Programming

Question

Int a[2][2][2]={{10,2,3,4},{5,6,7,8}};

int *p;

p=&a[2][2][2];

printf("%d",*p);

this gives garbage value,what should be the changes to get p value as 10

Answer 1

a[0]				a[1]
a[0][0]		a[0][1]		a[1][0]		a[1][1]
a[0][0][0]	a[0][0][1]	a[0][1][0]	a[0][1][1]	a[1][0][0]	a[1][0][1]	a[1][1][0]	a[1][1][1]
10	2	3	4	5	6	7	8

In 3D Arrays

a[i][j][k] = *(a[i][j] + k) = *(*(a[i] + j ) + k)  =  *(*(*(a + i) + j ) + k)

and pointer to 3D array is 

METHOD 1 :    int (*p)[2][2] = a;

To get value of 10 ie a[0][0][0]

i=0 , j=0 , k=0

*(*(*( a + 0) + 0) + 0) = ***a    OR    *(*(*( p + 0) + 0) + 0) = ***p

METHOD 2 :   int *p = (int*)(&a);

for (i=0;i<8;i++)
    {
        printf("%d\n",*(p + i));
    }
 

Comments

Int *p;

p=&a[2][2][2]; //this assignment is wrong? Please can u explain

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In method 1, considering everything mentioned to be the same and only printf changed to print **p and *p , what would the respective answers be

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a[2][2][2] - is outside the array bound of int a[2][2][2] as the array index in C starts from 0. So, *p is invalid memory access - probably should cause a segmentation fault and sometimes works fine. But "garbage value" is not the correct answer though we can get it if no segmentation error happens.

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p=&a[2][2][2]; as shown in 3D array diagram and also explained by Arjun Sir scope is out of array index as it varies from a[0][0][0]  to  a[1][1][1] .. so will print some garbage value

but at same time u can point  pointer to point

int *p;

 p=&a[1][1][1]; will print value = 8 - here the pointer will only point to cell a[1][1][1]

 

but if you write following code then

int *p;

    p=&a[0][0][0];

for (int i=0; i<8;i++)

{
    printf(" %d ",*(p+i));  // OUTPUT will be 10 2 3 4 5 6 7 8
}

 

its bcoz

   int *p = (int*)(&a);   is equal to      int *p=&a[0][0][0];  bcoz simply "a" is equal to writing a[0][0][0] bcz name of array is pointer to 1st element in the array

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In method 1,  **p and *p ,will point to address of memory location a[0][0][0] you try this code in programming...

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int(*p)[2][2]=&a;

int(*p)[2][2]=a;

int ***p=(int*)(&a);

int(**p)[2]=a;

among these which are the right ones and the wrong ones

if it's wrong please do tell why it is

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Only int(*p)[2][2]=a;  is Correct

1. &a means address of a  itself not address of array stored in a

3. one cannot declare pointer to 3D array in this fashion ( bcz ***p means pointer to pointer to pointer to address of array ) ie., int * cannot be used to initialize an entity of type int ***

4 . same as point 3  ie.,  value of type int (*)[2][2] cannot be used to initialize an entity of type int (**)[2]

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So int ***p=(int ***)(&a); is correct? If wrong then what's the correct way with the left side of the operator as int***p?

then in that case int *p=&a; should also be wrong? (Considering a to be a one dimensional array)

becz &a is address of a itself and not the array address

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int ***p=(int ***)(&a); is correct syntactically

int *p=&a; is not correct rather it should be int *p=a;

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but programs do run when given as int*p=&a; how?
or
is it correct because its given in different lines
int *p;
p=&a;  ??

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i think you know better now so try urself... as programming is never done by asking doubts its rather done by practising and asking your doubts to computer..dont waste time ur precious time in questioning

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okay :)

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for ur doubt  int*p=&a; is correct or not - read the previous comments carefully the answer there...

Answer 2

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

now printf("%d", *p);

will give u the 10