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.

Ah, I see that the parsing system has changed... Luckily, it's not anything bigger... Sorry about that

There was a FAQ here somewhere...

..edited..
Doesn't look much better now...

You'll either have to use iostream for output or an explicit cast. printf() doesn't do any type checking so won't call your conversion constructor.

So, going under the hood, what does
operator short int () { return bar; }
gives me (and what type)?

" what does
operator short int () { return bar; }
gives me (and what type)?"

It returns bar (no cast neccesary) by value as a short int, whenever a short int is acceptable (IE, function accepting int/short int, or less acceptably taking a char) but a Foo is passed.