trying to define even/odd numbers please help

Question

i am trying to define a number 55.15 or something like that i think this should be - double- am i right:) wether even or not. i`m going to use that in a calculation and if it`s odd i`ll need to round it up to closest decimal value. if you please give me some suggestions i`ll be more than happy... thanks...

Antti · Answer

Just figured one (obvious) way to check if the int is even/odd. This is more efficient way i think (for the computer that is). This method is based on a simple fact that all odd integers must end to bit 1 and all even numbers to the bit 0. For non-integer numbers you still have to use the previous way i think. Some examples: 2 => '10'. 3 => '11'. 121 => '111 1001' If we AND-mask the number 121 with 1 (that is ...0001) heres the result: 111 1001 &&& &&&& 000 0001 === ==== 000 0001 And as we all recall and operation simply checks if both of the operands are true(1). If both are true the result is true. 1 AND 1 is true. 1 AND 0 is false etc... using System; using System.Collections.Generic; using System.Text; namespace EvenOdd { class Program { static void Main( string [] args) { int number = -31; Console .WriteLine(number + :  + IsEven(number)); number = 235578; Console .WriteLine(number + :  + IsEven(number)); double number2 = 2332.1; Console .WriteLine(number2 + :  + IsEven(number2)); number2 = -2332.0; Console .WriteLine(number2 + :  + IsEven(number2)); } static bool IsEven( int number) { //this operation masks the number using bit mask 0x00000001 //if number's last bit is 1 the result of the bitwise-AND (&) operation is 1. //this result is converted to boolean and inverted using ! return ! Convert .ToBoolean(number & 1); } static bool IsEven( double number) { bool isEven = false ; double modulus = number % 2; if (modulus == 0) isEven = true ; else if (modulus == 1) isEven = false ; return isEven; } } }

atilla · Answer

thanks so much after little effort it worked.:))

Antti · Answer

First round your float/double/decimal to nearest integer using Math.Round -method. Then you can calculate the 2-modulus of the integer. If modulus is 0 integer is even. If modulus is 1 integer is odd. using System; using System.Collections.Generic; using System.Text; namespace EvenOdd { class Program { static void Main(string[] args) { double number = 55.15; int rounded = Convert.ToInt32(Math.Round(number)); Console.WriteLine(IsEven(rounded)); } static bool IsEven(int number) { bool isEven = false; int modulus = number % 2; if (modulus == 0) isEven = true; else if (modulus == 1) isEven = false; return isEven; } } }