c – Initialization from incompatible pointer type warning when assigning to a pointer

c – Initialization from incompatible pointer type warning when assigning to a pointer

• &arr gives an array pointer, a special pointer type int(*)[5] which points at the array as whole.
• arr, when written in an expression such a int *q = arr;, decays into a pointer to the first element. Completely equivalent to int *q = &arr[0];

In the first case you try to assign a int(*)[5] to a int*. These are incompatible pointer types, hence the compiler diagnostic message.

As it turns out, the array pointer and the int pointer to the first element will very likely have the same representation and the same address internally. This is why the first example works even though it is not correct C.

TL;DR Check the types.

• &arr is of type int (*) [5] (pointer to an array of 5 ints).
• arr is of type int [5], but not always.

Quoting C11, chapter §6.3.2.1, (emphasis mine)

Except when it is the operand of the sizeof operator, the _Alignof operator, or the
unary & operator, or is a string literal used to initialize an array, an expression that has
type ‘‘array of type’’ is converted to an expression with type ‘‘pointer to type’’ that points
to the initial element of the array object and is not an lvalue.

hence,

 int *q = arr;   // int[5] decays to int *, == LHS

and

 int *q = &arr[0];   // RHS == LHS

are same, whereas,

 int *q = &arr; // LHS (int *) != RHS (int (*) [5])

is a mismatched type expression.

Now, it works, because, as already mentioned in Lundins answer, the address of the array variable is likely to be the same as the address of the first element of the array, so despite the type mismatch, the value is same, so this seems to work.

c – Initialization from incompatible pointer type warning when assigning to a pointer

These are the ways to point to a (beginning of) array (without a warning), both work:

int *q = arr;
/* OR */
int *q = &arr[0];

This one is something in between, and will generate a warning:

int *q = &arr;