C Pointer Arrays Initializtion

Question

Im quite confused that what is difference between these two initializations:

    int (*p)[10];

and

    int *p[10]

I know they both can point to 2D array whose element count in row is 10....

Answer 1

The first is a pointer to array, the second is an array of pointers.

Answer 2

To elaborate a bit on the correct answers here already:

The first line:

 int (*p)[10];

declares that "p" is a pointer to the memory address of an array with the capacity of 10 ints. It can be read in English as: "integer-pointer 'p' points to 10 sequential ints in memory".

The second line:

int *p[10]

Declares that "p[]" is an array of 10 pointers to integers. This is an array of memory addresses that point to integers. In this case, "p" is a sequence of 10 pointers in memory (which happen to be the memory addresses of other ints).

Answer 3

int (*p)[10];


+------+                           +-------+-------+-------+-------+-------+-------+-------+-------+-------+-------+
|  p   | =========================>|(*p)[0]|(*p)[1]|(*p)[2]|(*p)[3]|(*p)[4]|(*p)[5]|(*p)[6]|(*p)[7]|(*p)[8]|(*p)[9]|
+------+                           +-------+-------+-------+-------+-------+-------+-------+-------+-------+-------+

sizeof p will return sizeof (void *) (4 on 32 bit systems, 8 on 64 bit systems)

sizeof *p will return 10*sizeof (int) (40 on most systems)

int *p[10];   is the same as     int* (p[10]);   

p
+------+------+------+------+------+------+------+------+------+------+
| p[0] | p[1] | p[2] | p[3] | p[4] | p[5] | p[6] | p[7] | p[8] | p[9] |
+------+------+------+------+------+------+------+------+------+------+

sizeof p will return 10*sizeof (void *) (40 on 32 bit systems, 80 on 64 bit systems)

sizeof *p will return sizeof (int *) (4 on 32 bit systems, 8 on 64 bit systems)