###### Database-Management-System

October 9, 2023###### Nielit Scientist-B IT 4-12-2016

October 9, 2023# Nielit Scientist-B CS 22-07-2017

Question 5 |

**Following is C like Pseudo code of a function that takes a number as an argument, and uses a stack S to do argument, and uses a stack S to do processing.**

**void fun(int n)**

**{**

**Stack s;//Say it creates an empty stack S**

**while(n>0)**

**{**

**// This line pushes the value of n%2 to**

**Stack S;**

**Push(&S,n%2);**

**n=n/2l**

**}**

**// Run while Stack S is not empty**

**while(!is Empty(&S))**

**Printf(%d”,pop(&S));//pop an element from S and print it**

**}**

**What does the above function do in general order**

Prints binary representation of n in reverse order | |

prints binary representation of n | |

Prints the value of Logn | |

Prints the value of Logn in reverse order |

Question 5 Explanation:

For any number, we can check whether its ‘i’th bit is 0(OFF) or 1(ON) by bitwise ANDing it with “2^i” (2 raise to i).,

1) Let us take number ‘NUM’ and we want to check whether it’s 0th bit is ON or OFF

bit = 2 ^ 0 (0th bit)

if NUM & bit == 1 means 0th bit is ON else 0th bit is OFF

2) Similarly if we want to check whether 5th bit is ON or OFF

bit = 2 ^ 5 (5th bit)

if NUM & bit == 1 means its 5th bit is ON else 5th bit is OFF.

Let us take unsigned integer (32 bit), which consist of 0-31 bits. To print binary representation of unsigned integer, start from 31th bit, check whether 31th bit is ON or OFF, if it is ON print “1” else print “0”. Now check whether 30th bit is ON or OFF, if it is ON print “1” else print “0”, do this for all bits from 31 to 0, finally we will get binary representation of number.

void bin(unsigned n)

{

unsigned i;

for (i = 1 << 31; i > 0; i = i / 2)

(n & i)? printf(“1”): printf(“0”);

}

1) Let us take number ‘NUM’ and we want to check whether it’s 0th bit is ON or OFF

bit = 2 ^ 0 (0th bit)

if NUM & bit == 1 means 0th bit is ON else 0th bit is OFF

2) Similarly if we want to check whether 5th bit is ON or OFF

bit = 2 ^ 5 (5th bit)

if NUM & bit == 1 means its 5th bit is ON else 5th bit is OFF.

Let us take unsigned integer (32 bit), which consist of 0-31 bits. To print binary representation of unsigned integer, start from 31th bit, check whether 31th bit is ON or OFF, if it is ON print “1” else print “0”. Now check whether 30th bit is ON or OFF, if it is ON print “1” else print “0”, do this for all bits from 31 to 0, finally we will get binary representation of number.

void bin(unsigned n)

{

unsigned i;

for (i = 1 << 31; i > 0; i = i / 2)

(n & i)? printf(“1”): printf(“0”);

}

int main(void)

{

bin(7);

printf(“\n”);

bin(4);

}

Correct Answer: B

Question 5 Explanation:

For any number, we can check whether its ‘i’th bit is 0(OFF) or 1(ON) by bitwise ANDing it with “2^i” (2 raise to i).,

1) Let us take number ‘NUM’ and we want to check whether it’s 0th bit is ON or OFF

bit = 2 ^ 0 (0th bit)

if NUM & bit == 1 means 0th bit is ON else 0th bit is OFF

2) Similarly if we want to check whether 5th bit is ON or OFF

bit = 2 ^ 5 (5th bit)

if NUM & bit == 1 means its 5th bit is ON else 5th bit is OFF.

Let us take unsigned integer (32 bit), which consist of 0-31 bits. To print binary representation of unsigned integer, start from 31th bit, check whether 31th bit is ON or OFF, if it is ON print “1” else print “0”. Now check whether 30th bit is ON or OFF, if it is ON print “1” else print “0”, do this for all bits from 31 to 0, finally we will get binary representation of number.

void bin(unsigned n)

{

unsigned i;

for (i = 1 << 31; i > 0; i = i / 2)

(n & i)? printf(“1”): printf(“0”);

}

1) Let us take number ‘NUM’ and we want to check whether it’s 0th bit is ON or OFF

bit = 2 ^ 0 (0th bit)

if NUM & bit == 1 means 0th bit is ON else 0th bit is OFF

2) Similarly if we want to check whether 5th bit is ON or OFF

bit = 2 ^ 5 (5th bit)

if NUM & bit == 1 means its 5th bit is ON else 5th bit is OFF.

Let us take unsigned integer (32 bit), which consist of 0-31 bits. To print binary representation of unsigned integer, start from 31th bit, check whether 31th bit is ON or OFF, if it is ON print “1” else print “0”. Now check whether 30th bit is ON or OFF, if it is ON print “1” else print “0”, do this for all bits from 31 to 0, finally we will get binary representation of number.

void bin(unsigned n)

{

unsigned i;

for (i = 1 << 31; i > 0; i = i / 2)

(n & i)? printf(“1”): printf(“0”);

}

int main(void)

{

bin(7);

printf(“\n”);

bin(4);

}

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