find a number whether it is even or odd without using any
control structures and relational operators?
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Answer / ramya
A number anded with the lower number that is n & (n - 1) =
0 then it is even if it is anything else it is odd
odd_even (int n)
{
if (!(n & (n -1))
number is odd
else
number is even
}
| Is This Answer Correct ? | 9 Yes | 29 No |
Answer / guest
#include<stdio.h>
#include<conio.h>
void main()
{
int x;
if(x%2==0)
printf("even");
else
printf("odd");
getch();
}
| Is This Answer Correct ? | 4 Yes | 25 No |
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You have an int array with n elements and a structure with three int members. ie struct No { unsigned int no1; unsigned int no2; unsigned int no3; }; Point1.Lets say 1 byte in the array element is represented like this - 1st 3 bits from LSB is one number, next 2 bits are 2nd no and last 3 bits are 3rd no. Now write a function, struct No* ExtractNos(unsigned int *, int count) which extracts each byte from array and converts LSByte in the order mentioned in point1.and save it the structure no1, no2, no3. in the function struct No* ExtractNos(unsigned int *, int count), first parameter points to the base address of array and second parameter says the no of elements in the array. For example: if your array LSB is Hex F7 then result no1 = 7, no2 = 2, no3 = 7. In the same way convert all the elements from the array and save the result in array of structure.
C 
