array as parameter of a function

问题:

There is a array of structures.

    static field fields[xsize][ysize];

I want to change it in function

    void MoveLeft(pacman *Pacman, field **fields,int **play)

But when I send it like this

     MoveLeft(&Pacman,fields,play);

I've got an error.

field - structure

     typedef struct
     {
    blossom blossoms;
    wall walls;
     }field;

where blossom & wall - another structures

回答1:

Although arrays and pointers are somewhat interchangeable in C, they're not exactly the same. In particular, an array of arrays and an array of pointers are laid out differently in memory.

Here's a way to make an array of pointers which refers to the same data as your existing array of arrays:

field* field_rows[xsize];
for (unsigned int i=0; i<xsize; i++) {
    field_rows[i] = fields[i];
}

Then a pointer to that field_rows array can be passed to MoveLeft:

MoveLeft(&Pacman,field_rows,play);

Another solution might be to change the declaration of MoveLeft instead, to take a pointer to array of arrays:

void MoveLeft(pacman *Pacman, field fields[xsize][ysize], int **play);

MoveLeft(&Pacman,fields,play);

回答2:

I guess the error is the following: two dimension array fields[xsize][ysize] is fixed size array (xsize/ysize are defines or consts) and in memory this is not look like field**, cause it's pointer to pointer to field, while fields[xsize][ysize] internally just one dimension fixed size array, where compiler handle double indexing for you.

So what you need is just define fields as field** and allocate it dynamically.

See picture for more explanation:

回答3:

While I'm not using Windows, I'm guessing your error is something similar to this:

error: cannot convert ‘field (*)[xx]’ to ‘field**’ for argument ‘2’ to ‘void MoveLeft(pacman*, field**,int**)’

A solution to this is to simply cast the fields parameter to the type the function wants:

MoveLeft(&Pacman, (field **) fields, play);