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TCS NQT Input/Output Quiz 1

Question 1

Time: 00:00:00

What will be the final value of the res when INPUT N1=100?

Function F(INPUT N1)

Static int res=0;

IF(N1>0)

    res=res+1

    F(N1/20)

END IF

return res

END Function

Error

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res is declared as a static int variable, it means that it will initialize only once.

In first call of function F,

N1 = 100

res =0

Since, N1 >0 => res will incremented by 1 (so res =1) and again function F will call by passing (N1/20 => 5)

In second call of function F,

N1 = 5

res =1 (not 0 because it is a static variable and can be initialize only once)

Since, N1 > 0 => res will incremented by 1(so res = 2) and again function F will call by passing (N1/20 => 0)

In third call of function F,

N1 = 0

res =2 (not 0 because it is a static variable and can be initialize only once)

Since, N1 is not greater than 0, function F will return res i.e, 2.

Hence, the output of the given code will be 2.

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Question 2

Time: 00:00:00

What will be the output of the following pseudo code

Integer x,y,z

x=2

y=1

z=5

Do

Print z--

while((x & y)|| (z+1))

End loop

5 4 3 2 1 0

5 4 3 2 1 1

4 3 2 1 0 0

4 3 2 1 1 1

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Initially, x=2 y=1 z=5

In first iteration, first z will be printed (i.e, 5) then decremented by 1, so z=4,
Now, in the condition, & is the bitwise & operator and will return 1 if and only if both bits are same, while || is the logical operator and will return 1 if any of the condition is true or both are true, (x&y)||(z+1) => (2&1) || (5) => 0 || 5 => 1
Since the condition is true,

So in Second iteration, first z will be printed (i.e, 4) then decremented by 1, so z=3,
Now, the condition, (x&y)||(z+1) => (2&1) || (4) => 0 || 4 => 1
Since the condition is true,

So in third iteration, first z will be printed (i.e, 3) then decremented by 1, so z=2,
Now, the condition, (x&y)||(z+1) => (2&1) || (3) => 0 || 3 => 1
Since the condition is true,

So in fourth iteration, first z will be printed (i.e, 2) then decremented by 1, so z=1,
Now, the condition, (x&y)||(z+1) => (2&1) || (2) => 0 || 2 => 1
Since the condition is true,

So in fifth iteration, first z will be printed (i.e, 1) then decremented by 1, so z=0,
Now, the condition, (x&y)||(z+1) => (2&1) || (1) => 0 || 1 => 1
Since the condition is true,

So in sixth iteration, first z will be printed (i.e, 0) then decremented by 1, so z=-1,
Now, the condition, (x&y)||(z+1) => (2&1) || (0) => 0 || 0 => 0
Since the condition is false, the loop gets terminated.

Hence the output will be ,5 4 3 2 1 0

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Question 3

Time: 00:00:00

What is the output of the following code?



Set integer _no1=0, _no2=1, _no3=1, _lower=6, _inc=1

while _inc is less than equals to _lower:

  _inc is increment by 1

  Print _no3

  _no3=_no1+_no2

  _no1 equals to _no2

  _no2 equals to _no3

End While

1 2 3 5 8

3 5 8

1 1 2 3 5 8

2 3 5 8

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Initially, _no1=0, _no2=1, _no3=1, _lower=6, _inc=1

Since,( _inc <= _lower ) => (1<=6), The condition of the while loop is true,
So, in first iteration, _inc is incremented by 1 => _inc = 2,
_no3 will gets printed (i.e 1), and _no = _no1 + _no2 => _no3 =(0+1) = 1,
And, _no1=1, _no2 = 1

Since,( _inc <= _lower ) => (2<=6), The condition of the while loop is true,
So, in second iteration, _inc is incremented by 1 => _inc = 3,
_no3 will gets printed (i.e 1), and _no = _no1 + _no2 => _no3 =(1+1) = 2,
And, _no1=1, _no2 = 2

Since,( _inc <= _lower ) => (3<=6), The condition of the while loop is true,
So, in third iteration, _inc is incremented by 1 => _inc = 3,
_no3 will gets printed (i.e 2), and _no = _no1 + _no2 => _no3 =(1+2) = 3,
And, _no1=2, _no2 = 3

Since,( _inc <= _lower ) => (4<=6), The condition of the while loop is true,
So, in fourth iteration, _inc is incremented by 1 => _inc = 5,
_no3 will gets printed (i.e 3), and _no = _no1 + _no2 => _no3 =(2+3) = 5,
And, _no1=3, _no2 = 5

Since,( _inc <= _lower ) => (5<=6), The condition of the while loop is true,
So, in fifth iteration, _inc is incremented by 1 => _inc = 6,
_no3 will gets printed (i.e 5), and _no = _no1 + _no2 => _no3 =(3+5) = 8,
And, _no1=5, _no2 = 8

Since,( _inc <= _lower ) => (6<=6), The condition of the while loop is true,
So, in first iteration, _inc is incremented by 1 => _inc = 7,
_no3 will gets printed (i.e 8), and _no = _no1 + _no2 => _no3 =(5+8) = 13,
And, _no1=8, _no2 = 13

Since,( _inc <= _lower ) => (7<=6), The condition of the while loop is false,
Hence, the loop will not executes further, so the output will be 1 1 2 3 5 8

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Question 4

Time: 00:00:00

Find the output of the following code

#include<stdio.h>

int main ()

{

   int num;

   printf ("enter number" );

   scanf ("%d", &num);

   no = (num * 2) + 3 ;

   (no & 1 && printf ("odd"))printf ("even") 

   return 0;

}

Enter a number: Odd

Odd

Error : ‘no’ undeclared

Even

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In the given code, no is not defined so, no = (num * 2) + 3 ; will give an error that no is not defined in the code.

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Question 5

Time: 00:00:00

What will be the output of the following code?

#include<stdio.h>

int main()

{

  while(){

   printf(“hello world”);

  }

return 0;

}

Compile error

No output will be printed

hello world

Infinite loop

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Explanation : Condition of the while loop is necessary, so the given code will give the compile error.

for(;;) //is correct

while() // is incorrect syntactically

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Question 6

Time: 00:00:00

What is the output of the following code

class A{

public int i;

public int j;

A(){

    i=1;

    j=2;

  }

}

class B extends A{

    int a;

    B(){

       super();

    }

}

class transform {

    public static void main(String,args[]){

        B obj = new B();

        System.out.println(obj.i+""+obj.j);

    } 

}

1 2

Compile-time error

Run-time error

No output

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Explanation : In the main function, the object of the B is created so it calls the constructor of the child class, but in child class constructor, super() is used so it calls the constructor of the parent class.
Hence, obj.i = 1 and obj.j=2 will be assigned.
And we get : 1 2 as output.

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Question 7

Time: 00:00:00

What will be the output of the following code

#include <stdio.h>

#define LowerLimit 10

int main(){

    auto int x=10;

    switch(x,x*2){

        case LowerLimit:

        printf("ABC");

        break;

        case LowerLimit*2:

        printf("XYZ");

        break;

        case LowerLimit*3:

        printf("PQR");

        break;

        default:

        printf("MNO");

        case LowerLimit*4:

        printf("www");

        break;

    }

}

ABC

XYZ

Compile time Error

No output

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Explanation : In switch x, x*2 is passed, and here (,) is used as a operator not separator so, x*2 is passed in switch as parameter,
LowerLimit is defined as macro with value 10,
Hence case LowerLimit*2 is the matched case, so XYZ will get printed and break will terminate the switch statement.

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Question 8

Time: 00:00:00

What will be the output of the following code ?

#include <stdio.h>

int main()

{

    int a=64;

    double b=256;

    int *x=&a;

    double *y=&b;

    printf("x and y are %d and %d",sizeof(x),sizeof(y));

    return 0;

}

Compilation error

x and y are 8 and 8

x and y are 1 and 2

Run-time error

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Explanation: x and y are the pointer variables and pointing to int and double value respectively. So, the sizeof(x) and sizeof(y) will not depend on the type of the pointer i.e, int and double.
Hence, the output will be: x and y are 8 and 8.

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Question 9

Time: 00:00:00

What would be the output of the below code?

#include <stdio.h>

int main ()

{

  int n, num = 8421;

  n = num;

  while (num >= 10)

  {

    num /= 10;

  }

  printf ("%d", num);

  return 0;

}

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num = 8421

Since, the condition of while loop (num>=10) i.e (8421>=10) is true,
So, In this iteration : num = num/10 => 842

Again, the condition of while loop (num>=10) i.e (842>=10) is true,
So, In this iteration : num = num/10 => 84

Again, the condition of while loop (num>=10) i.e (84>=10) is true,
So, In this iteration : num = num/10 => 8

Now, the condition of while loop (num>=10) i.e (8>=10) is false,
So, while loop gets terminated
And we get output as 8 (because num=8).

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Question 10

Time: 00:00:00

What is the output of the following code when inputChar = ‘y’

#include <stdio.h>

int main()

{

   char inputChar;

   printf("%d",scanf("%c",&inputChar));

   return 0;

}

Garbage value

Compile-time error

Run time error

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The output will be 1, because scanf(“%c”, &inputChar) will return the number of characters it read.

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["0","40","60","80","100"]

["Need more practice!","Keep trying!","Not bad!","Good work!","Perfect!"]

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