Is this code Right?

[xx1182]

Well:

  enum MyEnum{a, b, c, d, e};
int main()
{
 std::cout << a;

just works fine, it prints 0;

Is it also possible to declare an enumeration globally and just use it like this:

enum lewd{PLUS, MINUS};

struct lawd
{
    int a;
    int &operator()(lewd lw, int value)
    {
        if(lw == PLUS)
        {
            a+=value;
        }
    }
};

int main(int argc, char* argv[])
{
    lawd lel;
    lel.a = 1;
    lel(PLUS, 2);
    cout << lel.a << endl;
}

[Danny Toledo]

Unless otherwise specified, the default value for the first enumerated value should be 0: http://www.cprogramming.com/tutorial/enum.html

you can choose to use default values, which start at zero for the first constant

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#include <iostream> enum MyEnum {a, b, c, d, e}; int main() { std::cout<< "a: "<< a<< '\n'; return 0; }

a: 0

[xx1182]

But for istance if i do this

enum MyEnum{a, b, c}
int main()
{
 int a = 0;
 std::cout << a;
}

What variabile is going to use and why is going to use?

and also in the second code how the operator() doesn't need a declared type like

int main()
{
 lewd lw
 lawd lul = PLUS) // shouldn't the first code i posted uncompile?
 lewd(lul, 2);
}

[Danny Toledo]

What variabile is going to use and why is going to use?

Since you declare a variable a in the local scope of main, it shadows (hides) MyEnum::a:

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#include <iostream> enum MyEnum {a, b, c, d, e}; int main() { int a = 10; std::cout<< "a: "<< a<< '\n'; // local a std::cout<< "a: "<< MyEnum::a<< '\n'; // MyEnum::a std::cout<< "a: "<< ::a<< '\n'; // MyEnum::a return 0; }

a: 10 a: 0 a: 0

and also in the second code how the operator() doesn't need a declared type like

I'm not quite sure what it is you are asking, could you possibly rephrase the question?

shouldn't the first code i posted uncompile?

Maybe you mean not compile? For me it won't because your operator() function is missing a return statement.

[xx1182]

I mean

enum MyEnum{a, b, c, d}
struct MyStruct
{
 int a;
 int &operator()(MyEnum)
};

int main()
{
 MyStruct mystruct;
 MyStruct(b); //This shouldn't work right? Beacuse i've never declared a variabile of type MyEnum like MyEnum myenum = b;
}

[Danny Toledo]

I may be wrong, but I think what that line is doing is creating an instance of type MyStruct named b.

[JLBorges]

> Is it also possible to declare an enumeration globally and just use it like this:

Yes. (Provided a return statement is added to the overloaded operator).
http://coliru.stacked-crooked.com/a/46c25c65f41b49f2