Pointer (*p)[3] Issue

[Bob Kang]

I declare a pointer as
unsigned char (*p)[3];
does that mean a pointer point to a array of 3 int elements?

then I wrote codes as below:

unsigned char a[3]={1,2,3};
unsigned char b,c;
unsigned char (*p)[3];
p=a;
b=*p;

I supposed that the value of 'b' is 1
During my debugging, I found that the value of b is not 1 but equals to p (0x75).(the value of a is 0x75,the address of a[0],so is p)

Does anyone know why's that?

[phantomotap]

O_o

If you have a pointer to an integer, what do you get when you dereference? You will have yourself an integer.

You have a pointer to an array. What do you get when you dereference?

Soma

[Bob Kang]

sorry...I don't get it well......would u mind providing me a detailed example to better illustration?

[Satya]

I think p should also be like

p=&a;

[Bob Kang]

Hi Satya, why? the array name is the address of the first element of the array, isn't it? means a = &a[0].
if p=&a => p = &(&a)

[Satya]

&arr yields a pointer, of type pointer-to-array-of-T, to the entire array.
A simple reference (without an explicit &) to an array yields a pointer, of type pointer-to-T, to the array's first element.
Most important thing to remember is the type.
So for example the array is

int array[3]={1,2,3}

if you say

array or &array[0] gives the address and the type int  but if you say &array it gives address but the type is int [3], 3 being the size of array.

Hence if

int array[3]={1,2,3};
int *p;
p = array; // p is pointing to array first element
int (*q)[3] // q is pointing to int array of 3 
q = &array;
if you increment p then it will increment to one element of int or it points to the next element of array that is 2 here
but if you increment q it will increment beyond the entire array.

I am not sure if i am clear. If any mistakes in my explanation please correct me.

[Fiskker]

I've found his to be particularly useful to get your head around this stuff:
cdecl: C gibberish ↔ English

[cstryx]

type (*p)[3];

This is a pointer to an array of 3 elements of some type...