将平面指针转换为多维数组

Question

Is it possible to get the compiler to do the same pointer arithmetic on flat pointers as it does on multidimensional arrays?

With multidimensional arrays, the pointer arithmetic appears to work as follows:

int main(void)
{
    char ar[2][3][4];
#define At(a,x,y,z) printf("["#x"]["#y"]["#z"]=%ld\n", &(a)[x][y][z] - &(a)[0][0][0]);
    At(ar,0,0,0);
    At(ar,0,0,1);
    At(ar,0,0,2);
    At(ar,0,0,3);
puts("");
    At(ar,0,1,0);
    At(ar,0,2,0);
    At(ar,0,3,0);
puts("");
    At(ar,1,0,0);
    At(ar,2,0,0);
    At(ar,3,0,0);
/*
    [0][0][0]=0
    [0][0][1]=1
    [0][0][2]=2
    [0][0][3]=3

    [0][1][0]=4  // 1 * 4
    [0][2][0]=8  // 2 * 4
    [0][3][0]=12 // 3 * 4

    [1][0][0]=12 // 1 * 3 * 4
    [2][0][0]=24 // 2 * 3 * 4
    [3][0][0]=36 // 3 * 3 * 4 
*/
}

So I tried:

    char *blk;
    if(!(blk=malloc(1000))) return -1;
    char (*p)[2][3][4] = (char(*)[2][3][4])blk;

    At(p,0,0,0);
    At(p,0,0,1);
    At(p,0,0,2);
    At(p,0,0,3);
puts("");
    At(p,0,1,0);
    At(p,0,2,0);
    At(p,0,3,0);
puts("");
    At(p,1,0,0);
    At(p,2,0,0);
    At(p,3,0,0);

/*
    [0][0][0]=0
    [0][0][1]=1
    [0][0][2]=2
    [0][0][3]=3

    [0][1][0]=3
    [0][2][0]=6
    [0][3][0]=9

    [1][0][0]=6
    [2][0][0]=12
    [3][0][0]=18
*/
}

But as you can see, this doesn't give corresponding results. What exactly is going on here? Can the results be made to match?

Answer 1

You added an extra dimension, try this instead:

char (*p)[3][4] = (char(*)[3][4])blk;

Also note that your printf format is incorrect: the difference of 2 pointers has type ptrdiff_t , the corresponding format length modifier is t :

#define At(a,x,y,z) printf("["#x"]["#y"]["#z"]=%td\n", &(a)[x][y][z] - &(a)[0][0][0]);

If your C library is not fully C99 compliant, you should use a cast:

#define At(a,x,y,z) printf("["#x"]["#y"]["#z"]=%ld\n", (long)(&(a)[x][y][z] - &(a)[0][0][0]));

Answer 2

I think you are missing a dereference. You now have a pointer to a 3 dimensional array.

You need to do this:

At(*p, 0, 0, 1);