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C program to check whether all bits of a number are UNSET/LOW?
Learn: how we can check that whether all bits of a one byte (8 bits) number are UNSET/LOW using C program? Here, we are implemented the program for this using Bitwise AND (&) operator.
Problem statement
Give a number of one byte (8 bits) and we have to check whether all bits are UNSET/LOW.
Example -1
Input number: 0
Binary value: 00000000
Output: Yes, all bits are unset
Example -2
Input number: 50
Binary value: 00110011
Output: No, all bits are not unset
Checking whether all bits of a number are UNSET/LOW
To solved this program, we will use BITWISE AND (&) operator, we will traverse bits from 7 to 0 (in case of two bytes number, it would be 15 to 0 and so on...) and check if there is any SET/HIGH bit, then we will break the loop and output will be false that means "all bits are not UNSET/LOW".
C program to check whether all bits of a number are UNSET/LOW
#include <stdio.h>
//function to check whether all bits are
//UNSET/LOW or not?
int isAllBitsUnset(unsigned int num)
{
int loop, cnt=0;
for(loop=7; loop>=0; loop--)
{
//check, if there is any SET/HIGH bit
if( num & (1<<loop))
{
cnt =1;
break;
}
}
if(cnt==0)
return 1; //true
else
return 0; //false
}
//main function
int main()
{
unsigned int number;
//read number
printf("Enter an integer number (between 0-255): ");
scanf("%d",&number);
if(isAllBitsUnset(number))
printf("All bits are UNSET/LOW.\n");
else
printf("All bits are not UNSET/LOW.\n");
return 0;
}
Output
First Run:
Enter an integer number (between 0-255): 0
All bits are UNSET/LOW.
Second Run:
Enter an integer number (between 0-255): 50
All bits are not UNSET/LOW.
We wrote a function int isAllBitsUnset(unsigned int num), it is taking an unsigned integer type number as an argument and returning 0 if all bits are not UNSET/LOW and 1 if all bits are UNSET/LOW.
function:
int isAllBitsUnset(unsigned int num)
{
int loop, cnt=0;
for(loop=7; loop>=0; loop--)
{
//check, if there is any SET/HIGH bit
if( num & (1<<loop))
{
cnt =1;
break;
}
}
if(cnt==0)
return 1; //true
else
return 0; //false
}
To check the bits, we are traversing loop from 7 to 0 (in case of 8 bits) and checking each bit, whether it is HIGH/SET or not, if any bit is HIGH/SET then value of cnt will be 1 (which is 0 initially, cnt is using as a flag variable) and loop will break.
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