ostream showbase does not show "0x" for zero value

Question

PSPS: (a Pre-scripted Post-script)
It has just come to mind that a more prescient question would have included the notion of: Is this non-display of "0x"(showbase) for zero-value integers a standard behaviour, or is it just a quirk of my MinGW implementation?

It all began on a pleasant Sunday morning... I want to dump some Handles in their hex representation, and in a consistant, formatted way.
I want a leading 0x and a fixed width, but this is proving to be elusive using the expected stream manipulators.
The only way I've found to do this is to cast the Handles to an unsigned long. This seems a bit unreasonable, and I would expect that I'm not the only person to have ever wanted this..
Am I missing something in the standard hex manipulators? Is it because type void* (HANDLE) is simply defined outside of the normal hex-handling of ostream?

In summary: I don't want to have to cast HANDLE to something which it isn't.
and I don't want to hard code a "0x" prefix. Is there a way to do it using standard manipulators? or would I need to overload ostream's handling of HANDLE? (but that might overload me!)

Here is my test code (and its output).
I've used '.' as the fill char, for clarity, (I will actually be using '0')

HANDLE h; 
ULONG ul; 
int iH = sizeof(h); // how many bytes to this void* type.
int iW = iH*2;      // the max number of hex digits (width).
int iW2= iW+2;      // the max number of hex digits (+ 2 for showbase "0x").
int iX = 4;         // the number of bits per hex digit.
int ib = iH*8;      // the max number bits in HANDLE (exponent).
int i;
std::cout<<std::setfill('.'); // I actually want '0'; 
                              //   the dot is for display clarity
for( i=0; i<=ib; i+=iX )
{ ul = (pow(2,i)-1);
  h  = (HANDLE)ul;
  std::cout
  <<"//  ul "        <<std::setw(iW2)<<std::hex <<std::showbase  <<std::internal <<ul
  <<"     h "        <<std::setw(iW2) /* hex,showbase,internal have no effect */ <<h
  <<"      I want 0x"<<std::setw(iW) <<std::hex <<std::noshowbase<<std::right    <<(ULONG)h
  <<std::endl;
}

//  ul .........0     h .........0      I want 0x.......0
//  ul 0x.......f     h .......0xf      I want 0x.......f
//  ul 0x......ff     h ......0xff      I want 0x......ff
//  ul 0x.....fff     h .....0xfff      I want 0x.....fff
//  ul 0x....ffff     h ....0xffff      I want 0x....ffff
//  ul 0x...fffff     h ...0xfffff      I want 0x...fffff
//  ul 0x..ffffff     h ..0xffffff      I want 0x..ffffff
//  ul 0x.fffffff     h .0xfffffff      I want 0x.fffffff
//  ul 0xffffffff     h 0xffffffff      I want 0xffffffff

Epicarp

Have you tried casting your handle to a void*? Normal ostreams should print out the address when they're fed a pointer. Casting to unsigned long is bad because you can't always rely on it being the same size as a pointer type. – Scrubland

Thanks George.. A HANDLE is a typeded of void*. I've just now tried typcasting the HANDLE to (void*) explicitly, and the result was the same. The main thing I am trying to avoid here is casting.. I would prefer to use standard manpulators. (my third example column uses a cast). – Epicarp

Answer 1

I found this on https://bugzilla.redhat.com/show_bug.cgi?id=166735 - this is copy/paste straight from there.

Doesn't sound like a bug to me. ISO C++98, 22.2.2.2.2/10 says std::showbase means prepending # printf conversion qualifier. 22.2.2.2.2/7 says std::hex means the printf conversion specifier is %x. So the behaviour is IMHO required to be the same as printf ("%#x", 0); But http://www.opengroup.org/onlinepubs/009695399/functions/fprintf.html says: "For x or X conversion specifiers, a non-zero result shall have 0x (or 0X) prefixed to it." The same is in ISO C99, 7.19.6.1(6): "For x (or X) conversion, a nonzero result has 0x (or 0X) prefixed to it."

So it sounds like the C++98 standard (by saying 'make it like C's printf("%#x", 0)') requires this goofy behavior you're seeing. The only way to get what you want would be to get rid of the std::showbase and output 0x explicitly. Sorry.

Scrubland

Thanks again George. "goofy behaviour" :) I love it. It seems to be impossible to supress "0x". It is uneffected by hex/dec/oct and showbase/noshowbase. Also, operator<<(void*) is not responsive to positional-placement manips: left/right/internal. This means I can't get what I'm after without some fudging. So I've resorted to casting to an unsigned int, but before you get too alarmed; I am doing a sizeof() comparison first. It's not ideal, but is the only way I can see around it, short of overloading operator>>(void*) .. I'll look into it as time permits. C'est la vie... – Epicarp

@orthogonal: How about casting to intptr_t instead? – Kasi

Answer 2

The fact that std::showbase doesn't have an "0x" when the value is zero is sometimes annoying, but that is how it is defined and there is nothing you can do about it, except:

std::cout << "0x" << .....

The gap between the "0x" and the rest is fixable, but it is not very easy to guess. Try

std::cout << std::hex << std::showbase << std::setw(8) << std::setfill('0') << h;

Canzone

Thanks Michael... at least I now know that I hadn't missed something obvious and that its functionality is simply "as it appears"... I've managed to wangles my way around it (for now), by casting to an unsigned int, with sizeof() checks (I've commented in George's answer) – Epicarp