GATE CSE 2020 | Question: 48

Question

Consider the following C functions.

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<len ; i++) {          if (arr[i] ==1)              tot = tot * ex;         ex= ex*ex;     } return (tot) ; }

The value returned by $pp(3,4)$ is _______.

Answer 1

Answer $: 81$

$pp(3,4)$

• $a=3, b=4 , \text{tot} = 1;$

len = tob(4, array) will return 3 with array set as $001$ as array is updated only once when  b%2 != 0). The for loop actually iterates $3$ times for $b = 4, b = 2$ and $b=1,$ and only when $b=1,$ arr[i] is updated.

Now pp will run for loop 3 times:

1. $\text{arr}[0] = 0.$ So, $\text{ex} = 3\ast 3 = 9$
2. $\text{arr}[1] = 0.$ So, $\text{ex} = 9\ast 9 = 81$
3. $\text{arr}[1] = 1.$ So, $\text{tot} = 1\ast 81 = 81$

Comments

how Ara will become 001 it should  be 110  and why loop will run 3 time we got i = 2 they use len as varible not strlen function ? so i have two questions could you please explain  me ?

4%2 = 0  SO  a[0]=1  (i=0)

2%2=0 SO a[1]=1   (i=1)

1%2 != 0 so a[2]=0  (i=2)

so array i =2

there is len not strlen(function ) so how i =3?

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Why i=3 only?.... i should go to upto last index of the array so I'm getting i=19

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Milan1999 Anju Mehral

in tob() : initially i=0;

if(4%2) ---→ if(0) so, it execute else statement… a[0] = 0  then b=2, i =1;

if(2%2) ---→ if(0) so, it execute else statement… a[1] = 0  then b=1, i=2;

if(1%2) ---→ if(1) so, it execute if statement….. a[2] = 1 then b=0, i=3;

and for loop break and return(i=3) to the calling function…

i.e, len = 3;

 

 

Answer 2

THIS ILLUSTRATION MAY HELP😀

Answer 3

Answer: 81

Comments

@Bhupendra i will be 3 after the last iteration.

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rt.. !! thanks

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same i also got 27

Answer 4

https://ideone.com/Aedsbd