C++ int to enum class UB

Question

I have an enum which looks like this:

enum class MY_TYPE : int32_t{
    UNKNOWN = 0,
    A = 101,
    B = 102,
    C = 103,
    D = 104,
    E = 105
};

I want to cast given values from and to it.

First of all, is it guaranteed that the following code does not induce undefined behavior?

int32_t library_func(int* tp){ *tp = 275; return 0;}

int main(){
    MY_TYPE d = static_cast<MY_TYPE>(224);
    library_func(reinterpret_cast<int*>(&d));
    std::cout << "d is " << static_cast<int>(d);
}

Is it UB if I cast a value not inside the value range of MY_TYPE? About this, I found the following issue: What happens if you static_cast invalid value to enum class?

This prints 275 as expected. Do I understand the answer correctly that casting any value which fits in the enum's underlying type is fine?

For the most part, `reinterpret_cast` is a clear signal that you have UB. In this case, that signal is correct. A `MY_TYPE` is not an `int`, so you can not treat it as one. — NathanOliver, Jul 20 '21 at 19:44
related/dupe: https://stackoverflow.com/questions/33607809/can-an-out-of-range-enum-conversion-produce-a-value-outside-the-underlying-type — NathanOliver, Jul 20 '21 at 19:45
Another related Q/A: [Why can't C++11 strongly-typed enum be cast to underlying type via pointer?](https://stackoverflow.com/questions/24566576/why-cant-c11-strongly-typed-enum-be-cast-to-underlying-type-via-pointer) — alter_igel, Jul 20 '21 at 19:50
@Curunir I believe that is still not allowed, but [this](https://timsong-cpp.github.io/cppwp/basic.lval#11.2) might mean it is okay. I'm having trouble trying to figure out what it means. — NathanOliver, Jul 20 '21 at 20:01
@Curunir yes, absolutely. If it **were** an int, you could pass it as a pointer to int, and you'd not have to do any cast. — SergeyA, Jul 20 '21 at 20:03
@NathanOliver that means that following code is legal: `double* d= reinterpret_cast(new int); *reintepret_cast(d) = 12`. Dynamic type of `*d` is `int`, so you can legally reinterpret_cast it to int* and dereference. — SergeyA, Jul 20 '21 at 21:03

SergeyA · Answer 1 · 2021-07-20T20:08:45.077

6

Interestingly enough, UB in the example provided is strictly related to the aliasing by pointer, and not assigning enum to value which is not named. If it would not be aliasing by pointer, there would be no UB with using value outside of the enum, i.e. following is A-OK:

int32_t library_func(int32_t tp){ return tp * 10;}

int main(){
    MY_TYPE d = static_cast<MY_TYPE>(224);
    MY_TYPE v {library_func(static_cast<int32_t>(d))};

    std::cout << "v is " << static_cast<int32_t>(v);

}

edited Jul 20 '21 at 20:08

answered Jul 20 '21 at 19:59

SergeyA

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@MooingDuck good point, will rewrite my example w/ int32_t. – SergeyA Jul 20 '21 at 20:05
1

I had already upvoted your answer, but it would be worth adding an explicit note that `int` and `int32_t` might be different types. – Mooing Duck Jul 20 '21 at 20:07
@MooingDuck actually in this example both would be fine, since it would be non-narrowing. – SergeyA Jul 20 '21 at 20:09
`reinterpret_cast(&d)` would have hit the issue, but you're right that your code doesn't have that issue – Mooing Duck Jul 20 '21 at 20:50
hmm, unfortunately the library function needs an `int32_t*` parameter because it does not only return `void` but some `bool`. – Curunir Jul 21 '21 at 06:10
@Curunir can you add a second (output) integer parameter to the function than? – SergeyA Jul 21 '21 at 12:41

C++ int to enum class UB

1 Answers1