passing array of pointers from one function to another

How can I pass an array of pointers? Please help... here's a piece of my code...

Code:

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typedef struct node1 {
        int cell;
        struct node *prev;
        struct node *next;       
} box;

typedef struct node2 {
        char name[10];
        cell *move;
} player;

void play(box *tail) {
        player *p[i];
        int n = 5; //...n can be changed by the user..
 //..more codes...
        for(i = 0; i < n; i++) {
                p[i] = (player *)malloc(sizeof(player));
                }
        p[i]->move = tail;
//..more codes...

          displayBoard(tail,p[i]->move->cell, *p); //.. I will pass the array of pointers p to displayBoard
//..more codes...
}

void displayBoard(box *a, int cell, player *p){ //how can I pass p properly to this function?
        //..more codes...
}

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Code:

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          displayBoard(tail,p[i]->move->cell, p);

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Perhaps?

--
Mats

Passing the name of the array should do it. Can't pass *p as an array name is not a variable besides being a call by value instead of a call by reference.

Code:

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void displayBoard(box *a, int cell, player p)

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Well, p is an array of pointers, so this just won't work... without getting a warning or error.
If you put...
void displayBoard(box* a, int cell, player* (*p)[i])
...then it will work better (tricky type, I know!).
Also, pass the array with:
displayBoard(..., &p);
Basically, it's a pointer to an array of i pointers.

Code:

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displayBoard(..., &p);

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creates a pointer argument with three levels of indirection while

Code:

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displayBoard(..., p);

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keeps the level of indirection at two.

Try it, it compiles fine for me.
Other types do not compile fine.

I am on Unix...what platform are you on??
displayBoard(..., p) compiles fine for me while displayBoard(..., &p) errors out since the name of an array is not a variable.

Code:

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void foo(int* (* p)[10]) {}

int main()
{
        int* n[10];
        foo(&n);
}

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I'm using Visual Studio on Windows platform.
But it's really obvious, the type.
If we look at the middle, we see it's a pointer to an array (of 10 elements here), and the type of array is the type on the left, which in this case is int*.

An example with the OP's code:

Code:

---

typedef struct node1 {
        int cell;
        struct node *prev;
        struct node *next;       
} box;

typedef struct node2 {
        char name[10];
        //cell *move;
} player;

#define i 10

void displayBoard(box* a, int cell, player* (*p)[i]){} //how can I pass p properly to this function?

int main()
{
        player* p[i];
        displayBoard(0, 0, &p); //.. I will pass the array of pointers p to displayBoard
//..more codes...
}

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If I use
p
instead of
&p
I get an error (or warning in C):
error C2664: 'displayBoard' : cannot convert parameter 3 from 'player *[10]' to 'player *(*)[10]'

Now that you posted how you're defining and passing p to displayBoard() it makes sense. I'm passing p differently from the caller to the callee as in

Code:

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#define i 10

void displayBoard(box *a, int cell, player *p[]) {}

int main()
{
    player *p[i];
    displayBoard(0, 0, p);
}

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Aha.
I suppose...
player* p[]
...works too.