C++ Friend Function not accessing private members

Question

So I am trying to make a chess game ( bit of an ultimate challenge for me ), and I'm a stump for this part ..

So I made a piece object, and the idea is that in the main game code, I have an array of pieces and I pass the address of the array to the function "InitilisePieces" and the team ( Black or White ) and it shall assign all the pieces. So I made the function a friend to access all the private members and it comes up with an error saying "inaccessible", and I don't understand what's wrong with doing as I've done. Any help would be more than appreciated!

Side Note: things like State_ and _Location and structs and enums that are defined properly etc, not the problem ... ( I don't think )

Header File Contains:

class   __Piece
{
private:

    State_              e_state;
    Piece_Type_         e_type;
    Team_               e_team;
    _Location           st_location;

    friend void         InitilisePieces     ( __Piece(*)[16], Team_);

public:
    __Piece             ();

};

.cpp File Contains:

void                    InitilisePieces     ( __Piece * pao_piece[16], Team_ )
{
    int         n_count;

    for ( n_count = 0; n_count < 16; n_count++ )
    {
        pao_piece[ n_count ]->e_state;
    }
}

UPDATE:

Thankyou for the explanations and I get where I'm going wrong now ... so what is the parameter supposed to be to pass the address of an array of __Piece 's?

[Do not begin your identifiers with underscores](http://stackoverflow.com/a/224420/1938163) — Marco A., Oct 14 '15 at 11:30
As an aside, `__Piece` and `_Location` are reserved identifiers. — emlai, Oct 14 '15 at 11:30
I think that getters and setters are much more suitable than friend function here — David Haim, Oct 14 '15 at 11:30
@MarcoA. Beginning with an underscore is just fine, as long as the next character is not uppercase or another underscore, and the identifier is not in global scope. — emlai, Oct 14 '15 at 11:31
I like it, and I've never had a problem with it. and its not causing this error. — user3502489, Oct 14 '15 at 11:32
Not saying that it is the cause of your problem but it's a discouraged practice (read more from the link I posted). — Marco A., Oct 14 '15 at 11:32
@DavidHaim .. Getter Setters would be fine (ish, it would add more code ) and that was my next step, but regardless why doesn't ^^ that work — user3502489, Oct 14 '15 at 11:33

score 7 · Answer 1 · edited May 23 '17 at 11:44

7

Your friend function and the function you define later have the same name but different signatures. You have not defined the friend function.

This

void InitilisePieces( __Piece(*)[16], Team_);

is not the same as this

void InitilisePieces( __Piece * pao_piece[16], Team_ )

The former's first parameter is a pointer to an array of 16 __Pieces. The latter's first parameter is adjusted to __Piece** pao_piece, i.e a pointer to a pointer to a __Piece. In other words, it is this:

void InitilisePieces( __Piece** pao_piece, Team_ )

Also: watch out for reserved identifiers.

edited May 23 '17 at 11:44

Community

1
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answered Oct 14 '15 at 11:33

juanchopanza

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Great answer. It will be even better if you remove `pao_piece` from second line or add it to the first one (imho). – Lapshin Dmitry Oct 14 '15 at 11:36
Ohh okay, thats fine .. but when I pass the address of the array in it says, "__Piece(*)[16] incompatible with __Piece **". How do I get around it? – user3502489 Oct 14 '15 at 11:37
1

@user3502489 The address-of operator applied to an array yields a pointer to an array. That isn't a pointer to a pointer. You need to figure out what you want the code to do, then implement that. – juanchopanza Oct 14 '15 at 11:39
@juanchopanza I can't pass the address of "__Piece piece_array[16]" into a __Piece** though? – user3502489 Oct 14 '15 at 11:53
@user3502489 That is probably what the compiler is telling you. You can't convert pointer to array of X to pointer to pointer to X. – juanchopanza Oct 14 '15 at 11:55
@juanchopanza So you cant ever pass the address of an array? – user3502489 Oct 14 '15 at 12:02
@user3502489 You can pass to something that expects the address of an array. – juanchopanza Oct 14 '15 at 12:07
what would that be ...? – user3502489 Oct 14 '15 at 12:11
@user3502489 Read my answer carefully. The relevant information is in there. Otherwise ask another question (after searching for duplicates). This is unrelated to the question you posted. – juanchopanza Oct 14 '15 at 12:13

C++ Friend Function not accessing private members

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