What is the underlying type of a c++ enum?
Solution 1:
The type of a C++ enum is the enum itself. Its range is rather arbitrary, but in practical terms, its underlying type is an int
.
It is implicitly cast to int
wherever it's used, though.
C++11 changes
This has changed since C++11, which introduced typed enums. An untyped enum
now is defined as being at least the width of int
(and wider if larger values are needed). However, given a typed enum
defined as follows:
enum name : type {};
An enumeration of type name
has an underlying type of type
. For example, enum : char
defines an enum
the same width as char
instead of int
.
Further, an enum
can be explicitly scoped as follows:
enum class name : type {
value = 0,
// ...
};
(Where name
is required, but type
is optional.) An enum
declared this way will no longer implicitly cast to its underlying type (requiring a static_cast<>
) and values must be referenced with a fully-qualified name. In this example, to assign value
to a enum
variable, you must refer to it as name::value
.
Solution 2:
From N4659 C++ 7.2/5:
For an enumeration whose underlying type is not fixed, the underlying type is an integral type that can represent all the enumerator values defined in the enumeration. If no integral type can represent all the enumerator values, the enumeration is ill-formed. It is implementation-defined which integral type is used as the underlying type except that the underlying type shall not be larger than
int
unless the value of an enumerator cannot fit in anint
orunsigned int
. If the enumerator-list is empty, the underlying type is as if the enumeration had a single enumerator with value 0.
Solution 3:
IIRC its represented as int in memory. But gcc has switch -fshort-enum
to make it a shortest integer type that fits all the values, if you need to save space. Other compilers will have something similar.