Array of Pointers in C?

Question

int b[3][2] = { {0, 1}, {2, 3}, {4, 5} };
int (*bpp)[2] = b;
int *bp = b[0];

At the above code: Is *bpp a pointer to a two-dimensional array? Or an array of pointers with the length of 2? Why is *bpp surrounded with parenthesis? Is there a difference between *bpp[2] and (*bpp)[2] ?

Meantime, in the following code: (Changing the dimension of the array)

int i[4] = { 1, 2, 3, 4 };
int (*ap)[2] = (int(*)[2])i;

The second line is very confusing to me, especially the typecasting (int(*)[2]), what data type is it exactly casting to?

Thank you ^^

Mankarse · Accepted Answer · 2012-10-17 06:09:10Z

3

bpp is a pointer to an array of two int. *bpp is an array of two int. int *bpp[2] would declare bpp as an array of two pointers to int (this parentheses make it be a pointer to an array of two int).

(int(*)[2]) is a cast to a pointer to an array of two int.

These can be read by considering the "declaration follows use" rule (combined with knowledge of operator precedence):

  dereference (so bpp is a pointer)
     |
     v
int (*bpp)[2]
 ^         ^
 |         |
 |  array index (so the thing that bpp points to is an array)
 |
 the thing on the left is the final type... here it is int,
 so the array is an array of int

edited Oct 17, 2012 at 6:09

answered Oct 17, 2012 at 5:49

Mankarse

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