automatic Inf+NaN*i conversion

[Ben Abbott]

On Apr 25, 2008, at 7:11 PM, John W. Eaton wrote:

> On 25-Apr-2008, Ben Abbott wrote:
>
> | On Apr 25, 2008, at 12:42 PM, Jaroslav Hajek wrote:
> |
> | > No, you are just plain wrong. As you have seen earlier, Inf+NaN*i
> | > does not give
> | > complex(Inf, NaN), but rather complex(NaN, NaN), due to the fact  
> that
> | > it is calculated as
> | > complex(Inf + 0*NaN, 0 + 1*NaN). The behaviour of isinf is OK - it
> | > gives infinity if either part of the complex number is Inf.
> |
> | I thought I understood your point, but when I tested myself, the  
> test
> | did not meet my expectation :-(
> |
> | Specifically, It appears to me that the two examples below are in
> | contradiction.
> |
> | octave:45> NaN*(1+0i) + Inf*(0+1i)
> | ans = NaN + Infi
> | octave:46> NaN*complex(1,0) + Inf*complex(0,1)
> | ans = NaN - NaNi
> |
> | What is happening in this instance?
>
> Break it down into the components and look at each of them
> separately.  Note that 1+0i is 1 (i.e. not a complex value).  Although
> it looks like you are writing a complex number when you write 1+0i,
> you are really writing an addition expression, and 0i is automatically
> converted to 0, so you are really writing 1+0.  However, the result of
> complex(0,1) isn't automatically converted to real, so it does allow
> you to create complex values with zero imaginary components.
>
> Yes, NaN*complex(1,0) --> NaN - NaNi looks "wrong", but does the sign
> really matter here?  I mean, it's a NaN.  In any case, Jaroslav's
> patch is probably a reasonable thing to do (I haven't had a chance to
> try it yet).
>
> jwe

My thought is that NaN should trump Inf ... or Inf should trump a NaN.

octave:57> NaN+1i*Inf
ans = NaN + Infi
octave:58> Inf+1i*NaN
ans = NaN - NaNi
octave:59>

Although this behavior may be explained by examining the underlying  
algorithm. Is the function of the algorithm sensible or proper?

Should not both examples give consistent results? ... by consistent I  
mean that neither the real-part or the imag-part should be favored.

... or am I missing something?

Ben



This example doesn't make sense to me A