C Functions Questions 3

Option 1 boo(x1,x2,x3,....,xn) ==( (x1)&& boo(x1,x2,x3,....,xn) ) || ( (x1')&&boo(x1,x2,x3,....,xn) ) returns TRUE.

For case 1 where x1 is 0 the R.H.S =( (0)&& boo(x1,x2,x3,....,xn) ) || ( (1)&&boo(x1,x2,x3,....,xn) )

R.H.S=boo(x1,x2,x3,....,xn) )

Same thing happens is x1=1 R.H.S=boo(x1,x2,x3,....,xn) )

Thus R.H.S==L.H.S=TRUE

Similarly Option 2 is also correct.

Option three  boo (x1, x2, …, xn) = (xn&&)boo(x1, x2, …, 0) || (xn&&)boo(x1, x2, …,1)

Case 1: taking xn=0 RHS= 1&&boo(x1, x2, …, 0) || 0&&boo(x1, x2, …, 1) RHS =boo(x1, x2, …, 0)

Case 2: taking xn=1 RHS = 0&&boo(x1, x2, …, 0) || 1&&boo(x1, x2, …, 1) RHS =boo(x1, x2, …, 1)

In both cases RHS=LHS, so, option 3 is TRUE.

In option 4  boo (x1, x2, …, xn) = boo(0, x2, …, xn) + boo(1, x2, .., xn) Here, in either of the cases we can not have LHS == RHS=TRUE. Because no term depends on value of boo ‘x1’.

Note the i is a static value therefore it will only be initialized once and will retain it's value as it was in the previous recursion of the loop.

Note that for for every value of y i gets incremented i++ 9 times and this happens repeatedly until y=1 for which y needs to decremented 1023 times.

Thus the total number of time i is incremented is 9*1023=9207 times.

It is a trick question you'd think it might be a recursive loop but after the first if statement is executed i=95-10=85 and then control jumps else if and returns i.

In order to swap x and y we need to make the parameters call by reference like this declare the function swap(int *a,int *b) can call the function in main like this swap(&x,&y)

Currently there is no PrepInsta Explanation. Did you know?

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All a,b,c,d are side effect of scan conversion.

For the first iteration sin(pi * i/2)=sin(pi * 0/2)=0 which is not equal to a.

That's why else statement prints 0

For 2nd iteration sin(pi * i/2)=sin(pi * 1/2)=1 which is equal to a this if prints 1

Similarly for 3rd and 4th sin(pi * i/2)=0 and 1

Remember that i is a static variable it will only be set to 1 once and afterwards if the function is called it will retain it's previous value.

So the value of n in series goes like

n=n+i

1st time n=1+1=2

2nd time n=2+2=4

3rd time n=4+3=7

4th time n=7+4=11, now n>10 therefore if statement uses return this value of n.

 

The if condition is satisfied. Thus lol is first printed and return (x = x + y); ends the function which is \"This should get printed\" is not printed.