• bobef <bobef abv_nospam.bg> Sep 25 2007
• Nathan Reed <nathaniel.reed gmail.com> Sep 25 2007
• Don Clugston <dac nospam.com.au> Sep 25 2007
• Don Clugston <dac nospam.com.au> Sep 26 2007
• "Janice Caron" <caron800 googlemail.com> Sep 26 2007
• "Steven Schveighoffer" <schveiguy yahoo.com> Sep 25 2007
• BCS <ao pathlink.com> Sep 25 2007
• "Jarrett Billingsley" <kb3ctd2 yahoo.com> Sep 25 2007

bobef <bobef abv_nospam.bg> writes:

Janice Caron Wrote:

 On 9/25/07, bobef <bobef abv_nospam.bg> wrote:
 something is totally wrong. Just look at this. "!<>="


 Well consider, the imaginary number i is not less than one. It is also
 not greater than one. It is also not equal to one. It makes perfect
 sense.


I can't imagine such number, sorry :) But let aside my imagination. I don't
know what imaginary number is, so I am not commenting if it makes sense. All
I'm saying it that it is nonsense that if(double.nan!=double.nan) returns false
(IMHO).

Nathan Reed <nathaniel.reed gmail.com> writes:

bobef wrote:
 Janice Caron Wrote:
 
 On 9/25/07, bobef <bobef abv_nospam.bg> wrote:
 something is totally wrong. Just look at this. "!<>="


 not greater than one. It is also not equal to one. It makes perfect
 sense.


 I can't imagine such number, sorry :) But let aside my imagination. I don't
know what imaginary number is, so I am not commenting if it makes sense. All
I'm saying it that it is nonsense that if(double.nan!=double.nan) returns false
(IMHO).


I'm assuming you meant to say that it makes no sense if double.nan == 
double.nan returns false?  But if NaNs compared equal to each other, 
then this:

sqrt(-1) == acos(2)

would be true, since both return a NaN.  I bet you don't really want that.

Thanks,
Nathan Reed

Don Clugston <dac nospam.com.au> writes:

Nathan Reed wrote:
 bobef wrote:
 Janice Caron Wrote:

 On 9/25/07, bobef <bobef abv_nospam.bg> wrote:
 something is totally wrong. Just look at this. "!<>="


 not greater than one. It is also not equal to one. It makes perfect
 sense.


 I can't imagine such number, sorry :) But let aside my imagination. I 
 don't know what imaginary number is, so I am not commenting if it 
 makes sense. All I'm saying it that it is nonsense that 
 if(double.nan!=double.nan) returns false (IMHO).


 I'm assuming you meant to say that it makes no sense if double.nan == 
 double.nan returns false?  But if NaNs compared equal to each other, 
 then this:
 
 sqrt(-1) == acos(2)
 
 would be true, since both return a NaN.  I bet you don't really want that.


Yup, that's the reasoning.
Even so, I think it probably was a mistake by the IEEE standard to violate
x == x; I think the problems created by this decision are worse than the ones 
that were solved. (Note that they could even have used NaN payloads to detect 
the worst kinds of errors).

But it's built into the hardware everywhere, so it's Too Late Now.

Don Clugston <dac nospam.com.au> writes:

Janice Caron wrote:
 Think of NaN as meaning "I don't know the answer".
 
 With that understanding, when you compare two NaNs with ==, you are
 asking "Is one thing I don't know the answer to equal to another thing
 I don't know the answer to". The absolutely correct answer should
 really be "I don't know, but probably not", but since that can't be
 expressed in a bool, we go with the second best answer: no.


Yes, we're forcing 3 states (yes, no, maybe) into 2 states.
IMHO, it would have been better to retain the crucial identity x==x, rather
than 
try to detect obscure floating-point bugs. (and introduce a different operator 
to mean, equal-and-not-both-NaN).

I use NaNs a lot (probably more than anyone else on this ng), and have not
found 
many cases where the IEEE behaviour is helpful.

"Janice Caron" <caron800 googlemail.com> writes:

Think of NaN as meaning "I don't know the answer".

With that understanding, when you compare two NaNs with ==, you are
asking "Is one thing I don't know the answer to equal to another thing
I don't know the answer to". The absolutely correct answer should
really be "I don't know, but probably not", but since that can't be
expressed in a bool, we go with the second best answer: no.

"Steven Schveighoffer" <schveiguy yahoo.com> writes:

"bobef" <bobef abv_nospam.bg> wrote in message 
news:fdbedv$2pu2$1 digitalmars.com...
 Janice Caron Wrote:

 On 9/25/07, bobef <bobef abv_nospam.bg> wrote:
 something is totally wrong. Just look at this. "!<>="


 Well consider, the imaginary number i is not less than one. It is also
 not greater than one. It is also not equal to one. It makes perfect
 sense.


 I can't imagine such number, sorry :) But let aside my imagination. I 
 don't know what imaginary number is, so I am not commenting if it makes 
 sense. All I'm saying it that it is nonsense that 
 if(double.nan!=double.nan) returns false (IMHO).


From what I can recall reading on this message board, the behavior is 
exactly described as correct in the IEEE standard.  Therefore, the answer is 
that the behavior works as designed.  You need to address the rediculousness 
of it with the IEEE committee that designed the standard :)

-Steve 

BCS <ao pathlink.com> writes:

Reply to bobef,

 All I'm saying it that it is nonsense that
 if(double.nan!=double.nan) returns false (IMHO).
 


NaN is "Not A Number" and considering that we are talking about numeric
equality, 
it is reasonable for something that isn't a NUMber to not be NUMerically 
equal to anything. It is sort of like comparing to nothing (not the state 
of nothing or a set of nothing, but the absolute omission or the entity) 
this assertion "x is equal to" is syntactically broken so how can it be true?

"Jarrett Billingsley" <kb3ctd2 yahoo.com> writes:

"bobef" <bobef abv_nospam.bg> wrote in message 
news:fdbedv$2pu2$1 digitalmars.com...
 I can't imagine such number, sorry :) But let aside my imagination. I 
 don't know what imaginary number is


An imaginary number is an ifloat, idouble, or ireal.  ;)

(Everything I know I learned from D.)