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How can I input a state to a variable of enum type in C++?


By : a.b.r
Date : July 30 2020, 05:00 PM
this one helps. You can't use cin direct to enum type, but you can do it to an int and static cast it to your enum type like shown below:
code :
#include <iostream>

enum Day : uint16_t { Mon = 1, Tue, Wed, Thu, Fri, Sat, Sun };

void printa(Day day) {
     if (day > Fri)
        std::cout << "weekend" << std::endl;
     else
        std::cout << "weekdays" << std::endl;
}

int main() {
    uint16_t day = Sun;

    std::cin >> day;

    printa(static_cast<Day>(day));

    return 0;
}


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Can the methods in an Enum type change the state of the instances of an Enum?


By : David Hernandez
Date : March 29 2020, 07:55 AM
Hope this helps Java enums are really just classes with some special treatment by the compiler and runtime. So yes, method calls on an enum instance can certainly change its state. I'm not sure what you mean with "then the invariant of an Enum type ie some type that exports a few constants will break".
Who says enums can only be a bunch of constants? The essence of an enum type is that you have a predefined fixed number of instances which you can refer to by name - and Java enums implement that. But why would a language not be allowed to have enums that are more powerful than that?

Creating Enum variable from a dynamic type that I know is an Enum?


By : imcymon
Date : March 29 2020, 07:55 AM
may help you . Enum.ToObject(Type,object) solved this.
For example after I confirmed that type is a System.Enum then I can do (System.Enum)Enum.ToObject(type,0).

Can I use the pattern matching operator ~= to match an enum value to an enum type with an associated variable?


By : user3643340
Date : March 29 2020, 07:55 AM
will help you From the documentation for Swift 1.2 "Enumeration case patterns appear only in switch statement case labels". So yes, you need to define your ~= operator (as from the answer pointed in the comments).
In case you just need isA and isB you can implement them using switch combined with _. In your case ~= couldn't work out of the box anyway, since you use an Any associated type, which is not Equatable (i.e.: how can I say if two .B(any) are equal since I cannot say if two any are equal?)
code :
enum MyEnum {
    case A, B(object: String)
}

let myEnumA = MyEnum.A
let myEnumB = MyEnum.B(object: "Foo")


func ~=(lhs: MyEnum, rhs: MyEnum) -> Bool {
    switch (lhs, rhs) {
    case (.A, .A):
        return true
    case let (.B(left), .B(right)):
        return left == right
    default:
        return false
    }
}

myEnumA ~= .A   // true
myEnumB ~= .B(object: "Foo") // true
myEnumB ~= .B(object: "Bar") // false

func isA(value: MyEnum) -> Bool {
    switch value {
    case .A:
        return true
    default:
        return false
    }
}

func isB(value: MyEnum) -> Bool {
    switch value {
    case .B(_):     // Ignore the associated value
        return true
    default:
        return false
    }
}

Why does 'typeof enum constant' generate a warning when compared to a variable of enum type?


By : user3886605
Date : March 29 2020, 07:55 AM
I wish this help you Quoting directly from C11, chapter §6.7.2.2, Enumeration specifiers,

Why do I have to cast an enum element when assigning it to a same enum variable type in C?


By : Kavi Solitarius Puer
Date : March 29 2020, 07:55 AM
Any of those help (Hi, this is a new account so I cannot use the comments section yet to ask for further clarification, so, my answer may be broader than needed)
Based on the text of the warning message I assume you are talking about MISRA-C:2012 (the latest standard) which is a great improvement over the prior ones in that much more effort in stating the rationale along with many more compliant and non-compliant examples have been added. This being Rule 10.3, the rationale is: since C permits assignments between different arithmetic types to be performed automatically, the use of these implicit conversions can lead to unintended results, with the potential for loss of value, sign or precision.
code :
enum enuma { A1, A2, A3   } ena;
ena  = A1;
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