Why Array of Pointer and Array giving same output - c

Question

1)
How to access array element with array of pointers?
By pointer to an array we can access like this (∗a)[0]
,(∗a)[22]
,….. like this
right?
but how with array of pointer?
2)
Check these three codes , one of them use array and one of them use array of pointer, and 3rd one using double pointer, but all giving same output. Why??
#include <stdio.h>
int ptr[12];
int main(void) {
if(*(ptr+5)==*(ptr+3)){
printf("Equal");
}
else{
printf("Not Equal");
}
return 0;
}
#include <stdio.h>
int *ptr[12];
int main(void) {
if(*(ptr+5)==*(ptr+3)){
printf("Equal");
}
else{
printf("Not Equal");
}
return 0;
}
#include <stdio.h>
int **ptr[12];
int main(void) {
if(*(ptr+5)==*(ptr+3)){
printf("Equal");
}
else{
printf("Not Equal");
}
return 0;
}

Answer 1

In *(ptr+5):
ptr is the name of an array.
When an array is used in an expression and is not the operand of sizeof or unary & and is not a string literal used to initialize an array, it is converted to a pointer to its first element. So ptr points to element 0 of the array.
ptr+5 adds five, resulting in a pointer to element 5 of the array.
* dereferences the pointer, producing the contents of element 5 of the array.
Since the array is defined outside any function, it is initialized to zero. Therefore element 5 is zero, so *(ptr+5) is zero. Similarly, *(ptr+3) is zero, so comparing them shows they are equal.
The examples with int ptr[12];, int *ptr[12];, and int **ptr[12]; just change the types of the elements in the array. In each case *(ptr+5) and *(ptr+3) refer to elements 5 and 3 of the array. In each case, the values of the elements are zero (null pointers for the pointer types).
If you want to see differences between different elements of the array, then assign different values to the elements of the array.

Answer 2

Just keep incrementing the pointer as you go along reading the array. Analyze the
code below:
#include <stdio.h>
int main(void) {
int *ptr[3];
int a[] = { 0,1,2};
int b[] = { 3,4,5 };
int c[] = { 6,7,8 };
ptr[0] = a;
ptr[1] = b;
ptr[2] = c;
for (int i = 0;i < 3;i++)
printf("%d %d %d\n",*ptr[0]++,*ptr[1]++,*ptr[2]++);
return 0;
}
Answered in comments.