Write a program in c to find even and odd nos?

Answer 1

#include<iostream>

using namespace std;

void main() {

int odd,even,number;

odd=0;

even=0;

for(int x=1;x<=10;x=x+1){

cout<<"enter a number\n";

cin>>number;

if(number%2==0){

even++;

}

else {

odd++;

}

}

cout<<"even number has:"<<even;

cout<<"odd number has:"<<odd;

}

Answer 2

for persons who are wondering how to do this its really simple.

/* CLAUDE ANTHONY BURRELL*/

/* C CODE FOR EVENSUM*/

#include<stdio.h>

#include<stdlib.h>

int valChar(int*, char);

int main()

{

int values[11];

char alpha;

int evenSum;

int i;

printf("\t C PROGRAM FOR ALL EVEN NUMBERS AND FIND THE SUM\n\n\n\n");

system("color e");

for(i = 0; i < 10; i++)

{

printf("PLEASE ENTER %d VALUE\n",i); /* READING 10 DIFFERENT NUMBERS IN VALUES*/

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

}//end for

fflush(stdin);

printf("ENTER A CHARACTER\n\n");

scanf("%c",&alpha);

evenSum = valChar(values,alpha);/* RETURNS THE SUM OF EVEN NUMBERS IF THE CHARACTER IS 'A' OR 'a'*/

printf("THE SUM OF THE NUMBER IS = %d ",evenSum);

system("pause");

}//end main

int valChar(int * values, char alpha)

{

int i;

int evenSum = 0;

for(i = 0; i< 10; i++)

{

if (alpha 'A')

{

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

{

//even = values[i] % 2 = 0;

evenSum = evenSum + values[i];

}//end if

}//end if

else

{

evenSum = evenSum + values[i];

}//end else

}//end for

return evenSum;

}//end valChar

Answer 3

bool is_odd (unsigned n) {

return n & 1;

}

This function simply tests to see if the low-order bit is set using the bitwise-AND binary operator (&). If it is, the number must be odd because the low order bit represents 2^0 which is 1. All other bits represent increasing powers of 2 (2, 4, 8, 16, etc) so the low-order bit alone determines if a number is odd (is set) or even (is not set).

Note that the argument must be unsigned. This is because negative values are implementation-defined and may use ones-complement or twos-complement notation, depending on the implementation. The function will only work with signed integers using twos-complement notation. Twos-complement is more common than ones-complement, however portability is not guaranteed.

For a portable solution that caters for both ones-complement and two-complement signed integers, use the modus operator instead:

bool is_odd (signed n) {

return n % 2;

}

The modus operator returns the remainder after division. When we divide any signed integer by 2, the remainder can only be 0 (is even) or 1 (is odd).