Difference between array and pointer versions of external declarations

Question

Are the following same:

extern int a[];

and

extern int *a;

I mean, are they interchangable?

possible duplicate of [Changing the value of an external variable](http://stackoverflow.com/questions/11433953/changing-the-value-of-an-external-variable) — Oliver Charlesworth, Sep 01 '13 at 13:47
May be this help you: [Difference between `char *str` and `char str[]` and how both stores in memory?](http://stackoverflow.com/questions/15177420/what-does-sizeofarray-return/15177499#15177499) — Grijesh Chauhan, Sep 02 '13 at 04:34

score 2 · Accepted Answer · answered Sep 01 '13 at 13:47

2

No they are not. You'll see the difference when you try something like a++.

There's plenty of questions about the difference between pointers and arrays, I don't think it's necessary to write more here. Look it up.

answered Sep 01 '13 at 13:47

nickie

5,608
2
23
37

score 1 · Answer 2 · answered Sep 01 '13 at 14:06

They are not the same. The first:

extern int a[];

declares an array of int. The second

extern int *a;

declares a pointer to int. As the C FAQ states:

The array declaration "char a[6]" requests that space for six characters be set aside, 
to be known by the name "a". That is, there is a location named "a" at which six 
characters can sit. The pointer declaration "char *p", on the other hand, requests a 
place which holds a pointer, to be known by the name "p". This pointer can point 
almost anywhere: to any char, or to any contiguous array of chars, or nowhere.

This causes a difference in how the compiler behaves:

It is useful to realize that a reference like "x[3]" generates different code 
depending on whether "x" is an array or a pointer. Given the declarations above, when 
the compiler sees the expression "a[3]", it emits code to start at the location "a", 
move three past it, and fetch the character there. When it sees the expression "p[3]", 
it emits code to start at the location "p", fetch the pointer value there, add three 
to the pointer, and finally fetch the character pointed to. In other words, a[3] is 
three places past (the start of) the object named a, while p[3] is three places 
past the object pointed to by p.

You'll have unexpected behavior if you use extern int *a if a is actually an array. Given the expression a[3], the compiler will treat the first element of a as an address and try to get the element three places past that address. If you're lucky, the program will crash; if you're not, some data will be corrupted.

score 0 · Answer 3 · edited May 23 '17 at 10:26

0

extern int a[];

is an array object

extern int *a;

is a pointer that could be pointed to an array of int

See the above links for more information concerning difference between pointer and array object

*(a++) is giving error but not *(a+1)?? where a is array name?

edited May 23 '17 at 10:26

Community

1
1

answered Sep 01 '13 at 13:45

MOHAMED

41,599
58
163
268

You mean pointed to the integer that is first element of an array, right? Not array itself? Since array is an adress of int itself... – zubergu Sep 01 '13 at 14:03

score 0 · Answer 4 · answered Sep 01 '13 at 14:00

When you see extern int a[] it tells you that somewhere (may be in other file) is declared an array of integers, while extern int * a tells you that you have pointer to an integer declared somewhere else and you wish to use it here. There are more differences but that would be another topic.

Difference between array and pointer versions of external declarations

4 Answers4