Reputation: 25434
There is an:
enum SomeEnum
{
A = 0,
B = 1,
C = 2
}
Now compiler allows me to write:
SomeEnum x = SomeEnum.A;
switch(x)
{
case 0: // <--- Considered SomeEnum.A
break;
case SomeEnum.B:
break;
case SomeEnum.C:
break;
default:
break;
}
0
is considered SomeItems.A
. But I can't write:
SomeEnum x = SomeEnum.A;
switch(x)
{
case 0:
break;
case 1: // <--- Here is a compilation error.
break;
case SomeEnum.C:
break;
default:
break;
}
Why only implicit conversion exists for 0
?
Upvotes: 28
Views: 2118
Reputation: 223277
From ECMA-334 (C# Language Specification)
13.1.3 Implicit enumeration conversions
An implicit enumeration conversion permits the decimal-integer-literal 0 to be converted to any enum-type.
enum's default value is 0
and at compile time it is known that is why it is allowed in the switch statement. For value other than 0
, it can't be determine at compile time whether this value will exist in the enum or not.
Assigning additional values new versions of enums, or changing the values of the enum members in a new version, can cause problems for dependant source code. It is often the case that enum values are used in switch statements, and if additional elements have been added to the enum type, the test for default values can return true unexpectedly.
Upvotes: 20
Reputation: 18543
I would also add, that the syntax with 0
instead of the exact enum
in the switch
statement may become error prone. Consider the following code:
enum TestEnum
{
NA = 0,
A
}
and then
var e = TestEnum.NA;
switch(e)
{
case 0:
{
break;
}
case TestEnum.A:
{
break;
}
}
This compiles and works well. However if for some reason, enum
declaration changes to
enum TestEnum
{
NA = 1,
A
}
everything will get broken.
Though in the majority of situations the default value for enum
is 0
and for that reason this syntax may take place, I would use the exact enum
.
Upvotes: 4