Thread: extern variable throwing : "has both ‘extern’ and initializer" error

The code is throwing error despite 'i' being defined after initialization, why?

Code:

#include<iostream>


using namespace std;


int main()
{
        extern int i=898;
        cout<<i<<endl;
        return 0;


}

extern means that the variable defined elsewhere, possibly further down in the same file, but usually in another source file (and therefore object file). Since extern used on a variable does not define the variable, you cannot initialize it there. Here is one possibility that will compile.

Code:

#include <iostream>
using namespace std;

int main() {
    // tell compiler about existence of i that will be defined elsewhere
    extern int i;

    cout << i << '\n';

    return 0;
}

// define the variable (could be in another source file)
int i = 898;

Since extern used on a variable does not define the variable, you cannot initialize it there.

Initializing an extern variable defines that - the question is why definition - extern i =898; is different from definition - int i =898;

Originally Posted by gaurav#

Initializing an extern variable defines that - the question is why definition - extern i =898; is different from definition - int i =898;

I don't know what to tell you.
If you have a reference for your assertion, give it.
But obviously it's not the case here, hence the error.
Perhaps it has something to do with it being a local variable.

Check this out:

Code:

#include <iostream>
using namespace std;

extern int k =89;

int main() {
cout<<k;
	return 0;
}

This works fine

That throws a warning for me:

Code:

$ g++ -Wall -pedantic -O0 -g -std=c++14 main.cpp 
main.cpp:3:12: warning: ‘k’ initialized and declared ‘extern’
 extern int k = 898;
            ^

Source:

Code:

#include <iostream>

extern int k = 898;

int main (int argc, char* argv[])
{
    std::cout << k << std::endl;

    return 0;
}

Since it throws a warning, i would not count on it working the way you think it works.

This warning can only hide a linker error of multiple definition which doesn't stand in this case. focus on this: how these definitions: int K=898 & extern int K=898 are different at function scope.

In 2.2.2. Variable Declarations and Definitions of c++ Primer 5th ed Lippman et.al. I came across this rule:

It is an error to provide an initializer on an extern inside a function.

i.e. trying to define an external variable inside of a function is not allowed, which makes sense. But you can declare it there so that it can be used there.

The reason behind it is not given hence the question is asked.