The type of array by definition in C

Question

I've just read in the book "The C Programming Language" that array is not a variable and saw that assignment of array to pointers (and vice versa) can't be done as a result. So if array is not a variable then what is it?

Answer 1

Array is a data structure containing a number of values, all of which have the same type and array names are non-modifiable l-values(named memory locations)-- it is addressable, but not modifiable. It means that it can't be modified or can't be the left operand of an assignment operator.

int a[10] = {0};
int *p = a;     //OK
a++             // Wrong
a = p;          // Wrong

Answer 2

int numbers[] = {1,2,3}

numbers is not a variable, it is the array name which is nothing but the address of the first element in the array. To verifiy this, look at the address of numbers and the address of numbers[0] by doing this: printf("%p and %p and %p", &numbers, numbers, &numbers[0]); All the the three pointers will have the same values since numbers is nothing but the address of the first element in the array. Therefore numbers is not a variable that contain a pointer or a value since it does not have a dedicated address in the memory to store value in it.

However, look at this pointer variable:

int *pnumbers = numbers;
`printf("%p and %p and %p", &pnumbers, pnumbers, &pnumbers[0]);`

You will notice that &pnumbers has a different address in memory, and that's because pnumber has a dedicated address in the memory where it stores the address of the first element in the array numbers.

Putting the code all together:

#include <stdio.h>

main(){
    int numbers[] = {1,2,3};
    printf("%p and %p and %p\n", &numbers, numbers, &numbers[0]); // Address of numbers, value of numbers, first element of numbers

    int *pnumbers = numbers;
    printf("%p and %p and %p\n", &pnumbers, pnumbers, &pnumbers[0]); // Address of pnumbers, value of pnumbers, first element of the array pnumbers is pointing to
}

Output

0xbfb99fe4 and 0xbfb99fe4 and 0xbfb99fe4 // All three have the same address which is the address of the first element in the array
0xbfb99fe0 and 0xbfb99fe4 and 0xbfb99fe4 // The first one is different since pnumbers has been allocated a memory address to store a pointer which is the first element of the array numbers

Answer 3

It's a placeholder. A symbol to represent a commonly used method of referring to a sequential section of memory. It's not a variable all by itself.

int main(int argc, char** argv)
{
  int array[10];
  int value'
  int* pointer;

  value = array[0];   // Just fine, value and array[0] are variables
  array[0] = value;   // Just fine, value and array[0] are variables
  pointer = &array[0];   // Just fine, &array[0] is an address
  pointer = array;    // Also just fine 
  //Because the compiler treats "array" all by itself as the address of array[0]  
  //That is: array == &array[0]

  &array[0] = pointer   // ERROR,  you can't assign the address of something to something else.
  array = pointer;      // ERROR, array is not a variable, and cannot be assigned a value.

  //Also bad, but technically they compile and could theoretically have their use
  pointer = value;
  pointer = array[0];
  array[0] = pointer; 
  //Intermixing pointers and non-pointer variables is generally a bad idea.
}

array is often treated like a variable because it represents the adddress of (the first item in) that block of memory. But it's not a variable. It doesn't have it's own memory to store anything. People set pointers equal to 'array' because it's a handy convention, compilers know what that means, and it's pretty common.

Answer 4

array is a sequence of elements of the same type. if you assign a pointer to an array, pointer will point to the address of first variable from an array.

int a[10];
int *p;
int *p2;
p = a;
p2 = &a[0];
printf("%d\n", p == p2);

output:

1

Answer 5

 I've just read in the book "The C Programming Language" that array is not a 
 variable and saw that assignment of array to pointers (and vice versa)

Vice versa is allowed... An array is like a constant pointer (you cant change the address it is pointing to). However, you can assign that address to a pointer.

 #include <stdio.h>

 int main()
 {
    int x[3] = {1, 2, 3};

    int *p = x;
    p[0] = 50;
    printf("%d", x[0]);
 }