
So, we are not entering the “if” condition and come out of the loop. The count and “i” values remain “0”.
For(j = 1; j <= 3; j++) → Condition TRUE then enters into the loop.
if((j >= 0) && (i++)) → Condition fails because they are given logical AND.
So, we are not entering the “if” condition and come out of the loop. The count and “i” values remain “0”.
For(j = 0; j <= 3; j++) → Condition TRUE then enters into the loop.
if((j >= 0) && (i++)) → Condition TRUE then enters into the loop.
count=0+0 → Count=0
For(j = 1; j <= 3; j++) → Condition TRUE then enters into the loop.
if((j >= 0) && (i++)) → Condition TRUE then enters into the loop.
count=0+1 → Count=1
For(j = 2; j <= 3; j++) → Condition TRUE then enters into the loop.
if((j >= 0) && (i++)) → Condition TRUE then enters into the loop.
count=1+2 → Count=3
For(j = 3; j <= 3; j++) → Condition TRUE then enters into the loop.
if((j >= 0) && (i++)) → Condition TRUE then enters into the loop.
count=3+3 → Count=6
For(j = 4; j <= 3; j++) → Condition FALSE we are not entering the loop.
count=6+4 → We are given a condition as a post increment. So, “i” updates the next instruction.
The above code segment executes successfully and will print value=10.
Question 3 
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;
A  10 
B  20 
C  10 
D  None 
X = 10
Question 4 
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. <>
A  I 
B  II 
C  III 
D  All of the above 
Question 5 
Assume that X and Y are nonzero 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);
A  Computes the LCM of two numbers 
B  Divides the larger number by the smaller number 
C  Computes the GCD of two numbers 
D  None of the above 
1^{st} pass : X=3 and Y=2
2^{nd} pass : X=1 and Y=2
3^{rd} pass : X=1 and Y=1
Write(X), which writes 1. Which is nothing but GCD of 3 & 5.
Question 6 
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  2 
B  √2 
C  Run time error 
D  None of the above 
X in the procedure FIND is a local variable. No change will be reflected in global variable X.
Question 7 
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?
A  2800 
B  2400 
C  2000 
D  1200 
Question 8 
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(bAb); W :=AB Z :=A*B End;
A  Theory Explanation. 
Question 9 
(a) Consider the following Pascal function where A and B are nonzero 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(A1,B)+GET(A1,B1) 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  Theory Explanation. 
Question 10 
(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);
A  Theory Explanation. 
Question 11 
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;
A  Theory Explanation. 
Question 12 
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 (n1); fun2 (n1); } } return (i); return (i); } }
The return value of fun2 (5) is _______.
A  55 
int fun1(int n) {
printf("fun1 call\n");
static int i = 0;
if(n>0){
++i;
printf("fun1(%d1)\n",n);
fun1(n1);
}
printf("fun1(%d)= %d\n",n, i);
return(i);
}
int fun2(int n) {
printf("\n******* fun2 call ********\n");
static int i = 0;
if(n>0){
printf("%d + fun1(%d)\n", i,n);
i=i+fun1(n);
fun2(n1);
}
printf("fun2(%d)= %d\n",n, i);
return(i);
}
void main()
{
printf("final = %d\n", fun2(5));
}
Check step by step hand run of the code to understand the recursion:
******* fun2 call ********
0 + fun1(5)
fun1 call
fun1(51)
fun1 call
fun1(41)
fun1 call
fun1(31)
fun1 call
fun1(21)
fun1 call
fun1(11)
fun1 call
fun1(0)= 5
fun1(1)= 5
fun1(2)= 5
fun1(3)= 5
fun1(4)= 5
fun1(5)= 5
******* fun2 call ********
5 + fun1(4)
fun1 call
fun1(41)
fun1 call
fun1(31)
fun1 call
fun1(21)
fun1 call
fun1(11)
fun1 call
fun1(0)= 9
fun1(1)= 9
fun1(2)= 9
fun1(3)= 9
fun1(4)= 9
******* fun2 call ********
14 + fun1(3)
fun1 call
fun1(31)
fun1 call
fun1(21)
fun1 call
fun1(11)
fun1 call
fun1(0)= 12
fun1(1)= 12
fun1(2)= 12
fun1(3)= 12
******* fun2 call ********
26 + fun1(2)
fun1 call
fun1(21)
fun1 call
fun1(11)
fun1 call
fun1(0)= 14
fun1(1)= 14
fun1(2)= 14
******* fun2 call ********
40 + fun1(1)
fun1 call
fun1(11)
fun1 call
fun1(0)= 15
fun1(1)= 15
******* fun2 call ********
fun2(0)= 55
fun2(1)= 55
fun2(2)= 55
fun2(3)= 55
fun2(4)= 55
fun2(5)= 55
final = 55
Question 13 
int tob (int b, int* arr) {
int i;
for (i=0; b>0; i++) {
if (b%2) arr [i] = 1;
else arr [i] = 0;
b = b/2;
}
return (i);
}
int pp (int a, int b) {
int arr [20];
int i, tot = 1, ex, len;
ex = a;
len = tob (b,arr);
for (i=0; i
tot = tot * ex;
ex = ex * ex;
}
return (tot);
}
The value returned by pp(3,4) is ________.
A  81 
a=3,b=4
tot=1
ex=a=3
len=tob(b,arr) which is 3
[
tob(4,arr)==>
b=4
b%2 =4%2=0 Condition is false then arr[0]=0
=> b=b/2 =4/2 =2
b=2
b%2 =2%2=0 condition is false then arr[1]=0
=>b=b/2=2/2=1
b=1
then b%2=1%2 condition is true then arr[2]=1
=>b=b/2=1/2=0
The i value is 3 [length is 3]
]
i=0,
arr[0] ==1 condition is false
ex=3*3=9
i=1
arr[1]==1 condition is false
then
ex=9*9=81
i=2
then arr[2]==1 condition is true
tot=tot*ex=1*81=81
ex=81*81
Finally it returns tot value which 81.
Question 14 
Given the following Pascallike 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
A  x or A, y, x of B and z in S1 and x of B, y and i in S2 
B  x or B, y and z in S1 and x of B, i and z in S2 
C  x or B, z and y in S1 and x of A, i and y in S2 
D  None of the above 
Question 15 
Match the pairs in the following questions:
(a) Small talk (p) Logic programming (b) LISP (q) Data flow programming (c) Prolog (r) Functional programming (d) VAL (s) Objectoriented programming
A  (a)  (s), (b)  (r), (c)  (p), (d)  (q) 
Question 16 
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;
A  89 
B  90 
C  91 
D  92 
fun(95) = fun(fun(106))
= fun(96)
= fun(fun(107))
= fun(97)
= fun(fun(108))
= fun(98)
= fun(fun(109))
= fun(99)
= fun(110)
= fun(100)
= fun(fun(111))
= fun(101)
= 91
Question 17 
What is the result of the following program?
program sideeffect (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;
A  5 
B  25 
C  36 
D  42 
If it is call by value then answer is 36.
Question 18 
Given the programming constructs (i) assignment (ii) for loops where the loop parameter cannot be changed within the loop (iii) ifthenelse (iv) forward go to (v) arbitrary go to (vi) nonrecursive 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.
A  (ii), (iii), (iv) 
B  (v), (vii), (viii) 
C  (vi), (vii), (viii) 
D  (iii), (vii), (viii) 
Question 19 
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:
A  Finds the maximum of a, b, and c 
B  Finds the minimum of a, b and c 
C  Finds the middle number of a, b, c 
D  None of the above 
Question 20 
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 (N1)
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.
A  Theory Explanation. 
Question 21 
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.
A  Theory Explanation. 
Question 22 
Consider the following pascal program skeleton:
program sort(...);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.
A  Theory Explanation. 
Question 23 
The following C declarations
struct node { int i; float j; }; struct node *s[10];
define s to be
A  An array, each element of which is a pointer to a structure of type node 
B  A structure of 2 fields, each field being a pointer to an array of 10 elements 
C  A structure of 3 fields: an integer, a float, and an array of 10 elements 
D  An array, each element of which is a structure of type node 
Question 24 
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:
A  X – 1 Y – 3 Z – 2 
B  X – 2 Y – 1 Z – 3 
C  X – 3 Y – 2 Z – 1 
D  X – 3 Y – 1 Z – 2 
Y → n is pointer to invalid memory, a making it as a dangling pointer.
Z → p is not initialized.
p = malloc (size of(char))p = malloc (size of(char)); should have been used before assigning 'aa' to ∗p.
Question 25 
Aliasing in the context of programming languages refers to
A  multiple variables having the same memory location 
B  multiple variables having the same value 
C  multiple variables having the same identifier 
D  multiple uses of the same variable 
Question 26 
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
A  22 bytes 
B  14 bytes 
C  18 bytes 
D  10 bytes 
max[float, long] = max [4, 8] = 8
Total = short[5] + max[float,long] = 10 + 8 = 18
Question 27 
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
A  10 
B  4 
C  6 
D  7 
i=1; count=1
i=2; count=3
i=3; count=6
i=4; count=10
It return count value is 10.
Question 28 
Consider the following program is pseudoPascal 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 callbyvalue and the scope rule is static scooping?
(b) The parameter passing mechanism is callbyreference and the scope rule is dynamic scooping?
A  Theory Explanation is given below. 
Question 29 
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; }
A  10, 3 
B  31, 3 
C  27, 7 
D  None of the above 
And variable y and z of func1 points to address of variable x.
Therefore, y = y+4 ⇒ y = 10+4 = 14
and z = x+y+z ⇒ z = 14+14+3 = 31
z will be stored back in k.
Hence, x=31 and y will remain as it is (y=3).
Hence, answer is (B).
Question 30 
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?
A  10 
B  11 
C  3 
D  None of the above 
W(n)=W(3)
Procedure W(var x; int)
begin
x = x+1 = 3+1 = 4
Print x → Print x=4
end
Question 31 
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 nullterminated string s of length n characters (not
counting the null (‘\0’) character), how many characters will be printed by abc(s)?
A  Theory Explanation is given below. 
Question 32 
In the C language
A  At most one activation record exists between the current activation record and the activation record for the main 
B  The number of activation records between the current activation record and the activation record for the main depends on the actual function calling sequence. 
C  The visibility of global variables depends on the actual function calling sequence. 
D  Recursion requires the activation record for the recursive function to be saved on a different stack before the recursive fraction can be called. 
Question 33 
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?
A  (∀x) [(∃y) [B(x,y) ∧ C(y)] ⇒ A(x)] ∧ ¬(∃x)[B(x,x)] 
B  (∀x) [(∀y) [B(x,y) ∧ C(y)] ⇒ A(x)] ∧ ¬(∃x)[B(x,x)] 
C  (∀x) [(∃y) [B(x,y) ∧ C(y)] ⇒ A(x)] ∨ ¬(∃x)[B(x,x)] 
D  (∀x) [(∀y) [B(x,y) ∧ C(y)] ⇒ A(x)] ∧ (∃x)[B(x,x)] 
Question 34 
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
A  115, 220 
B  25, 220 
C  25, 15 
D  115, 105 
P(100+5) = P(105)
→void P(105)
{
int i=10;
print (x+10); ⇒ 105+10=115 prints
i=200;
j = 20;
print (x); ⇒ x=105 prints
}
115, 105 prints
Question 35 
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:
A  115, 220 
B  25, 220 
C  25, 15 
D  115, 105 
In void P(x)
{ int i = 10;
print(x + 10); ⇒ 105+10 = 115 prints
print (x); ⇒ print x=220;
Question 36 
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
A  1 2 1 
B  2 1 1 
C  2 1 2 
D  2 2 2 
Note: The given question is not in the present syllabus
Question 37 
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*⌈log_{2}(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
A  {mm ≤ n, (∃i)[m=i!]} 
B  {mm ≤ n, (∃i)[m=i^{2}]} 
C  {mm ≤ n, m is prime}

D  { } 
Now Trace the code,
for (k=3; k<=n; k++)
A[k]=0; // A[3]=0
A[4]=0
for (k=2; k<=Two log_n; k++)
for(j=k+1; j<=n; j++)
A[j] = A[j] // (j%k); // A[3] = 0 // I=1
A[4] = 0 // I=1
for (j=3; j<=n; j++)
if (!A[j]) printf("%d", j);
// if (!1) means if (0), so printf will never execute
Hence, Option (D) is the answer.
Question 38 
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
A  12 7 6

B  22 12 11 
C  14 6 6 
D  7 6 6 
p(&x) it goes to P( ) function
y=5
x=5+2=7;
Q(x)
z=7
z=7+5=12(Print+z→I)
comes to P( )
*y=71=6
x=7(Print x→II)
comes to main ( ),
print x=*y=6 (print x→III)
Output: 12 7 6
Question 39 
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.
A  call swap (x, y)

B  call swap (&x, &y)

C  swap (x,y) cannot be used as it does not return any value

D  swap (x,y) cannot be used as the parameters are passed by value

Here parameters passed by value in C then there is no change in the values.
Option B:
Here values are not swap.
Here parameters are passed by address in C.
Option C:
It is false. Return value is not valid for exchanging the variables.
Option D:
It is correct.
We cannot use swap(x,y) because parameters are passed by value.
Only exchanging the values (or) variables are passing their address and then modify the content with the help of dereferencing operator(*).
Question 40 
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
A  5 
B  6 
C  7

D  8 
The value return by f(1) = 7
Question 41 
Consider the following program fragment for reversing the digits in a given integer to obtain a new integer. Let n = d_{1}d_{2}…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:
A  n = d_{1}d_{2}…d_{mi} and rev = d_{m}d_{m1}…d_{mi+1}

B  n = d_{mi+1}…d_{m1}d_{m} or rev = d_{mi}…d_{2}d_{1}

C  n ≠ rev 
D  n = d_{1}d_{2}…d_{m} and rev = d_{m}…d_{2}d_{1} 
Question 42 
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
A  gnirts 
B  string

C  gnirt 
D  no output is printed 
P[0] = S[71] = S[6] = \0.
In P[ ], the first character is '\0'. Then it will results a empty string. If P[0] become '\0', then it doesn't consider about next values in sequence.
Question 43 
It is desired to design an objectoriented 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
A  (i), (iv), (vi), (viii) 
B  (i), (iv), (vii) 
C  (i), (iii), (v), (vi), (viii) 
D  (ii), (v), (viii) 
Question 44 
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
A  x + y using repeated subtraction 
B  x mod y using repeated subtraction 
C  the greatest common divisor of x and y 
D  the least common multiple of x and y 
Question 45 
Choose the best matching between the programming styles in Group 1 and their characteristics in Group 2.
Group1 Group2 P. Functional 1. Commandbased, procedural Q. Logic 2. Imperative, abstract data type R. Objectoriented 3. Sideeffect free, declarative, expression evaluation S. Imperative 4. Declarative, clausal representation, theorem proving
A  P  2, Q  3, R  4, S  1 
B  P  4, Q  3, R  2, S  1 
C  P  3, Q  4, R  1, S  2 
D  P  3, Q  4, R  2, S  1 
Q) Logic is also declarative but involves theorem proving.
R) Object oriented is imperative statement based and have abstract data types.
S) Imperative programs are made giving commands and follow definite procedure.
Question 46 
What function of x, n is computed by this program?
Function what (x, n:integer): integer: Var value : integer; begin value:=1 if n>0 then begin if n mod 2 = 1 then value:=value*x; value:=value*what(x*x, n div 2); end; what:value end;
A  Theory Explanation. 
Question 47 
Consider the program below:
Program main; var r:integer; procedure two; begin write (r) end; procedure one; var r:integer; begin r:=5 two; end begin r:=2; two; one; two; end.
What is printed by the above program if
(i) Static scoping is assumed for all variables;
(ii) Dynamic scoping is assumed for all variables.
Give reasons for your answer.
A  Theory Explanation. 
Question 48 
An array a contains n integers in nondecreasing order, A[1] ≤ A[2] ≤ ... ≤ A[n]. Describe, using Pascal like pseudo code, a linear time algorithm to find i, j, such that A[i] + A[j] = a given integer M, if such i, j exist.
A  Theory Explanation. 
Question 49 
(a) Draw a precedence graph for the following sequential code. The statements are numbered from S_{1} to S_{6}
S_{1} read n S_{2} i:=1 S_{3} if i>n goto next S_{4} a(i):=i+1 S_{5} i:=i+1 S_{6} next : Write a(i)
(b) Can this graph be converted to a concurrent program using parbeginparend construct only?
A  Theory Explanation. 
Question 50 
Program PARAM (input, output); var m, n : integer; procedure P (var, x, y : integer); var m : integer; begin m : = 1; x : = y + 1 end; procedure Q (x:integer; vary : integer); begin x:=y+1; end; begin m:=0; P(m,m); write (m); n:=0; Q(n*1,n); write (n) end
The value of m, output by the program PARAM is:
A  1, because m is a local variable in P 
B  0, because m is the actual parameter that corresponds to the formal parameter in p

C  0, because both x and y are just reference to m, and y has the value 0 
D  1, because both x and y are just references to m which gets modified in procedure P 
E  none of the above 