初学C＃麻烦

Question

Im new to c# and Im having a little problem. I want to make an easy program to ask the user for a integer number between 1-50, and then to display on the console if its a odd number or not. So, what i tried is this:

 Console.WriteLine("Skriv ut ett heltal: ");
 int x = int.Parse(Console.ReadLine());

 if (x == 1,3,5,7,9,11,13,15,17,19)
 {
     Console.WriteLine("The number is odd");
 }
 else 
 {
     Console.WriteLine("The number is not odd");
 }

Now i get an error at my if statements condition. How can i fix this?

Answer 1

C# does not allow you specify multiple values to check a variable against using a single if statement. You would need to check each value (1, 3, 5, etc) individually if you wanted to do it this way, and that would be a lot of redundant typing.

In this particular example, an easier way to check if something is odd or even is to check the remainder after dividing by 2, using the modulus operator % :

if (x % 2 == 1)
{
   Console.WriteLine("The number is odd");
}
else 
{
    Console.WriteLine("The number is even");
}

However, if you really do need to check against a list, then the easy way is to use the Contains method on an array (an ICollection<T> , really). To make it nice and easy, you could even write an extension function that lets you check against a list in a syntactically pretty fashion:

public static class ExtensionFunctions
{
    public static bool In<T>(this T v, params T[] vals)
    {
        return vals.Contains(v);
    }
}

Then you could say:

if (x.In(1,3,5,7,9,11,13,15,17,19)) 
{
    Console.WriteLine("The number is definitely odd and in range 1..19");
}
else 
{
    Console.WriteLine("The number is even, or is not in the range 1..19");
}

Voila! :)

Answer 2

if(x % 2 == 0)
{
// It's even
}
else
{
// It's odd
}

Answer 3

If you want to test whether x is a number in a particular list:

int[] list = new int[]{ 1,3,5,7,9,11,13,15,17,19};
if(list.Contains(x)) 

The common way to check to see if an integer is odd is to check if it divides evenly by 2:

if(x % 2 == 1)

Answer 4

x == 1,3,5,7,9,11,13,15,17,19 is not valid syntax for expressing multiple options. If you really want to do this then you can use a switch statement:

 switch(x) {
     case 1:
     case 3:
     case 5:
     case 7:
     case 9:
     case 11:
     case 13:
     case 15:
     case 17:
     case 19:
          // is odd
          break;
     default:
          // is even
          break;
 }

The correct way would be to use the modulo operator % to determine if a number is exactly divisible by 2 or not, rather than trying every odd number, like so:

if( x % 2 == 0 ) {
   // even number
}  else {
   // odd number
}

Answer 5

That's not valid C#. You can't test set inclusion like that. In any case, it's not practical to test for all the numbers in the world.

Why don't you just do this instead;

if (x &1 == 1) // mask the 1 bit

Bitwise operations are pretty quick so that code should be pretty fast.

Answer 6

While, as others have pointed out, this is not the best way to solve this problem, the reason you're getting an error in this case is because you can't have multiple values like that in an if statement. You have to word it like this:

if (x == 1 || x == 3 || x == 5)

If you don't know, || is the symbol for 'or'

Answer 7

Your if statement should be like this if you are having multiple conditions:

if any 1 of conditions is true:

if(x == 1 || x == 3 || x == 5) 
{
    //it is true
}

if all of the condition must be true:

if(x == 1 && y == 3 && z == 5) 
{
    //it is true
}

But if you are only looking for odd/even numbers. Use the % operator as the other answer says.

Answer 8

Try the following:

Console.WriteLine("Skriv ut ett heltal: ");
int x = int.Parse(Console.ReadLine());

Console.WriteLine(x % 2 == 1 ? "The number is odd" : "The number is not odd");

x % 2 == 1 does a modulus of 2 on the input (takes as many '2's off as possible until the number is between 0 and 2 - so 0 or 1 in this case)

Answer 9

One way to do that is:

if (x == 1 || 3 || 5){ 
Console.writeLine("oddetall");
}

or so it is possible to create an Array []

int[] odd = new int[3]; // how many odd to be tested
if(x=odd){
Console.WriteLine("Oddetall");
}