Declaration and definition of an external variable.

**Mr.Lnx** · 01-11-2014

Hello to all

Why I take warning on this code :

Code:

#include<stdio.h>
extern int v  = 0;  // Declaration and definition of an external

int main(void)
{
    v=5;

    printf("%d" , v);

    return 0;
}

Is there any error? Why I take a warning: 'v' initialized and declared 'extern'|
||=== Build finished: 0 errors, 1 warnings ===| ???

and what is the meaning of :

An extern declaration that initializes a variable serves as a definition of the variable. This rule prevents multiple extern declarations from initializing a variable in different ways. ??? any example???? of this??? in order to understand it better.

Thank you in advance

**oogabooga** · 01-11-2014

You don't need the keyword "extern" to declare an object with external linkage. Usually you don't use the word extern in the object definition, only in the declarations (usually in other files) that you want to link to the original object. So your program could be written like this:

Code:

int v = 1;

void func(); // declaration of function in other file

int main() {
    v = 5;
    func();
    return 0;
}

// in another file
#include <stdio.h>

extern int v;   // this connects v to the v in the other file

void func() {
    printf("%d\n" , v);
}

**smokeyangel** · 01-11-2014

Originally Posted by Mr.Lnx

An extern declaration that initializes a variable serves as a definition of the variable.

This bit is straightforward. Initializing v to 0 makes it a definition, so you can't define any other global variables named 'v' in your program. 'extern' doesn't actually make any difference in this case.

Code:

extern int v;  // no initialisation, no definition

Here 'extern' means "don't define v, I'll define it somewhere else". If you have only 'extern v' declarations in your programs, you can't use 'v', you'll get an indefined symbol error.

Code:

int v;

This is an annoying one (I can't remember the term for it) -- but it's a declaration in that you can have as many as you like so long as they're all the same type. However unlike extern declarations, having these is sufficient for one of them to become the definition at compiler/link time.
If you have a single

Code:

int v = 0;

definition anywhere in your program, that will take precedence over the others.

This rule prevents multiple extern declarations from initializing a variable in different ways. ??? [/I] any example???? of this??? in order to understand it better.

I don't know what this means. You can only ever initialise the variable once. There's nothing wrong with what you've done -- but it is unusual to see 'extern' on a definition.

**c99tutorial** · 01-12-2014

You can think of "extern" as the variables counterpart to function prototypes. You place all function prototypes in .h files, and you define the actual functions in the .c files, right? You can do something similar with the variables

foo.h file

Code:

extern const char PROGNAME[100];

foo.c file

Code:

#include "foo.h"

const char PROGNAME[100] = "Calculator";

Any file which now #include's that .h file will then have access to the symbol PROGNAME, which in this example is a 100 byte character buffer.

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