How to define extern variable along with declaration?

问题

Wiki says:

The extern keyword means "declare without defining". In other words, it is a way to explicitly declare a variable, or to force a declaration without a definition. It is also possible to explicitly define a variable, i.e. to force a definition. It is done by assigning an initialization value to a variable.

That means, an extern declaration that initializes the variable serves as a definition for that variable. So,

/* Just for testing purpose only */ 
#include <stdio.h>
extern int y = 0;
int main(){
    printf("%d\n", y);
    return 0;
}

should be valid (compiled in C++11). But when compiled with options -Wall -Wextra -pedantic -std=c99 in GCC 4.7.2, produces a warning:

[Warning] 'y' initialized and declared 'extern' [enabled by default]

which should not. AFAIK,

extern int y = 0;

is effectively the same as

int i = 0;

What's going wrong here ?

回答1:

All three versions of the standard — ISO/IEC 9899:1990, ISO/IEC 9899:1999 and ISO/IEC 9899:2011 — contain an example in the section with the title External object definitions (§6.7.2 of C90, and §6.9.2 of C99 and C11) which shows:

EXAMPLE 1

int i1 = 1;        // definition, external linkage
static int i2 = 2; // definition, internal linkage
extern int i3 = 3; // definition, external linkage
int i4;            // tentative definition, external linkage
static int i5;     // tentative definition, internal linkage

The example continues, but the extern int i3 = 3; line clearly shows that the standard indicates that it should be allowed. Note, however, that examples in the standard are technically not 'normative' (see the foreword in the standard); they are not a definitive statement of what is and is not allowed.

That said, most people most of the time do not use extern and an initializer.

回答2:

This code is perfectly valid.

But any compiler is free to issue additional (informative or not) diagnostics:

(C99, 5.1.1.3p1 fn 8) "Of course, an implementation is free to produce any number of diagnostics as long as a valid program is still correctly translated."

What a compiler cannot do is not emitting a diagnostic when there is a constraint or syntax violation.

EDIT:

As devnull put in the OP question comments, Joseph Myers from gcc team explains in a bug report questioning this diagnostic:

"This is a coding style warning - the code is valid, but extremely unidiomatic for C since "extern" is generally expected to mean that the declaration is not providing a definition of the object."

来源：https://stackoverflow.com/questions/24436770/how-to-define-extern-variable-along-with-declaration

标签

c99

extern

gcc-warning

initializer