nan or -nan?

[Iain Buclaw]

On 16 November 2011 04:35, Ali Çehreli <acehreli at yahoo.com> wrote:
> Some floating point operations produce .nan:
>
> import std.stdio;
>
> void main()
> {
>    double zero = 0;
>    double infinity = double.infinity;
>
>    writeln("any expression with nan: ", double.nan + 1);
>    writeln("    zero / zero        : ", zero / zero);
>    writeln("    zero x infinity    : ", zero * infinity);
>    writeln("infinity / infinity    : ", infinity / infinity);
>    writeln("infinity - infinity    : ", infinity - infinity);
> }
>
> Last time I checked, I think the program above would produce nan for all of
> those operations. Now some are -nan when compiled with 64-bit dmd 2.056 on
> Linux:
>
> any expression with nan: nan
>    zero / zero        : -nan
>    zero x infinity    : -nan
> infinity / infinity    : -nan
> infinity - infinity    : -nan
>
> Is that expected? Does it matter?
>
> Ali
>

This behaviour may be due to the libraries rather than the compiler.
But whether the bit that controls signed-ness is on or off, doesn't
stop the value being NaN.  So I would not give much concern to the
result.

-- 
Iain Buclaw

*(p < e ? p++ : p) = (c & 0x0f) + '0';