operator declaration within namespace in a header?

Question

Please excuse me but I didn't know to give a name to the title in a short way.

Why do I need to declare an overloaded operator inside the header to make it work in this example:

HEAD.H

#pragma once

namespace test {
    class A {
    public:
        A() : x(0) {}
        int x;
    };
    A& operator++(A& obj);  //this is my question
}

HEAD.CPP

#include "head.h"

namespace test {
    A& operator++(A& obj) {
        ++obj.x;
        return obj;
    }
}

MAIN.CPP

#include <iostream>
#include "head.h"
using namespace std;
using namespace test;

int main() {
    A object;
    ++object;  //this won't work if we delete declaration in a header
    return 0;
}

operator++ is defined and declared in a namespace inside "head.cpp" so why do I need to declare it one more time in a header? Thank you.

@Alex : Defining non-assignment operators outside of your class is generally considered good practice. However, the operator could be defined inline inside of the header... — ildjarn
– ildjarn, Commented Nov 18, 2011 at 22:11
@Alex I defined it outside of a class cos of this example on which I'm strugling currently, it could be the same question on any other function. thanks. — codekiddy
– codekiddy, Commented Nov 18, 2011 at 22:16
Just found a good answer here: stackoverflow.com/questions/4652932/… :) — AlexTheo
– AlexTheo, Commented Nov 18, 2011 at 22:22

Vlad · Accepted Answer · 2011-11-18 22:21:39Z

5

The CPP files are compiled independently of each other, and they only see the header files they've included (which are in fact textually added to the source code of the CPP before compilation). As such you'll use the header file to inform the compiler that a function exists with that signature (be it an operator overload).

Then the output from your CPP files is put together by the linker, which is when you'd find out if for instance you had declared a function in a header file but never taken the trouble to implement it.

Simple example with namespaces:

#include <iostream>

namespace test{
    int f() { return 42; }
    int g() { return -1; }
}

namespace other{
    int f() { return 1024; }
}

using namespace other;

int main(){

    //error: 'g' was not declared in this scope
    //std::cout << g() << std::endl;

    std::cout << test::f() << std::endl; //42
    std::cout << f() << std::endl; //1024

    return 0;
}

edited Nov 18, 2011 at 22:21

answered Nov 18, 2011 at 22:02

Vlad

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5 Comments

bames53 Over a year ago

Yeah, during compilation head.cpp has no effect on main.cpp, they just don't see each other. For example, try removing the operator in head.cpp but leaving your declaration in head.h. main.cpp will still compile just fine. (there will be an error in the linking stage after compilation.) This is a result of the C++ compilation model.

codekiddy Over a year ago

I know I sound noobish but I just dont get the point, I've included everything from namespace in my "MAIN.CPP" by "using namespace test" so operator function is defined and visible and why it doesnt work then? header is just a declaration.

Vlad Over a year ago

Sadly that's not how C++ namespaces work. See here for a quick introduction. Essentially they're about grouping things logically and eliminating naming conflicts. Among other things, using namespace std allows you to say cout << ... instead of std::cout << ..., but it doesn't mean you don't have to include <iostream>.

Werner Henze Over a year ago

When you are compiling main.cpp the compiler will NOT see what is inside head.cpp. "using namespace test" will not pull in any stuff declared in head.cpp, it will just take the namespace test and make every symbol, that is in namespace test, also available without explicitly preceedig it with the namespace; after "using namespace test" you do not need to write test::A, you can just write A.

codekiddy Over a year ago

OK!! I got it now, I thought namespaces have similar behaviour as headers.(sily me) thanks a bunch to everyone.

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