Thread: About conversion operator

What does a conversion operator excatly does, more precisely what does it return? Before going any further read this

Code:

#include <stdio.h>

#define BUG

class Foo
{
  private:
	short int bar;
#ifdef BUG
  char c;
#endif
  public:
	Foo() :
		bar(100)
#ifdef BUG
		,c(12)
#endif
		{}
	operator short int () { return bar; }
};

void foobar(short int n)
{
	printf("%d\n", n);
}

int main()
{
	Foo foo;
	printf("%d\n", foo);
	foobar(foo);
	return 0;
}

output while compiling:
G:\source\test>gxx foobar.cpp -Wall
foobar.cpp: In function `int main()':
foobar.cpp:32: warning: int format, Foo arg (arg 2)

output while BUG is not defined:
100
100

and when BUG is defined
786532
100

So, all works fine if I don't declare `c' but that's not the case. It's also quite surprising how `foobar()' works just fine in both times and direct `printf()' call doesn't. So it must be because stdarg.h that printf is using in it's declaration, right?

Anyway, is there any way to retrieve the value of `bar' so that a) it works if `c' is declared and most important b) it doesn't give me a warning ie. to do it as supposed? All this without ``short int GetBar() const { return bar; }'' of course.