TCS NQT Input/Output Quiz- 6

n=10
n=stringofLength(s);//here the length of the string is 9

9= 1001//binary of 9
1= 0001//binary of 1
n^1=1000 //the xor of 9 and 1
And decimal value of 1000 = 8
Hence the answer is 8

Same
If we evaluate the value of x1*x2 + x3/x4
=1*3 + ⅖
=3+⅖

And value of x3/x4 + x1*x2
=⅖ +1*3
=3+⅖

99 99 100
Here first the value of num=100
printf("%d\t",--num); // this is pre-decrement operator hence the first the value will be decremented and then printed on the console this will print 99
printf("%d\t",num);// now the value of naum is 99
printf("%d\t",++num);// again this is a pre-increment operator here first we will increment the value of num = 100 and then print it on the console.

() has associativity of left to right, so in the given expression q=(p+q)-(p=q) so,

q=(p+q)-(p=q) => q=(50+100)-(100) => 50

as, p=q will change p to 100. So (p=q) gives 100 not 50. 

So we get output as 50 100

Here the value of k-- is pre decrement operator so after execution of this line the value of k=-1

Here the value sum = a++|| b++ || c++ ;
Initially a=0, b=5, c=3 since the precedence of the ++ operator is greater than || operator so first a++ b++ and c++ increments the value of a, b and c(a=1, b=6, c=4) then logical OR will work and it return true if any of the condition is true, 1|| 6 || 4 leads to 1, so the sum becomes 1 and we get output as 1 1 6 4.

initially i=0;num=1122;

while (i<3)

n[0]=num/10 --> (ie 1122/10=112 since array is a int type)

num=num%10 -->(ie 1122%10 =2 and the remainder is assigned to num)

 

n[1]=num/10 --> (ie 2/10=0 since array is a int type)

num=num%10 -->(ie 2%10 =0 and the remainder is assigned to num)

 

n[2]=num/10 --> (ie 0/10=0 since array is a int type)

num=num%10 -->(ie 0%10 =0 and the remainder is assigned to num)

 

n[3]-->fail (as i<3 i.e, 3<3 becomes false)

Ans= 112

res = fun(a,b) this will call fun function with parameter (10,6)
In fun function 10<6 condition is false this will check condition in else if block (ie) 6!=0 condition is true
Again this will call fun function with (10,5) as parameter and this process of recursion continues till x And execute 6 times so we get output as 60.

Here first the loop will enter the value of i=0 and the condition is true it will print
In for loop 1 time(s) as i+1=0+1=1

Explanation : i = 8, i>>1 => i/(2^1) => 8/2 =>4
i<<2 => i*(2^2) = 8*4 => 32
So the output will be 432.