How to change the integer type used by an enum (C++)?
Solution 1
You cannot do this in C++98/03. C++11 does allow you to do it, and without enum class
the way everyone else seems to keep telling you:
enum EnumType : uint16_t
{
Bar,
Baz,
Bork,
};
Again, you don't have to use enum class
. Not that it's a bad idea, but you don't have to.
Does GCC support this feature in its implementation of C++11?
Which version of GCC? It looks like GCC 4.4 added this functionality, but you should probably look to more recent versions, just for the sake of stability.
Solution 2
In C++11, you can do that:
enum class Foo : uint16_t
{
Bar,
Baz,
Bork,
};
Later on, you can also know the underlying type of the enum as:
#include <type_traits>
std::underlying_type<Foo>::type v = 10; //v is uint16_t
Solution 3
With pre-C++2011 you can force a minimum storage by using a suitable range of values:
enum foo {
v0 = 0,
vmax = 32767
};
I think the compiler is free to choose either a sign or an unsigned integer type as the underlying type. The above range enforced that the representation uses at least short
as it underlying integer. Making it even one bigger may cause it to use long
instead. Of course, this only forces a minimum range and the compiler is free to choose a bigger range. Also, with the above definition you are not allowed to stray outside the range [0, 32767]
: if you really need a 16 bit range (at least) you need to use corresponding values).
Solution 4
With c++11
you now have enum class
, that allows you to set underlying type explicitly:
enum class Foo: uint16_t
{
Bar,
Baz,
Bork,
};
Linuxios
Updated on March 25, 2020Comments
-
Linuxios about 4 years
If I have an C++ enum:
enum Foo { Bar, Baz, Bork, };
How do I tell the compiler to use a
uint16_t
to actually store the value of the enum?EDIT: Does GCC support this feature in its implementation of C++11?
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Lol4t0 about 12 years@YochaiTimmer, standard changed and topic on the link does not tell about
enum class
-
Linuxios about 12 years@YochaiTimmer: Not at all. I want to know if I can change enum to store its values internally with a
uint16_t
, not whether it is the same size asint
.
-
-
chris about 12 yearsI was wondering that myself. The only difference is that
enum class
creates a new scope, whereasenum
does not. -
Nicol Bolas about 12 years@chris: No, there are several differences between
enum class
andenum
. Both do create scopes (the aboveBar
can be accessed withEnumType::Bar
), but only theclass
version forces you to use the scoped enum. Also,enum class
variables do not implicitly convert integers to the enumeration type. So you must either pass an actual enumerator of the correct type, or you must perform an explicit cast. This adds a lot more type safety. -
chris about 12 yearsOh, I got my point from another question on SO: stackoverflow.com/questions/441552/… As for your second point, I simply never heard about that one. Thanks for the free knowledge :)
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Matthieu M. about 12 yearsActually, there is a restriction on the size of the underlying type as well. [dcl.enum]/6 precises: except that the underlying type shall not be larger than
int
unless the value of an enumerator cannot fit in anint
orunsigned int
.