Nice

VERY nice. You wouldn't believe how I have been doing it in the past. Thanx!

more efficient
int Num = 5;

if ((Num & 1) == 0)
{
// It's even
}
else
{
// It's odd
}

only work with bits not division...

Thanx a bunch buddy... that really helped..

more more efficient (save one more CPU cycle)

int Num = 5;

if (Num & 1)
{
// It's odd
}
else
{
// It's even
}

How to find whether a number is even or odd without using conditional operator using C language??

int nNum = 5;

bool bOdd = (bool)(nNum & 0x1);

\"bool bOdd = (bool)(nNum & 0x1);\"

Casting is not efficient.

How to find whether a number is even or odd using bit fields using C language?

/// <summary>
/// Checks if the number if odd or even.
/// </summary>
/// <param name=\"number\">Number to check</param>
/// <returns>Returns true if Odd</returns>
public static bool OddOrEven(int number)
{
return ((Num & 1) != 0) ? true : false;
}

bool bOdd = (bool)(nNum & 0x1);

Casting is highly efficient - the compiler performs the cast not the CPU. The \'anding\' operation above takes exactly one clock cycle.

Casting is not efficient. At runtime the CLR checks casting operations to ensure that casts are always to the object's actual type or any of its base type - if the CLR simply allowed the casting, there would be no type-safety (google 'type spoofing'). It's very easy to write code that will compile yet fail at runtime with an InvalidCastException.

The question was is reference to the C language, which has a native compiler; i.e. there is no runtime manager.

A managed system such a .Net won\'t even compile this:
bool bOdd = (bool)(nNum & 0x1);

It will complain about an invalid cast right from the get go. A native compiler such as C or C++ will allow this cast all day long without any complaints. In fact, older compilers didn\'t even require the cast. They figured you knew what you were doing. The cast just tells the compiler that \'yes, I know I\'m about to change the variable type\'.

The following is the compiled code for the above test (including the disassembly):

int _tmain(int argc, _TCHAR* argv[])
{
bool bOdd = (bool)(argc & 0x1);

return bOdd;
00401000 movzx eax,byte ptr [esp+4]
00401005 and eax,1
}
00401008 ret

In terms of execution, it\'s not going to get any faster than this.

You're discussing C/C++ in a C# FAQ.

int num = 000;
cout<<"Enter the number:";
cin>>num;
if(num%2==0)
{
cout<<"This number is a Even Number.";
}
else
{
cout<<"This number is a Odd Number";
}
getch();
}

Dude! Get a room!
You guys are Geek indeed!
It\'s just Even or Odd... Like child play!
If I ever need a \"clock\" it\'d hang one on my kitchen wall.

There\'s a world out there. Full of girls.
Go get it tiger!
^^

int num;
cout<<"Enter the number:";
cin>>num;
clscr()
if(num%2==0)
{
cout<<"This number is a Even Number:/n";
}
else
{
cout<<"This number is a Odd Number:"/n;
}
getch();
}

int num;
cout<<"Enter the number:";
cin>>num;
clscr()
if(num%2==0)
{
cout<<"This number is a Even Number:/n";
}
else
{
cout<<"This number is a Odd Number:"/n;
}
getch();
}

int num;
cout<<"Enter the number:";
cin>>num;
clscr()
if(num%2==0)
{
cout<<"This number is a Even Number:/n";
}
else
{
cout<<"This number is a Odd Number:"/n;
}
getch();
}

int num;
cout<<"Enter the number:";
cin>>num;
clscr()
if(num%2==0)
{
cout<<"This number is a Even Number:/n";
}
else
{
cout<<"This number is a Odd Number:"/n;
}
getch();
}

how about if you're also writing the code for not just odd or even, but if the number is zero as well?

if(num ==0)

dsgdthfth6r

Now do it with an array!

int m;
if(m/2*2==0)
cout<<"no. is even";
else
cout<<"no. is odd";

i need similar program but using only IF STATEMENT.

how to determine a number is even or not without dividing or without taking modulus operator?

one way is to store all the even numbers in array and proceed by comparison is number with the number given by the user. But it cannot be efficient method to store all the even numbers in the array.

So what is the more efficient method to determine the number as an even or odd without using modulus operator?

how to determine a number is even or not without dividing or without taking modulus operator?

one way is to store all the even numbers in array and proceed by comparison is number with the number given by the user. But it cannot be efficient method to store all the even numbers in the array.

So what is the more efficient method to determine the number as an even or odd without using modulus operator?

u fuck

how do you do this is visual basic?

thanks alot its very helpfull

find odd or even using simple if in c?

#include "stdafx.h"

int main(int argc, char* argv[])
{
printf("Hello World!\n");
int a,b;
scanf("%d",

find odd or even using simple if in c?

#include "stdafx.h"

int main(int argc, char* argv[])
{
printf("Hello World!\n");
int a,b;
scanf("%d",

find odd or even using simple if in c?

#include "stdafx.h"

int main(int argc, char* argv[])
{
printf("Hello World!\n");
int a,b;
scanf("%d",

find odd or even using simple if in c?

#include "stdafx.h"

int main(int argc, char* argv[])
{
printf("Hello World!\n");
int a,b;
scanf("%d",

how to use the for loop and while loop in simple programs of C language?

using system;

namespace Even_Odd
{
class Program
{
int x;
Console.Write("Enter number");
x=int.Parse(Console.ReadLine());
if(x%2==0)
{
Console.WriteLine("Even");
Console.ReadLine();
}
else
{
Console.WriteLine("Odd");
Console.ReadLine();
}
}
}

i want to find even and odd nos. in vb pl help i am new to programing language...

wap for accept ten int into an array find totalof all even

#include<stdio.h>
#include<conio.h>
#include<math.h>
main()
{
int x;
clrscr();
printf("Enter the number which you want to know even or odd");
scanf("%d",

#include<stdio.h>
#include<conio.h>
#include<math.h>
main()
{
int x;
clrscr();
printf("Enter the number which you want to know even or odd");
scanf("%d",

quit simple in conditional operator

int Num = 6;
Console.WriteLine(Num % 2 == 0 ? "even":"odd");

conditional operator use full links :
http://msdn.microsoft.com/en-us/library/ty67wk28(VS.80).aspx

http://msdn.microsoft.com/en-us/library/ms173224(VS.80).aspx

thanks its so useful

#include<iostream>
usinng namespace std odd or even;

using system;

namespace Even_Odd
{
class NumEO
{
int x;
Console.Write("Enter number");
x=int.Parse(Console.ReadLine());
if(x%2==0)
{
Console.WriteLine("That's Even");
Console.ReadLine();
}
else
{
Console.WriteLine("That's Odd");
Console.ReadLine();
}
}
}

using system;

namespace Even_Odd
{
class NumEO
{

Console.Write("Enter number");
int x = Convert.ToInt32(Console.ReadLine());
if(x%2==0)
{
Console.WriteLine("That's Even");
Console.ReadLine();
}
else
{
Console.WriteLine("That's Odd");
Console.ReadLine();
}
}
}

#include<stdio.h>
#include<conio.h>
void main()
{ clrscr() ;
int subject,topic;
printf("enter the number to find odd or even\n");
scanf("%d",

#include<stdio.h>
#include<conio.h>
void main()
{ clrscr() ;
int subject,topic;
printf("enter the number to find odd or even\n");
scanf("%d",

#include<stdio.h>
#include<conio.h>
void main()
{ clrscr() ;
int subject,topic;
printf("enter the number to find odd or even\n");
scanf("%d",

you're all on crack

why do num

why do num

why do num

I have an 8 digit sequence of numbers, how can I get the numbers in the odd and even position?
ex:

int Number = 19345678;
while (Sin > 0)
{
int t = Number % 10;
Console.WriteLine(t);
}
console.read();

I want to get the number in the first, third, fifth and seventh position.

I have an 8 digit sequence of numbers, how can I get the numbers in the odd and even position?
ex:

int Number = 19345678;
while (Number > 0)
{
int t = Number % 10;
Console.WriteLine(t);
}
console.read();

I want to get the number in the first, third, fifth and seventh position.

Write a program, with a function called odd() that tells whether a number is odd or not. The function odd() should take an int value as an argument and returns the result as a boolean value. The values should be passed on by the user.
someone know what is the code of this program?

continue

@farwa, yes, yes, I know. :-)

It's this:
public static bool IsOdd(this double number)
{
return !number.IsEven();
}

public static bool IsEven(this double number)
{
bool result = false;

if (number > Int64.MaxValue)
{
result = number % 2 == 0;
}
else
{
result = (Convert.ToInt64(number)

) Write a program that reads in an integer that represents the number of iterations. The program then uses a counter-controlled loop to determine all the odd numbers between 1 and the number entered and sums those values.

PLZ any one can help me to get this solution in C

) Write a program that reads in an integer that represents the number of iterations. The program then uses a counter-controlled loop to determine all the odd numbers between 1 and the number entered and sums those values.

PLZ any one can help me to get this solution in C

Simple Syntax in C language

int a=5;
if(a%==0)
{printf("the value of a is %d even ",a);
}
else
{printf("Value is Odd : ",a);
}

Simple Syntax in C language

int a=5;
if(a%==0)
{printf("the value of a is %d even ",a);
}
else
{printf("Value is Odd : ",a);
}

you can try this dear
int a,i,j=0,G=0;
scanf("%d",

I came across http://rindovincent.blogspot.com/p/javascript.html where there was a simple Javascript program to find whether the number is odd or even. I am pasting the same code with permission here.

<html>
<head>
<script type="text/javascript">
var n = prompt("Enter a number to find odd or even", "Type your number here");
n = parseInt(n);
if (isNaN(n))
{
alert("Please Enter a Number");
}
else if (n == 0)
{
alert("The number is zero");
}
else if (n%2)
{
alert("The number is odd");
}
else
{
alert("The number is even");
}
</script>
</head>
<body>
</body>
</html>

loop for numbers then when you press "9" it close the loop and calculate how many odd and even numbers?

can any one help me to get the solution of this task in c# ,Thanks:)

int a=5;
if(a%2==0)
{printf("the value of a is %d even ",a);// for 'C'
Cosole.Write("Value is even " a);// C Sharp
System.out.println("Value is even " a);//Java
}
else
{printf("Value is Odd : ",a);
}

button press is not allow when loop is continue dear mohamed i can do for jst this that when 9 is come then loop was stop and show you how many odd

How to create a program that would determine if the number is either positive or negative integer?

if it is greater than 0
example:
n>o it is positive else it is negative

int a=-3;
if(a>=0)
{
print("The number is Positive");
}
else
{
print("The number is Negative");
}
Simple

int a=-3;
if(a>=0)
{
print("The number is Positive");
}
else
{
print("The number is Negative");
}
Simple

Determine if a number is odd or even and
Determine if a number is
positive even
negative even
positive odd
negative even

Determine if a number is odd or even and
Determine if a number is
positive even
negative even
positive odd
negative even

Determine if a number is odd or even and
Determine if a number is
positive even
negative even
positive odd
negative even

int a=-2;
if(a%2==0)
{ if(a>0) { printf(positve even) }
else {printf(negative even) }
}
else
{ if(a>0)
{ printf(positve Odd) }
else{printf(negative Odd) }
}

you can also contact me with mail
skygocomputer@yahoo.co.in

if you face some problem send me a mail INSHA ALLHA i will try to solve that

THANX SOHAIL SHAIKH
skygocomputer@yahoo.co.in

mod
doesnt work with negative number!

i want to more c program than you can send to me program to my email id

i want to more c program than you can send to me program to my email id

i want to more c program than you can send to me program to my email id

It was very useful for me, thank you very much!

write a program to perform addition of two matrix.

write a program to perform addition of two matrix.

how to check if a number is odd or even number?

how to check if a number is odd or even number?

how to check if a number is odd or even number?

how to check if a number is odd or even number?

int a=5;
if(a%2==0)
{printf("the value of a is %d even ",a);// for 'C'
Cosole.Write("Value is even " a);// C Sharp
System.out.println("Value is even " a);//Java
}
else
{printf("Value is Odd : ",a);
}

#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int a[3][3],i,j,sum ;
printf("THIS IS A 3*3 MATRIX PROGRAM\n");
for (i=0;i<3;i )
{ printf("The row no is%d\n",i);
for(j=0;j<3;j )
{
scanf("%d",

Part 1 Of Matrix
int a[3][3],i,j,sum ;
printf("THIS IS A 3*3 MATRIX PROGRAM\n");
for (i=0;i<3;i )
{ printf("The row no is%d\n",i);
for(j=0;j<3;j )
{
scanf("%d",

send me email if you want matrix add sub or multi program

Your blog is very informative. I really appreciate your hardwork. Thanks You for such good information. Keep up good work !!

Here is a very way to find odd or even.

using System;
class MainClass
{
public static void Main(string[] args)
{
int i;
Console.WriteLine("Enter the Number:");
i=Convert.ToInt32(Console.ReadLine());
if((i%2)==0)
Console.WriteLine( "The Number is Even" );
else
Console.WriteLine( "The Number is Odd" );
}
}

It is surely very workable.

Thanks for the blog.
<a href="http://www.allaboutdivorce.net">click here</a>

Thanks for the blog. The blog is very helpful and informative. Keep up the good work.

Thanks for the blog. The blog is very helpful and informative. Keep up the good work.

int Num = 5;

if(Num % 2 == 0)
{
// It's even
}
if(Num % 2 != 0)
{
// It's odd
}

int Num = 5;

if(Num % 2 == 0)
{
// It's even
}
if(Num % 2 != 0)
{
// It's odd
}

int Num = 5;

if(Num % 2 == 0)
{
// It's even
}
if(Num % 2 != 0)
{
// It's odd
}

int Num = 5;

if(Num % 2 == 0)
{
// It's even
}
if(Num % 2 != 0)
{
// It's odd
}

int Num = 5;

if(Num % 2 == 0)
{
// It's even
}
if(Num % 2 != 0)
{
// It's odd
}

int Num = 5;

if(Num % 2 == 0)
{
// It's even
}
if(Num % 2 != 0)
{
// It's odd
}

int Num = 5;

if(Num % 2 == 0)
{
// It's even
}
if(Num % 2 != 0)
{
// It's odd
}

int Num = 5;

if(Num % 2 == 0)
{
// It's even
}
if(Num % 2 != 0)
{
// It's odd
}

int Num = 5;

if(Num % 2 == 0)
{
// It's even
}
if(Num % 2 != 0)
{
// It's odd
}

int Num = 5;

if(Num % 2 == 0)
{
// It's even
}
if(Num % 2 != 0)
{
// It's odd
}

int Num = 5;

if(Num % 2 == 0)
{
// It's even
}
if(Num % 2 != 0)
{
// It's odd
}

int Num = 5;

if(Num % 2 == 0)
{
// It's even
}
if(Num % 2 != 0)
{
// It's odd
}

int Num = 5;

if(Num % 2 == 0)
{
// It's even
}
if(Num % 2 != 0)
{
// It's odd
}

int Num = 5;

if(Num % 2 == 0)
{
// It's even
}
if(Num % 2 != 0)
{
// It's odd
}

int Num = 5;

if(Num % 2 == 0)
{
// It's even
}
if(Num % 2 != 0)
{
// It's odd
}

int Num = 5;

if(Num % 2 == 0)
{
// It's even
}
if(Num % 2 != 0)
{
// It's odd
}

There are many people who simply prefer to use and read a brief guide books, which certainly will provide accurate and correct information for you.

If some one enters a number and requires the answer as if i enter 12340 , compiler should tell me that how many even ,odd and zeroes are present in 12340

If some one enters a number and requires the answer as if i enter 12340 , compiler should tell me that how many even ,odd and zeroes are present in 12340

If some one enters a number and requires the answer as if i enter 12340 , compiler should tell me that how many even ,odd and zeroes are present in 12340

If some one enters a number and requires the answer as if i enter 12340 , compiler should tell me that how many even ,odd and zeroes are present in 12340

If some one enters a number and requires the answer as if i enter 12340 , compiler should tell me that how many even ,odd and zeroes are present in 12340

If some one enters a number and requires the answer as if i enter 12340 , compiler should tell me that how many even ,odd and zeroes are present in 12340

If some one enters a number and requires the answer as if i enter 12340 , compiler should tell me that how many even ,odd and zeroes are present in 12340

If some one enters a number and requires the answer as if i enter 12340 , compiler should tell me that how many even ,odd and zeroes are present in 12340

I am having issues with this assignment in VB 2010:

Code the application. Use the InputBox function to get the input. Save the solution and then start the application. Test the application using the following integers: 45,2,34,7,55,90 and 32. The sum of the odd and even integers should be 107 and 158 respectively.

Can someone please help me code this?