Pointers to array of pointers in c

Question

Can someone please explain me what is the difference between the two following declarations:

char (*arr_a)[5];
char arr_b[20];

and why:

sizeof (*arr_b) = sizeof (char)
sizeof (*arr_a) = 5*sizeof(char)

Answer 1

arr_a is a pointer to an array of 5 char while arr_b is an array of 20 chars. arr_b is not a pointer unlike arr_a.

sizeof (*arr_b) equals to sizeof (char) because *arr_b is of type char (equivalent to arr_b[0]). For

sizeof (*arr_a) equals to 5*sizeof(char) because *arr_a refers to an array of 5 chars and sizeof returns the size of array which is 5.

Answer 2

char (*arr_a)[5];

declares a pointer to a 5-element array of char.

char arr_b[20];

declares just a 20-element array of char.

So, the output of

sizeof (*arr_a)

should be straight forward -- dereferencing the pointer to an array yields the array and it's size is 5.

The following:

sizeof (*arr_b)

gives 1, because dereferencing the identifier of an array yields the first element of that array, which is of type char.

---

One thing you need to know to fully understand this is how an array evaluates in an expression:

• In most contexts, the array evaluates to a pointer to its first element. This is for example the case when you apply indexing to the array. a[i] is just synonymous to *(a+i). As the array evaluates to a pointer, this works as expected.

• There are exceptions, notably sizeof, which gives you the storage size of the array itself. Also, _Alignof and & don't treat the array as a pointer.