Interview questions on Bitwise Operators

1)Which bit wise operator is suitable for turning off a particular bit in a number?

2)what will be printed out when the following code is executed:

main()
{
printf("%x",-1<<4);
}

3)Which one is equivalent to multiplying by 2?

* Left shifting a number by 1
* Left shifting an unsigned int or char by 1?

4)Write a function which gets the n bits from an unsigned integer x, starting from position p .(the right most digit is at position 0)

5)Write a function using bitwise opetators to check whether an integer is a power of 2 or not?

6)Write a Program that swaps the contents of two variables without
using any other variable,using bitwise operators?

7) Which bit wise operator is suitable for checking whether a particular bit is on or off?

8) Which bit wise operator is suitable for putting on a particular bit in a number?

9) Which bit wise operator is suitable for checking whether a particular bit is on or off?

10)Write a function setbits(x,p,n,y) that returns x with the n bits that begin at position p set to the rightmost n bits of y,leaving the other bits changed.

11)Write a function invert(x,p,n) that returns x with the n bits that begin at position p inverted leaving other unchanged.

Please post your answers in the comments section.Answers Shall soon be put.

16 comments:

  1. Answer for Q1:

    #include "stdafx.h"

    #include "stdio.h"
    void SetNthBit(int *number, int pos)
    {
    int i;
    int x = 1;

    x = x << (pos-1);

    *number = *number | x;
    }

    void ResetNthBit(int *number, int pos)
    {
    int i;
    int x = 1;

    //First get the Nth bit value to know whether it is 1 or 0

    int temp;

    temp = (*number) >>(pos-1);
    if( (temp % 2) == 0)
    {
    //The Nth bit is 0. Hence it is already Reset
    }
    else
    {
    x = x << (pos-1);
    *number = (*number) ^ x;
    }
    }
    int main(int argc, char* argv[])
    {
    int number = 2;
    printf("Number before setting the bit %d\n",number);
    // Set the 3rd bit to 1 from Right most bit
    SetNthBit(&number,3);
    printf("Number after setting the bit %d\n",number);
    ResetNthBit(&number,3);
    printf("Number after Re-setting the bit %d\n",number);
    return 0;
    }

    ReplyDelete
  2. find the no. of 1's in a given number using bitwise operations

    ReplyDelete
  3. main()

    {
    //Turn off a particular bit
    int n = 4;
    int x = 3; // Switch off 3rd bit
    int i = 1 << x - 1;
    i = ~i;
    printf("%d", n & i);


    //Test if << 1 time multiplies the number by 2
    //Doesnt work for char
    unsigned int b = 2147;
    char c = 2147;
    int d = 2147;

    printf ("\nUnsigned %d",b << 1);
    printf ("\nchar %d",c << 1);
    printf ("\nint %d",d << 1);

    //Swap
    int x = 202;
    int y = 409;
    x = x ^ y;
    y = x ^ y;
    x = x ^ y;

    printf("\nswapped %d %d",x,y);

    }

    ReplyDelete
  4. answer to Q1)
    //: create a mask where everywhere is 1 except that position and AND the number with the mask
    int turn_off(int number, int pos) {
    int mask = ~(1 << pos-1);
    return number & mask;
    }

    ReplyDelete
  5. please provide answers for c puzzles

    ReplyDelete
  6. 1.OR
    2.FFF0
    3.LEFT SHIFTING AN UNSIGNED INT OR CHAR BY 1
    4.GETTING X AND POSITION P
    R1=X>>P;
    TEMP=~R1;
    TEMP=R1^X;
    R1=X&TEMP;

    5.GET X AS INPUT
    FOR(I=0;I>I&10?PRINTF("YES"):PRINTF("NO");

    6.GET A AND B ARE INPUTS
    A=(A^B)^B;
    B=(B^A)^A;

    7.&
    8.|
    9.&
    10.DONT KNOW
    11.GET X ,POS ,N
    WHILE(N>0)
    {
    IF(((X>>POS+1-N)&01)==0)
    X=X|(01>>POS+1-N);
    ELSE
    {
    TEMP=~X;
    TEMP=TEMP|01;
    TEMP=X^TEMP;
    X=TEMP&X;
    }
    N--;
    }

    SAMPLE INPUT:
    10101010
    N=4;P=3;
    OUTPUT: 1010 0101

    ReplyDelete
  7. 5. To check if it is a power of 2,

    If the given number is n,

    if n&(n-1) == o , power of 2 else not a power of 2

    ReplyDelete
  8. 1.and operator
    2. for 16 bits ans is ffff for 32 bits ans is ffffffff
    3. left shift the number by 1
    4.

    ReplyDelete
  9. 11)
    assuming rightmost digit is at pos 0
    x = x XOR (pow(2, p+n) - pow(2,p))

    ReplyDelete
  10. Question #2 is implementation dependent. How the computer stores a signed number is not defined in the C spec and can change from compiler to compiler.

    ReplyDelete
  11. send answer plz

    ReplyDelete
  12. Q10.
    void trunc(byte& x, byte p, byte n, byte y){
    printf("y = %x\n",y);
    y = y>>p;
    byte t = 0xffffffff;
    t = ~(t<<n);
    printf("t = %x\n",t);
    x = y & t;
    printf("x = %x\n",x);
    }

    ReplyDelete
  13. Q11.
    void invert(byte& x, byte y, byte p, byte n){
    printf("y = %x\n", y);
    byte bitmask = 0;
    byte y_ = 0;
    trunc(y_, p, n, y);

    for (int i=0; i>= 1;
    if (i!=0) x<<= 1;
    x |= bitmask;
    }
    printf("x = %x\n", x);
    }

    ReplyDelete
  14. void invert(byte& x, byte y, byte p, byte n){
    printf("y = %x\n", y);
    byte bitmask = 0;
    byte y_ = 0;
    trunc(y_, p, n, y);

    for (int i=0; i>= 1;
    if (i!=0) x<<= 1;
    x |= bitmask;
    }
    printf("x = %x\n", x);
    }

    ReplyDelete
  15. send answers pls

    ReplyDelete