Thread: about extern declaration

you will have linker error due to duplicate external symbol, I think

If you only initialize one "int a", then any declaration "int a;" (without initialization) will be equivalent to "extern int a;".

Multiple initializations should generate a linker error for multiple symbols of the same name (or whatever the linker calls it).

--
Mats

Originally Posted by jam_fiveface

as we compile each source files seperately and link them after that. while compiling each file seperately compiler will allocate seperate memory if we declare the variable as just int a (not with extern prefix).. so how will compiler handles this when linking the files..

Because it has a special directive to the linker to say "This variable is called a, and it takes up 4 bytes - make one space for it". You'd get very interesting effect, by the way if you do something like this:

Code:

// main.c
#include <stdio.h>
#include <stdlib.h>

int a;

int main()
{
    printf("%d (%08x)\n", a, a);
    return 0;
}

// b.c
float a = 1.0;

It will print some huge number in the 1 billion range (and 3f800000 in hex - I know that one).

Note that g++ (instead of gcc) causes the above code to generate an error for "multiple defined symbols".
--
Mats