Why is C++ numeric_limits<enum_type>::max() == 0?

浪尽此生 提交于 2019-11-30 17:47:16

std::numeric_limits is specialized in the Standard Library "for each arithmetic type, both floating point and integer, including bool" (§18.3.2.1/2).

Your enumeration test is not one of these types, so the primary template is used. Its behavior is specified by §18.3.2.3/1: "The default numeric_limits<T> template shall have all members, but with 0 or false values."

If you want to know the traits of the underlying type of test, you can use underlying_type:

std::numeric_limits<std::underlying_type<test>::type>::max()

Alternatively, you can specialize numeric_limits for test and have it return the values you want. This is not a particularly good idea, though.

For non-specialized versions of the template, max returns T(). You have not written a numeric_limits specialization for your test type, so you get the default implementation.

The numeric_limits<T> is a regular class template, it is not connected to the compiler in any special way as to find out about user-defined enum types. If you look at the <limits> file, it has the default template definition that returns zeros for everything, and a whole bunch of type-specific specifications for the individual types, returning the right constants.

You can "plug in" your enum into numeric_limits by providing a specification of numeric_limits<test> by yourself. You can copy the one for int from the <limits>, and modify it to suit your needs.

From the C++11 draft:

In 18.3.2.1, about numeric_limits:

Non-arithmetic standard types, such as complex (26.4.2), shall not have specializations.

And an enum is not an arithmetic standard type.

Then, in the non-specialized template:

template<class T> class numeric_limits {
    public:
    [...]
    static constexpr bool is_specialized = false;
    static constexpr T max() noexcept { return T(); }
};

That is, the non-specialized max() function returns the default initialized value for that type, that is 0.

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