array of integers vs. pointer to integer in c++

Question

If I have int x[10] and int *y, how can I tell the difference between the two?

I have two ideas:

1. sizeof() is different.

2. &x has different type --- int (*p)[10] = &x works but not int **q = &x.

Any others?

In some template library code, I need to determine whether a pointer is a "real" pointer or degenerated from an array. I can't look at source code as the library user does not exist until I write the library. ... I can work around this by rewriting the code, so now this is only a theoretical exercise.

Answer 1

There is no general method - you either already know the type because you have just declared the object, or the type will have decayed to a pointer and been lost. Please explain what problem you are trying to solve by differentiating between them.

Answer 2

The sizeof idea is not very good, because if the array happens to have a single element, and the element type happens to be the same size as a pointer, then it will be the same size as the size of a pointer.

The type matching approach looks more promising, and could presumably be used to pick a template specialization (if that's what you're up to).

Answer 3

int x[10] will be allways allocated on the stack at the place were it is called. x value can never be changed.

int *y simply declare a pointer to a memory , the pointer value can be changed at any time without any restrictions.

int *y can have any arbitrary value. Meaning , it can point to stack-allocated Memoryor heap allocated memory. Technicly it can also point to invalid memory but that won't make any sense.

int x[10] guarntee you are allways pointing to a valid memory and you don't have to worry about memory deallocation.

when using int* y , you have to worry about the memory it is pointing to .

Also keep in mind that the lot your porgram has dynamically allocated memory the more it will be exposed to errors , leaks , performece issues , allocation\deallocation assimetry and many other kinds of problem.

Answer 4

They are different types.

How do you tell the difference between an int32 and a uint32, or a uint32 and a char[4]?

The right way to think of it is that the only way that they are similar is that in certain contexts (including array indexing!) an array promotes into a pointer.

Answer 5

Assuming that you aren't trying to do this for a type declared at function scope:

struct yes { char pad; };
struct no { yes pad[2]; };

template <typename T, size_t N> yes is_array_test(T (&arr)[N]);
no is_array_test(...);

#define IS_ARRAY(x) (sizeof(is_array_test(x))==sizeof(yes))