class won't accept enum in a function

I have this big problem with using enumerators in a class function

Code:

class Automat
{
    public:
        void print();
        int AddCoins(enum Coins, int MoreCoins);
    private:
        typedef enum {One=1, Two, Three} Coins;
        int MoreCoins;

--snip--

Automat::AddCoins(Automat::Coins, int MoreCoins)
{
    if (Coins == ONE)
        CoinStack += MoreCoins;
}
etc...

I get this error from Eclipse

Code:

error: 'enum Automat::Coins' is private

I have tried moving it around to public: and outside the class entirely.
I also get an error on the function:

Code:

error: 'Automat::AddCoins(int, int)'
error: expected identifier before ',' token
error: C++ forbids declaration of 'AddCoins' with no type

can someone point me in the right direction, as I am a bit lost on the syntax, as to why it doesn't accept my enums as parameters to a function

Code:

int AddCoins(enum Coins, int MoreCoins);

The name "Coins" is a type, so you can't specify it as a variable name as you are here. You need to do something like this:

Code:

int AddCoins(Coins coins, int MoreCoins);

The function definition should have the same parameter list, and "coins" is what you reference inside the function. Also, "Coins" should be in the public section of the class.

Thank you, solved my problem entirely.

I was on track with the enum, but I just couldn't see that the enum-name "Coins" was a type and not a name to a type.

Originally Posted by Roffemuffe

Thank you, solved my problem entirely.

I was on track with the enum, but I just couldn't see that the enum-name "Coins" was a type and not a name to a type.

A "type" and a "name to a type" are the same. Well, maybe not strictly speaking: an "int" is a name to the type "int", but can also be said to be of the type "integer". Anyways, when using typedef you make a new TYPE (yes, you make a new name/alias for a specific type, really). That doesn't mean you make a variable, as your usage seemed to imply. But then again, of course you can't use a variable as a type as you try by specifying it as a type for a first parameter in the function call.

What's the deal with this?

Code:

typedef enum {One=1, Two, Three} Coins;

wouldn't you prefer the following instead?

Code:

enum Coins {One, Two Three};

Because the default integer values for an enumeration start at 0.
Example:

Code:

enum Coins {One, Two, Three}

Sets One = 0, Two = 1, and Three = 2. Setting One = 1 means each successive option will have an integer one higher than the previous, e.g. in this case the same numerical value as they are written.

Originally Posted by iamthesgt

Because the default integer values for an enumeration start at 0.
Example:

Code:

enum Coins {One, Two, Three}

Sets One = 0, Two = 1, and Three = 2. Setting One = 1 means each successive option will have an integer one higher than the previous, e.g. in this case the same numerical value as they are written.

I suppose. Personally I'm averse to acknowledging (let alone using) the integer association with enum elements. They're meaningless to me.

typedef enum ... name;

That's C style. Just write enum name.

Your out-of-line definition of AddCoins is missing the return type.

Finally, put the enum definition above the declaration of AddCoins. If you write this:

Code:

class X {
public:
  void foo(E e);
  enum E { e1, e2 };
};

then a strictly conforming compiler will reject it, saying that E is unknown.

Originally Posted by CornedBee

then a strictly conforming compiler will reject it, saying that E is unknown.

Nice pickup. I totally missed that.