Programming Forums > Application Development > C++

Jimbo · May 12th, 2006, 6:28 AM

I'm writing Chess in what little free time I have these days, and I'm currently working on getting the pieces to calculate their possible moves. My specific problem is with pawns, which can only move in a single direction, determined by where they started (i.e. by color). Anyways, I'm trying to create a forward pointer that will get the square "in front" of where the pawn is now.

I have the line in pawn.cpp:

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Square (*fwd)() = (getColor() == White) ? &Square::up : &Square::down;

where Square::up() and Square::down() are public functions declared in square.h as

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public: /* ... */
     Square* up();
     Square* down();

The error I get is:

Quote:

pawn.cpp:16: error: cannot convert `Square*(Square:

)()' to `Square (*)()' in initialization

Is there a good way to get this to work? For that matter, what exactly does this error message mean? I'm thinking that it's complaining about making a pointer to a function specifically inside the Square class. Should I perhaps rethink how I'm doing this?

BTW, I'm compiling using g++ and running Linux, not that it should matter for this...

nnxion · May 12th, 2006, 7:38 AM

What do you mean by forward pointer?

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Square (*fwd)() = (getColor() == White) ? &Square::up : &Square::down;

I believe you are saying that you are assigning a pointer to function returning Square, to a pointer to a Square. I think that's the problem you are having.
I'm not sure as I haven't done function pointers in a while but here goes anyway:

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Square* (Square::*fwd)() = (getColor() == White) ? &Square::up : &Square::down;

Should fix the errors, I haven't thought about what it would solve though.

I'd be hesitant on the function pointer in the first place. It's almost always possible to get the same functionality in a different manner. Oh and don't forget the following either:

Quote:

Originally Posted by wikipedia

Unlike other pieces, the pawn does not capture in the same way as it moves. A pawn captures diagonally, one square forward and to the left or right. Any piece directly in front of a pawn, friend or foe, blocks its advance.

DaWei · May 12th, 2006, 7:47 AM

Missing an asterisk.

Quote:

Originally Posted by MSDN

A pointer to a member of a class differs from a normal pointer because it has type information for the type of the member and for the class to which the member belongs. A normal pointer identifies (has the address of) only a single object in memory. A pointer to a member of a class identifies that member in any instance of the class. The following example declares a class, Window, and some pointers to member data.

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class Window
{
public:
    Window();                               // Default constructor.
    Window( int x1, int y1,                 // Constructor specifying
        int x2, int y2 );                   //  window size.
    BOOL SetCaption( const char *szTitle ); // Set window caption.
    const char *GetCaption();               // Get window caption.
    char *szWinCaption;                     // Window caption.
};

// Declare a pointer to the data member szWinCaption.
char * Window::* pwCaption = &Window::szWinCaption;

In the preceding example, pwCaption is a pointer to any member of class Window that has type char*. The type of pwCaption is char * Window:

. The next code fragment declares pointers to the SetCaption and GetCaption member functions.

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const char * (Window::*pfnwGC)() = &Window::GetCaption;
BOOL (Window::*pfnwSC)( const char * ) = &Window::SetCaption;

Jimbo · May 12th, 2006, 5:53 PM

Thanks for the replies, that seems to have fixed it.

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May 12th, 2006, 5:53 PM	#4
Jimbo Battle Programmer Join Date: Feb 2006 Location: Bellevue, WA, USA Posts: 769 Rep Power: 3	Thanks for the replies, that seems to have fixed it.

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Programming Forums > Application Development > C++

Function pointer to a member function