[PD] 1 / 0 = 0?

I have a function which blows up at a certain point, moves to infinity. In this case, it is basically 1/0.

I was a little surprised to find that 1/0 = 0 (according to Pd). I was expecting NaN, or an overflow condition would could be trapped.

So what I have to do is examine the *input* for the values which will blow up my function, and set a flag. While this is easy in this single case, it is awkward in the general case.

This seems somewhat broken to me. Is this a bug?

Thanks BH

-- William Huston: WilliamAHuston@gmail.com Binghamton NY

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