pointers and arrays

[dasarisrikar]

can anybody explain me the concept of multi dimensional arrays that deals with pointers in detail....

i hav declared a 3 dimesional array like a[2][3][2]={{{2,4},{7,8}, {3,4}},{{2,2},{ 2,3},{3,4}}};

then
a,*a,**a,***a,a +1,*a+1,**a+1,* **a+1...etc what it means?

can any body explain me how to represent arrays in pointer notation for 3 dimensional arrays
( like in one dimensional arrays &a[1] can be writen as (a+1) and a[1] as *(a+1)....)

[Banfa]

Taking this, slightly modified example

int a[2][3][2]={{{2,4},{7,8}, {3,4}},{{2,2},{ 9,3},{3,4}}};

Declares a 3 dimensional array of int that contains a total of 12 integers 2 x 3 x 2

a has type int (*)[3][2]
a[0] has type int (*)[2]
a[0][0] has type int *

taking

int *p = &a[0][0][0];

and treating p as an int array you should find that

p[0] == 2
p[1] == 4
p[2] == 7
p[3] == 8
...
p[8] == 9
...
p[11] == 4

notice that a[1][1][0] is equivilent to p[8]

8 = (((1 * 3) + 1) * 2) + 0

this may not be clear but each step is take the array index used and multiply by the size of the next index. Noteated in algabra as

Declarations

int a[SX][SY][SZ];
int *p = &a[0][0][0];

then

a[X][Y][Z] is equivilent to p[ (((X * SY) + Y) * SZ) + Z ]

[dasarisrikar]

Hello friend,
I am a basic learner....i dint understand the words like
.. a has type int (*)[3][2]
a[0] has type int (*)[2]
a[0][0] has type int *
for the above question......
actually what int (*) syntax mean...pls explain the above three statements meanings deeply........

[dasarisrikar]

i was asked this proble in an exam.....

main()
{
int a[2][3][2]={{{2,4},{7,8}, {3,4}},{{2,2},{ 9,3},{3,4}}};
printf("%u %u %u %d\n",a,*a,**a, ***a);
printf("%u %u %u %d\n",a+1,*a+1, **a+1,***a+1);
}

could u explain the solution for above problem......

Thanks & Regards

[Banfa]

OK well lates take a step back given these declareations

int x;
int *y;
int **z;

then

x has type int, i.e. the expresion x evaluates to value of type int
y has type int * i.e. the expresion y evaluates to value of type int *, a pointer to an integer
z has type int ** i.e. the expresion z evaluates to value of type int **, a pointer to a pointer to an integer

YOU NEED TO UNDERSTAND THIS PARAGRAPH
Now the type "int (*)[2]" can be read as a pointer to an integer array of size 2, "int (*)[3][2]" is a pointer to a 2 dimensioned integer array with dimension sizes 3 and 2.

So what is the value of a in the question?
well a is an array int a[2][3][2] so the expression a returns the value &a[0] which has type int (*)[3][2], this is a pointer to the first item in the array a[2] (&a[0]).

So what is the value of *a in the question?
well a has type "int (*)[3][2]" from above so *a has type int [3][2], since this is an array it is converted to a pointer by the compiler as &(*a)[0] type "int (*)[2]" which is quite complex but basically it becomes a pointer to the first item in the array a[2][3] (&a[0][0]),

What is the value of **a in the question?
well *a has type "int (*)[2]" from above so **a has type int [2], since this is an array it is converted to a pointer by the compiler as &(**a)[0] type int * which is quite complex but basically it becomes a pointer to the first item in the array a[2][3][2] (&a[0][0][0]),

Note that it is always converted to a pointer to the first entry in the array so ignoring type

a == *a == **a

Finally what is the value of ***a
well **a has type "int *" from above so ***a has type int, this is not an array it is an int value reference by pointer through it's array name. This will be the first entry in the array a[2][3][2], a[0][0][0].


Now on to a+1
now a has type "int (*)[3][2]" from above, so a+1 is this pointer moved on by 1. In C pointer arithmatic garuntees that if you add 1 to a pointer it moves on by the size of the thing pointed to. That is for any type T (int, float, double, struct etc.)

T *pT;

(pT+1) - pT = sizeof(T) = sizeof(*pT)

so a+1 is a moved on by the sizeof(*a) a has type "int (*)[3][2]" therefore sizeof(*a) == sizeof(int *(*)[3][2]) == sizeof(int [3][2]) or the size of a 2 dimensioned array of integers dimension 3 and 2. So where a = &a[0], a+1 = &a[1].

Similarly for *a+1
*a has type "int (*)[2]" so *a+1 is a moved on by sizeof(int [2]), the size of what *a points to. Where *a == &a[0][0], *a+1 == &a[0][1]

And **a+1
**a has type "int *" so **a+1 is a moved on by sizeof(int), the size of what **a points to. Where **a == &a[0][0][0], **a+1 == &a[0][0][1]

But ***a+1
***a has type int, suddenly we are no longer doing pointer arithmatic we are doing integer arithmatic, take the value of the integer ***a and add 1. Since ***a == a[0][0][0] == 2 then ***a+1 == 2+1 == 3

[dasarisrikar]

Thankyou BANFA....thanks for your reply.....i am getting new things and i got one more doubt..pls clarify it......

you said....

So what is the value of a in the question?
well a is an array int a[2][3][2] so the expression a returns the value &a[0] which has type int (*)[3][2], this is a pointer to the first item in the array a[2] (&a[0]).

my doubt is....
How the type of a be int (*)[3][2]..and *a of int (*)[2]...so on.
How this is a pointer to the first item in the array a[2] and how a[2] is related to (&a[0])
could you tell me the detailed explanation of the above wt u said.

Thankyou & Regards.

[dasarisrikar]

waiting for your reply.....

[dtimes6]

have you ever seen code like this:

int a[20];
5[a] = 5;
printf("%d",a[5]);

in fact a is address(pointer ) to the array, a[5] is just *(a + 5), and this tells 5[a] is just *(5 + a). The two are same in result, but they are different: in a[b] , a is the base address, b is the address to the base address.