C, What does the following declaration mean? int (*ptr)[10];

int (*ptr)[10] - pointer to an array of integers

int* ptr[10] - an array of int pointers

Array of pointers
pointer may be arrayed like any data type. To assign the address of an integer variable called var to third element of the pointer array, write
ptr[2] = &var;

to find the value of var, write
*ptr[2].

Pointer to array
pointer to an array is a single pointer, that hold the address of an array of variables. in above case it is an integer array.

hence these two are different concept. not same

It declares a variable ptr which is a pointer to an array of 10 ints

ptr is a pointer to an array of 10 int.
Try

int (*ptr)[10];
int a[10]={0};
ptr= &a;
*ptr[0] = 5;
printf("%d\n", a[0]);

Hi,

as many suggests here, I don't think ptr is an array of 10 ints. Instead, it is an array of integer pointers.

ptr is an array[sized 10] of integer pointers

In fact it creates an uninitialized set of pointers. You can understand it from the code below.

#include<stdio.h>
 
int main()
{
    int (*ptr)[10];
    *ptr[0]=1;
    printf("%d", ptr[0][0]);
    return 0;
}

The aforesaid code when compiled shows a warning:
"warning: 'ptr' is used uninitialized in this function"

But when executed it outputs '1'.

If you desire, you can initialize ptr[0],ptr[1] etc with different sized integer arrays.

I think I have given enough explanations. Your comments are welcome. :-)