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

Question

For the following code:

    int (*ptr)[10];
    int a[10]={99,1,2,3,4,5,6,7,8,9};
    ptr=&a;
    printf("%d",(*ptr)[1]);

What should it print? I'm expecting the garbage value here but the output is 1.
(for which I'm concluding that initializing this way pointer array i.e ptr[10] would start pointing to elements of a[10] in order).

But what about this code fragment:

int *ptr[10];
int a[10]={0,1,2,3,4,5,6,7,8,9};
*ptr=a;
printf("%d",*ptr[1]);

It is giving the segmentation fault.

Answer 1

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

Answer 2

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.

Answer 3

int (*p)[10] is a pointer to an array of 10 integers in each row i.e there can be any number of rows. basically it can be used to point to a 2D array and the dimensions can be accessed by incrementing i for *(p+i)[10] which is same as a[i][10], here 'i=1,2,3...' to access 1,2,3.. rows.

where as, int *p[10] is an array of 10 integer pointers.

If array is two dimesional i.e, for example

b[2][10]={{0,1,2,3,4,5,6,7,8,9},{10,11,12,13,14,15,16,17,18,19}}; basically, (*ptr)[10] // where the higher dimension index is omitted i.e, 2 can be used as two dimensional pointer to an array(containing 10 elements in each row i.e, the 1st dimension) like (*ptr)[10] = &b. In the printf("%d",(*ptr)[1]); as provided (*ptr) is same as *(ptr+0)evaluates to the first dimension i.e, b[0][0] which value is 0. like wise, to access the 2nd dimension of the array b[1][0] the expression will be **(ptr+1) or *(ptr+1)[0] or *(*(ptr+i)+j); // where i=1 and j=0 gives the first element of the 2nd dimension i.e, 10.

I've answered it long so to understand it easy.

Answer 4

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;
}

Answer 5

int (*p)[10] means that p is now a pointer to an integer array of size 10.

int *p[10] means that p is an array of 10 integer pointers .