Re: SPAM Re: ieee754 -0.0f?

then it's your compiler. When it translates the string/token "-0.0f" 
apparently it's smart enough to generate a 'real' zero. I don't know of 
any requirement that a compiler supports explicit generation of -0 in 
any form; only that its underlying math behaves correctly.
  So if you try this:
union { unsigned int asUInt; float asFloat; } val;  val.asUInt = 
0x80000000; printf("%f",val.asFloat);
  you -may- see -0.0f depending on whether your printf implementation 
cares to show the difference!  Personally I've frequently been bitten by 
quirks in the way a debugger's windows accept input or display output - 
you need to look at a hex or binary display of memory and registers, and 
use an assembly (rather than source-level) debug, to be sure of what's 
really happening.

Peter

Andras Balogh wrote:

> I run this code:
>
> float f = -0.0f;
> void* p = *(reinterpret_cast<void**>(&f));
> ::printf("%f = %p\n", f, p);
>
> And it prints all zeroes. For any other number I see the correct  
> hexadecimal representation of the float number.
>
>
> Thanks,
>
> Andras
>
>
>
> On Wed, 01 Feb 2006 10:32:28 -0700, Peter <[email protected]> wrote:
>
>> what is the context here?  If you're interpreting 0x80000000 (32 bit  
>> int) as a 32-bit IEEE float it's -0.0f   and 0x00000000 is 0.0f.
>>   Where are you seeing the -0.0f? In a debugger? I think you must be  
>> seeing some idiosyncrasy of your development environment or toolchain.
>>
>> Andras Balogh wrote:
>>
>>> I know that -0.0f == +0.0f is true, but the floating point  
>>> representation  of the two numbers should be different, no? 
>>> Shouldn't  the negative zero  have the sign bit set? Well, on my 
>>> machine -0.0 and  0.0 are bitwise  equal!! What's going on here?
>>>
>>> thx,
>>>
>>>
>>> Andras
>>
>
>
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