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Guy
Guy

Reputation: 50949

Enum with specified values for only some of the members

I have an enum with custom value for only part of the list

public enum MyEnum
{
    FirstValue,
    SecondValue,
    ThirdValue,
    ForthValue = 1,
    FifthValue = 2
}

When I tried strina name = (MyEnum)2; name was ThirdValue.

But when I changed the enum to

public enum MyEnum
{
    FirstValue = 3,
    SecondValue,
    ThirdValue,
    ForthValue = 1,
    FifthValue = 2
}

In strina name = (MyEnum)2; name was FifthValue.

Does the compiler (I'm using Visual Studio 2012) initialize custom values only if the first has custom values?

And if ThirdValue got default value 2 in the first example how come there wasn't any error in FifthValue = 2?

Upvotes: 9

Views: 1674

Answers (3)

Moho
Moho

Reputation: 16553

More than one enumerator can have the same value. If you don't want to specify the value for each enumerator, simply switch the order you define the enumerators:

public enum MyEnum
{
    FourthValue = 1,
    FifthValue,
    FirstValue,
    SecondValue,
    ThirdValue
}

Upvotes: 0

Wai Ha Lee
Wai Ha Lee

Reputation: 8815

When you assign values for an enum member, the compiler increments the value by one for the next member, unless it's defined. If no members have values, the numbering starts at 0.

Your first example, with what the compiler is doing is:

public enum MyEnum
{
    FirstValue,    // == 0
    SecondValue,   // == 1
    ThirdValue,    // == 2
    ForthValue = 1,
    FifthValue = 2
}

so you have two members with value 2.

Either give them all values, or give them no values. Anything else will likely lead to confusion.

The C# standard, section 14.3 says (emphasis mine):

The associated value of an enum member is assigned either implicitly or explicitly. If the declaration of the enum member has a constant-expression initializer, the value of that constant expression, implicitly converted to the underlying type of the enum, is the associated value of the enum member. If the declaration of the enum member has no initializer, its associated value is set implicitly, as follows:

  • If the enum member is the first enum member declared in the enum type, its associated value is zero.
  • Otherwise, the associated value of the enum member is obtained by increasing the associated value of the textually preceding enum member by one. This increased value must be within the range of values that can be represented by the underlying type; otherwise, a compile-time error occurs.

Upvotes: 8

Lasse V. Karlsen
Lasse V. Karlsen

Reputation: 391664

The corresponding integer values that enum values map to always start at 0 unless you specifically change the values.

As such, your first piece of code is equivalent to this:

public enum MyEnum
{
    FirstValue = 0,
    SecondValue = 1,
    ThirdValue = 2,
    ForthValue = 1,
    FifthValue = 2
}

So you can see that 2 maps to both ThirdValue and FifthValue.

Your second example is equivalent to this:

public enum MyEnum
{
    FirstValue = 3,
    SecondValue = 4,
    ThirdValue = 5,
    ForthValue = 1,
    FifthValue = 2
}

Upvotes: 2

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