Extern code not working

Question

Why aren't the codes below working ? Please explain.

#include<stdio.h>
#include<stdlib.h>

int main(int number, char arg[])
{
    extern int i;
    i = 5;
    printf("%d",i);
    return 0;
}

#include<stdio.h>
#include<stdlib.h>

int main(int number, char arg[])
{
    extern int i;
    i = (int) malloc(sizeof(int));
    i = 5;
    printf("%d",i);
    return 0;
}

You need to say what you mean by not working. A link error? Where do you define `i`? — CB Bailey, Oct 21 '12 at 14:50
By `extern int i;` you are telling the compiler that `i` *exists* elsewhere. Since, you define it nowhere, it complains. — P.P, Oct 21 '12 at 14:53
Incidentally, the arguments of `main()` are `int argc, char *argv[]`. You might be surprised if you start accessing `arg` in your program. — Pascal Cuoq, Oct 21 '12 at 15:23

score 2 · Accepted Answer · answered Oct 21 '12 at 14:51

extern is used to specify that a variable exists, but is not yet defined. You do not create the variable, only specify to the compiler that it exists. If it does not, you will have an error at linking time.

I suggest you read more about extern keyword

A simple example of use would be two .c files, one with your extern variable as global, and one which prints this variable

file.c

int value = 5;

main.c

int main() {
  extern int value;

  printf("%i\n", value);
  return 0;
}

compiling this using gcc file.c main.c will output 5

score 0 · Answer 2 · answered Oct 21 '12 at 14:50

0

Extern variable's values must be defined from outside the function in which they are defined.

answered Oct 21 '12 at 14:50

Afaq

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Armen Tsirunyan · Answer 3 · 2012-10-21T15:35:43.527

0

When writing

extern int i;

you're simply declaring (but not defining) a variable named i. You need to define i somewhere in your program. For example, it could be just after main or in another .c file

int main(int number, char* arg[])
{
    extern int i;
    i = 5;
    printf("%d",i);
    return 0;
}
int i; //here you define i

edited Oct 21 '12 at 15:35

answered Oct 21 '12 at 14:51

Armen Tsirunyan

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score 0 · Answer 4 · edited May 23 '17 at 10:34

The extern keyword provides, by default (whitout an initializer for instance), a declaration of the variable. You need to define your variable. To have an internal linkage:

#include <stdio.h>
#include <stdlib.h>

static int i;

int main(int number, char arg[])
{
    extern int i;
    i = 5;
    printf("%d",i);
    return 0;
}

And an external linkage:

#include <stdio.h>
#include <stdlib.h>

int i;

int main(int number, char arg[])
{
    extern int i;
    i = 5;
    printf("%d",i);
    return 0;
}

Extern code not working

4 Answers4