Does C compiler allocates memory for variable name used for declaring an array?

Question

Let us take an example :--

int arr[10] = {1,2,3,4,5,6,7,8,9,10};
int *p = NULL;

For the variable arr, will there be any memory allocation ??

Now, what will happen if =>

p = arr;
p = &arr;
p = &arr[0];

Please help me out to understand this.

Answer 1

This declaration:

int arr[10] = {1,2,3,4,5,6,7,8,9,10};

causes 10 * sizeof (int) bytes to be allocated to hold the 10-element array object. (No space for any pointer object is allocated.) The array is initialized as specified.

int *p = NULL;

creates a single pointer object and initializes it to contain a null pointer value.

p = arr;

arr is an expression of array type. In most contexts, its value is implicitly converted to a pointer to its first element. So this assignment causes p to point to the first (0th) element of arr; *p == 1.

p = &arr;

This is invalid. &arr is of type int(*)[10], or pointer to array of 10 ints. p is of type int*. The types are incompatible. Any conforming C compiler must diagnose this error. (The diagnostic may be a non-fatal warning, but don't let that fool you; it's still an error, what the C standard calls a "constraint violation".)

p = &arr[0];

This is identical to p = arr;. arr[0] is the first (0th) element of the array, an int object with the value 1. &arr[0] is the address of that int object, and is of type char*. So this also causes p to point to the initial element of the array arr.

After this assignment, you can use either arr or p as the prefix for an indexing operator. The indexing operator is actually defined to take a pointer, not an array, as its prefix, so arr[0] uses the result of the array-to-pointer conversion, making it identical to p[0].

But arr and p still cannot always be used interchangeably. For example, sizeof arr gives you the size of the array object (10 * sizeof (int)), while sizeof p gives you the size of a pointer (sizeof (int*)).

Suggested reading: Section 6 of the comp.lang.c FAQ.

(To answer the question in your title, the compiler doesn't, or at least isn't required to, allocate memory at run time for the name of an array. It won't allocate 3 bytes of memory at run time because you named your array arr, or 22 bytes because you called it array_with_a_long_name. It might do so for debugging purposes, but then any such allocated space isn't accessible to your program.)