how to get enum value from enum type?

Question

The question in the title may sound trivial, so I better explain with some code what I want to do...

In C++11 I can do this:

#include <iostream>

namespace X {
    enum class  FOO { A,B };
}

template <typename T> void foo(T t) { 
    if (t == T::A) { std::cout << "A"; }
}

int main() {
    foo(X::FOO::A);
}

The important point here is that the template foo does not need to know in what namespace the enum is declared. I could as well call foo(Y::FOO::B) (provided there is a enum class called FOO in namespace Y having members A and B).

Now the question is: How to get the same with plain old enums (and only C++98 stuff)?

This works:

#include <iostream>

namespace X {
    enum FOO { A,B };
}

template <typename T> void foo(T t) { 
    if (t == X::A) { std::cout << "A"; }
}

int main() {
    foo(X::A);
}

but only because foo knows in what namespace the enum is declared. And it wont work for a Y::FOO::B ! (In C++11 it also works if I replace the line with if (t == T::A) ..., even with a plain enum)

Is there a way to get this working in C++98/03 without refering to X in the template explicitly?

For the sake of completeness, in C++98, this

template <typename T> void foo(T t) { 
    if (t == T::A) { std::cout << "A"; }
}

results in

error: ‘A’ is not a member of ‘X::FOO’

PS: I am not allowed to change the enum and the template has to live in a different namespace than the enum.

PPS: a simple if (t == 0) would probably work, but thats something I would like to avoid

Nicol Bolas · Accepted Answer · 2017-07-25 15:39:24Z

6

Until C++11, there was no way to say "the enumerator name within this enumeration". That's why it was added to C++11, even to unscoped enumerations.

answered Jul 25, 2017 at 15:39

Nicol Bolas

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6 Comments

463035818_is_not_an_ai Over a year ago

and there is no way to somehow get the namespace from T?

Nicol Bolas Over a year ago

@tobi303: That would require reflection, which we don't have, even in C++17.

463035818_is_not_an_ai Over a year ago

...but doesnt this mean that enums were pratically unusable as template parameters prior to C++11 ?

Nicol Bolas Over a year ago

Unusable for what purpose? I can certainly make a vector<EnumType>. I can pass an enum value to any number of algorithms that take a value and do something with that value. What you can't do is pass an enum value to a function that is then going to compare it to a static enumerator. But that's quite rare; I've never had to do that.

463035818_is_not_an_ai Over a year ago

unusable in the sense that you cannot refer to a value of the enum without knowing exactly where it is declared. Anyhow, thanks for the answer. I dont really like it, but I guess I will have to accept it ;)

|

AndyG · Accepted Answer · 2017-07-25 16:00:21Z

3

To add to Nicol Bolas' answer, you can kind of hack your way to a solution using ADL:

namespace X {
    enum FOO { A,B };
    bool IsA(FOO t)
    {
        return t == A;
    }
}

template <typename T> void foo(T t) { 
    if (IsA(t)) { std::cout << "A\n"; }
    else{std::cout << "not A\n";}
}

Live Demo

answered Jul 25, 2017 at 16:00

AndyG

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3 Comments

463035818_is_not_an_ai Over a year ago

unfortunately I cant add anything to X. However, I didnt expect it to be so diffucult, so I am happy about any suggestions / workarounds

Nicol Bolas Over a year ago

@tobi303: "unfortunately I cant add anything to X." Of course you can. C++ doesn't close namespaces (except for std, which is reserved). You can always inject whatever you want into whatever namespace you want.

463035818_is_not_an_ai Over a year ago

@NicolBolas oh yes sure ;). The problem is that what is X in my question in reality is a couple of hundreds different namespaces, so I ruled that out without really thinking about it. Anyhow for my real problem I found a different workaround so there is no point in trying to explain it further. For the question as asked here, this is a nice solution.

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