Question

I am unable to understand the weird behaviour of the program. It throws a warning but the output is surprising. I have 3 files, file1.c, file2c, file1.h

file1.c :

#include "file1.h"

extern void func(void);

 int main(void)
{
func();
printf("%d\n",var);
return 0;
}

file2.c:

int var;

 void func(void)
{
var = 1;
}

and file1.h is :

#include<stdio.h>

extern double var;

When I compile and run, it shows a warning as expected but on printing prints 1. But how is it possible when I am changing the var declared in file2.c

Answer 1

Just setting the type on an extern definition doesn’t change the value stored in memory – it only changes how the value is interpreted. So a value of 0x0001 is what is stored in memory no matter how try to interpret it.

Once you’ve called func() you’ve stored the value 0x0001 in the location where a is stored. Then in main when a is passed to printf the value stored at location a plus the next 4 bytes (assuming 32bits) is pushed onto the stack. That is the value 0x0001xxxx is pushed onto the stack. But the %d treats the value passed in as an int and will read the first 4 bytes on the stack, which is 0x0001 so it will print 1.

Try changing your code as follows:

file1.c :

#include "file1.h"

extern void func(void);

int main(void)
{
    func();
    printf("%d %d\n",var);
    return 0;
}

file2.c:

int var;
int var2;

void func(void)
{
    var = 1;
    var2 = 2;
}

and file1.h is :

#include<stdio.h>

extern double var;

You will probably get the output 1 2 (although this depends on how the compiler stores the variables, but it’s likely it will store them in adjacent locations)