TCS NQT Input/Output Quiz 1

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.

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

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

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.

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

for(;;)  //is correct

while() // is incorrect syntactically

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.

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.

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.

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).

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