Does it make a difference to specify its size when declaring an array using extern?

Question

e.g. Is there any difference between these two statements:

extern char a[];
extern char a[4];

What if the real definition of a (in another source file) is

char a[5];

but not

char a[4];

Have a look at http://techblog.rosedu.org/c-extern-internals.html — Mihai Maruseac, Aug 14 '13 at 05:32
possible duplicate of [C the same global variable defined in different files](http://stackoverflow.com/questions/17800187/c-the-same-global-variable-defined-in-different-files) — jxh, Aug 14 '13 at 05:37

score 1 · Answer 1 · answered Aug 14 '13 at 06:03

extern int a[] declares a to be an array of int with an unspecified size, and is considered an "incomplete type" (C.11 §6.7.6.2 ¶4). An incomplete type is one that for which there is insufficient information to determine its size (C.11 §6.2.5 ¶1). The use of extern means the name has "external linkage" (C.11 §6.2.2 ¶4). All references in the program to the same name with external linkage refer to the same object (C.11 §6.2.2 ¶2).

If you have extern int a[4], but it is defined elsewhere as int a[5], then this will lead to undefined behavior (C.11 §6.2.7 ¶2):

All declarations that refer to the same object or function shall have compatible type; otherwise, the behavior is undefined.

score 0 · Answer 2 · edited Aug 14 '13 at 05:51

The following:

extern char a[ ]; // (1)

... means: "go look somewhere for an array of characters called a, it exists.", whereas:

extern char a[ c ]; // (2), where c is some constant.

... means: "go look somewhere for an array of characters of size ( char * c ) called a, it exists.".

Practical examples of both declarations: You should be doing the second declaration if the size of the array is known. If your array is an VLA then it should be declared using the first declaration.

score -2 · Answer 3 · edited Aug 14 '13 at 06:07

-2

I've understood, the "extern" statement is just "declaration statement" for the c compiler.

"extern char a[4]"'s "4" is nothing. C compiler doesn't use "4";

Please note the following:

source a : extern char a[4]
source b:  char[5]

gcc 4.6.3 compile result : ok
running result : ok

edited Aug 14 '13 at 06:07

NREZ

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answered Aug 14 '13 at 05:42

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Source b won't compile. – alk Aug 14 '13 at 06:30
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-1. The compiler may well use the `4` if you use `sizeof a` somewhere. – Jens Gustedt Aug 14 '13 at 07:11

Does it make a difference to specify its size when declaring an array using extern?

3 Answers3