pointers

Question

char ** array [12][12][12]

what is valid initialization of p ?

ANS = char * * * * *p=array;

but i feel it should be char * * * * * *p =array;

please help

Comments

Debashish, i had checked it, no problem...take this code snippet and run in ideone...it is compiling without any error

#include <stdio.h>

int main(void) {
char a[100];
char *p1=a;
char *b[100];
char **p2=b;
char **c[100][100];
char ****p3=c;
return 0;
}

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ok. I stated my view already. using c99 will give the warning and using -Wall will also produce warning regarding 

initialization from incompatible pointer type. 

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plz anyone provide a good resource for  multidimensional array(2D 3D).

Answer 1

char ** a[12][12][12] ;
Array 'a' is a 3-Dimensional array, where every element is a pointer to a char pointer(pointing to char). Type of 'a' is char ** (*)[12][12], whereas the type of every element of 'a', is char **. Array 'a' will be pointing to the first(among the 12) two-dimensional array.

What is the valid initialization of p?

If p is also pointing to the same as 'a'. Means p is also pointing to the first two-dimensional array of a. Then
char ** (*p)[12][12] = a;

Type of p will be char ** (*)[12][12].
Note: p is a pointer to a two-dimensional array.

If p is pointing to array 'a'. Means p is a pointer to a three-dimensional array.
char ** (*p)[12][12][12] = &a;

Type of p is char ** (*p)[12][12][12] as it is pointing to three-dimensional array 'a'.

If p is pointing to an element of array 'a'. The type of every element of array 'a' is char **. And we need a pointer to this. Therefore,
char ***p = (char ***)a;
Here we do type casting because the type of 'a' is char ** (*)[12][12]. By this action, p will point to the first element of array 'a'.

Here is the code.

 

#include<stdio.h>

int main() {
	int a1[2] = {1, 2}; // Type of a1 is int *
	//becuase a is pointing to first element whose type is int.
	int *p1; // Type of p1 is int *.
	p1 = a1; // As both type is int *. This conversion is compactible.
	printf("%d\n", *p1); // This will print the first element of a.
	
	int a2[2][3] = {{1, 2, 3}, {4, 5, 6}}; // Type of a2 is int (*)[3];
	// a2 is pointing to one dimensional array consisting of 3 elements.
	int (*p2)[3]; // p2 is a pointer to one dimenstional array consisting 3 elements.
	// Type of p2 is int (*)[3].
	p2 = a2; // As both type is int (*)[3]. This conversion is compactible.
	
	//Printing the second element of array 2.
	printf("%d\n", a2[0][1]);
	printf("%d\n", (*(*p2) + 1));
	
	
	// When we should use the run of * * * * to initialize p.
	int a3 = 5; 
	int *p3 = &a3; 
	int **pp3 = &p3;
	int ***ppp4 = &pp3;
	
	printf("%d\n", a3);
	printf("%d\n", *p3);
	printf("%d\n", **pp3);
	printf("%d\n", ***ppp4);
	printf("===================\n");
	
	/* ============ */
	int ** array[12][12][12];
	// Every element of array is pointer to int pointer(pointer pointing to an int).
	// Type of array is int ** (*)[12][12]. It is three dimensional array.
	
	array[0][0][0] = pp3;
	// pp3 has type int ** and every element of array has also the same type int **.
	printf("%d\n", **array[0][0][0]); // This will output 5.
	
	//Now if we want a pointer to array named array.
	int ** (*pointer)[12][12] = array;
	printf("%d\n", **(*(*(*pointer)))); // This will also output 5.
	
	//If you want a pointer to the element of array named array.
	int ***parray = (int ***)array;
	// Here we are doing the type casting because type of array is int ** (*)[12][12].
	// We want a pointer to the first element of array.
	// Type of every element of array is int **.
	// And we want to pointer to that element That's why type of parray is int ***.
	printf("%d", ***parray);
}

Comments

so u mean to say char * * * * *p =array is correct?

or u mean  to say both are wrong??

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@Aish, Both are wrong. Run the code, it will help you to understand better.

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@DebashishSir, Thank you after a long,i get this.