Difference between *ptr[10] and (*ptr)[10]

int *ptr[10];

This is an array of 10 int* pointers, not as you would assume, a pointer to an array of 10 ints

int (*ptr)[10];

This is a pointer to an array of 10 int

It is I believe the same as int *ptr; in that both can point to an array, but the given form can ONLY point to an array of 10 ints

int (*ptr)[10];

is a pointer to an array of 10 ints.

int *ptr[10];

is an array of 10 pointers.

Reason for segfault:

*ptr=a; printf("%d",*ptr[1]);

Here you are assigning the address of array a to ptr which would point to the element a[0]. This is equivalent to: *ptr=&a[0];

However, when you print, you access ptr[1] which is an uninitialized pointer which is undefined behaviour and thus giving segfault.

int(*)[10] is a pointer to an int array with 10 members. i.e it points to int a[10].

where as int *[10] is array of integer pointers

#include <stdio.h>
int main()
{

int *ptr[10];
int a[10]={0,1,2,3,4,5,6,7,8,9};

printf("\n%p  %p", ptr[0], a);

*ptr=a; //ptr[0] is assigned with address of array a.

printf("\n%p  %p", ptr[0], a); //gives you same address

printf("\n%d",*ptr[0]); //Prints zero. If *ptr[1] is given then *(ptr + 1) i.e ptr[1] is considered which is uninitialized one.

return 0;
}