How to write a static C++ library and link it to an executable using g++ on Windows 10?

Question

main.cpp

#include <iostream>

int main() {
    std::cout << "main()" << std::endl;
    foo();
    return 0;
}

foo.cpp

#include <iostream>

extern "C" {
  void foo() {
      std::cout << "bar" << std::endl;
  }
}

Compile static library:

$ g++ foo.cpp -static

Error:

undefined reference to `WinMain'

But this compiles:

$ g++ foo.cpp -shared -o foo.lib

Now I have a static library named foo.lib (supposedly).

I try to compile an executable that links to it:

$ g++ -L -lfoo main.cpp -o main.exe

And get this error:

'foo' was not declared in this scope

But foo is declared in the static library that I'm linking with. If the link works, I don't think I need to declare it in main.cpp also. So why isn't the link working?

---

Update.

I added void foo(); to main.cpp so it doesn't complain that foo needs to be declared.

#include <iostream>

void foo();

int main() {
    std::cout << "main()" << std::endl;
    foo();
    return 0;
}

So I try to compile again and I get this new error:

undefined reference to `foo()'

Why would I need to define foo in main.cpp? It's already defined in foo.cpp which is the static library.

If I have to define foo in main.cpp that defeats the entire purpose of linking to the library foo.lib.

UPDATES

• Removing all the extern "C" { ... } lines doesn't make the "foo is undefined" errors go away.

Answer 1

What follows are the magical incantations you seek:

• main.cpp

#include <iostream>
extern void foo();
int main() { 
  std::cout << "main()" << std::endl; 
  foo();
} 

• foo.cpp

#include <iostream>
void foo() { 
  std::cout << "bar" << std::endl; 
}

Console commands:

$ g++ -o foo.obj -c foo.cpp

$ ar rcs foo.lib foo.obj

$ g++ main.cpp foo.lib -o main.exe

These spells conjure up the static lib foo with the executable main statically linked to it.