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cc99externgcc-warninginitializer

How to define extern variable along with declaration?


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 ?


Solution

  • 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.