A c program to interchange odd and even components of an array?

Answer 1

#include<stdio.h>

#include<conio.h>

void main()

{

int a[10],i,evc=0,odc =0;// sum=0;

clrscr();

printf("enter 10 no.s in the array\n");

for(i=0;i<=9;i++)

{

scanf("%d",&a[i]);

}

for(i=0;i<=9;i++)

{

if(a[i]%2==0)

{

evc=evc+1;

}

else

{

odc=odc+1;

}

}

printf("Even nos %d\n",evc);

printf("odd nos %d\n",odc);

getch();

}

0"). This is a perfectly valid and correct approach, but in many implementations, a simple test for the least significant bit ("a[i] & 1") can be more efficient. This also allows to replace logic with arithmetic, which generally is more efficient (i.e. "odc += a[i] & 1")

Answer 2

... int number_to_check = 0;

...

cin >> number_to_check;

if ((number_to_check % 2))

{

printf("The number %d is even\n", number_to_check);

}

else

{

printf("The number %d is odd\n", number_to_check);

}

...

Answer 3

Write a c program that interchanges the odd and even elements of an array?

Answer:

#include

int main()
{
int num[]={1,2,3,4,5,6};
int i,temp;

for(i=0;i<=5;i=i+2)
{
temp=num[i];
num[i]=num[i+1];
num[i+1]=temp;
}
for(i=0;i<=5;i=i+2)
printf(''%d",num[i]);
return 0;
}

Answer 4

Full source code for counting even and odd elements of an array is available at:

http://bitsbyta.blogspot.com/2011/02/counting-even-odd-elements-in-array.html