You need a simple method to test a numeric value to determine whether it is even or odd.
Solution
The solution is actually implemented as two methods. To test for an even integer value, use the following method:
public static bool IsEven(int intValue)
{
return ((intValue & 1) == 0);
}
To test for an odd integer value, use the following method:
public static bool IsOdd(int intValue)
{
return ((intValue & 1) == 1);
}
Discussion
Every odd number always has its least-significant bit set to 1. Therefore, by checking whether this bit is equal to 1, we can tell whether it is an odd number. Conversely, testing the least-significant bit to see whether it is 0 can tell you whether it is an even number.
To test whether a value is even we AND the value in question with 1 and then determine whether the result is equal to zero. If the result is zero, we know that the value is an even number; otherwise, the value is odd. This operation is part of the IsEven method.
On the other hand, we can determine whether a value is odd by ANDing the value with 1, similar to howthe even test operates, and then determine whether the result is 1.If the result is set to 1, we know that the value is an odd number; otherwise, the value is even. This operation is part of the IsOdd method.
Note that you do not have to implement both the IsEven and IsOdd methods in your application, although implementing both methods might improve the readability of your code.
The methods presented here accept only 32-bit integer values. To allowthis method to accept other numeric data types, you can simply overload it to accept any other data types that you require. For example, if you need to also determine whether a 64bit integer is even, you could modify the IsEven method as follows:
public static bool IsEven(long longValue)
{
return ((longValue & 1) == 0);
}
Only the data type in the parameter list needs to be modified.
1.6 Obtaining the Most- or Least-Significant Bits of a Number
Problem You have a 32-bit integer value that contains information in both its lower and upper 16 bits. You need a method to get the 16 most-significant bits and/or the 16 least-significant bits of this value.
Solution To get the most-significant bits (MSB) of an integer value, perform a bitwise and between it and the value shown in the following method:
public static int GetMSB(int intValue)
{
return (intValue & 0xFFFF0000);
}
To get the least-significant bits (LSB) of a value, use the following method:
public static int GetLSB(int intValue)
{
return (intValue & 0x0000FFFF);
}
This technique can easily be modified to work with other sizes of integers (e.g., 8-bit, 16-bit, or 64-bit); this trick is shown in the Discussion section.
Discussion
In order to determine the values of the MSB of a number, use the following bitwise AND operation:
uint intValue = Int32.MaxValue;
uint MSB = intValue & 0xFFFF0000;
// MSB == 0xFFFF0000
This method simply ANDs the number to another number with all of the MSB set to 1. This method will zero out all of the LSB, leaving the MSB intact.
In order to determine the values of the LSB of a number, use the following bitwise AND operation:
uint intValue = Int32.MaxValue;
uint LSB = intValue & 0x0000FFFF;
// LSB == 0x0000FFFF
This method simply ANDs the number to another number with all of the LSB set to 1, which zeroes out all of the MSB, leaving the LSB intact.
The methods presented here accept only 32-bit integer values. To allowthis method to accept other numeric data types, you can simply overload this method to accept any other data types that you require. For example, if you need to also acquire the least-significant byte or most-significant byte of a 16-bit integer, you could modify the GetMSB method as follows:
public static int GetMSB(short shortValue)
{
return (shortValue & 0xFF00);
}
The GetLSB method is modified as shown here:
public static int GetLSB(short shortValue)
{
return (shortValue & 0x00FF);
}
