automatic Inf+NaN*i conversion

[John W. Eaton]

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