What are the differences between these type of Pointers? [duplicate]

Question

I was programming a code in C++ when I accidentally put brackets to my pointer and my programs output changed.

Since I am new to programming I wanted to know the difference between these type of pointers:

int* A[n]; 
int (*A)[n];
int *(A[n]);

I have read in my textbook that arrays are also a type of pointers.

Answer 1

int* A[n];

A first and foremost is an array no matter what type the element is. After applying pointer *, we know A is an array of int pointers.

int (*A)[n];

By applying brackets, pointer * has higher precedence over array [] in this case. Then A is first and foremost a pointer no matter what it is pointing to. After applying array [], we know A is a pointer to an array of int.

int *(A[n]);

Brackets won't change any precedence order that would affect the array [], therefore removing brackets would yeild int* A[n] same as your 1st case.

Are array pointers?

No. Array is a datastructure that allocates memory pool and stores the data sequentially where as Pointer points to a particular index in memory pool and references the data stored at that memory location.

Answer 2

This article contains good examples in reading type declarations in C. http://www.unixwiz.net/techtips/reading-cdecl.html

Basically, you can read out types according to the following precedence:

1. (Often parenthesized) Inner-most type

2. Right-most types (Mostly array types: [])

3. Left-most types, except the outer-most one (Mostly pointer types: *)

4. Outer-most types (Mostly primitive types: int, char..)

For example, the types you presented can be read out as follows:

int* A[n];  // An array ([n]) of pointer (*) of int.
int (*A)[n];  // A pointer (*) of array ([n]) of int.
int *(A[n]);  // An array ([n]) of pointer (*) of int.

So basically, the first and third type are identical.