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 Re: distinguishing -0 in l3fp Andrew Parsloe <[log in to unmask]> Fri, 14 Apr 2017 20:14:45 +1200 text/plain (57 lines) ```On 14/04/2017 2:28 p.m., Bruno Le Floch wrote: > On 04/02/2017 12:09 AM, Andrew Parsloe wrote: >> Is there a simple way to distinguish -0 from 0 in l3fp? >> >> I had been aligning left in a table a column of sines, +0 or -0 followed >> by decimal point & digits. By adding an \hphantom{-} before the positive >> numbers, they aligned nicely on the decimal point in the column, but >> then along came -0. The simple conditional, "if !(number < 0) add >> phantom", added the phantom and the column aligned like >> 0 >> 0.5878 >> -0 >> -0.5878 >> etc. >> In the absence of formatted printing of numbers, -0 is a problem. Even a >> built-in test for -0 would ease matters, since in any kind of formatted >> printing of numbers 0 and -0 will generally require different handling. >> >> As it is, I've had to test each fp for whether it is zero; if it is I >> convert to a tl variable and test whether that is -0. This is clumsy. >> Hence my opening question. >> >> Andrew > > Sorry for the delay. > > One option is to normalize -0 to +0 by adding +0. Namely, number+0 is > equal to number for any non-zero number and is +0 for both +0 and -0. > > Another option is to compute 1/number after disabling the > "division-by-zero" trap, and test whether that's +inf or -inf, but > that's more complicated than the tl test you're doing. > > A better option is that I should add a "copysign" function, that is in > the IEEE standard: copysign(1,x) gives +1 or -1 depending on the sign > bit of x (so in particular this distinguishes +0 and -0). > > For your specific use case I would use \fp_to_tl:n then test whether the > first character (\str_head:n) is "-" or not. This also covers nan properly. > > Bruno > Ah, the "add +0" suggestion is helpful, and certainly simpler than what I was doing. You implemented, very promptly, an earlier suggestion of mine about adding "sign" to l3fp. Thank you. In the course of dealing with the above problem I noticed that sign(-0) = -0. I presume this is the expected result. (I was rather hoping sign(x) might evaluate to one of the *three* values -1, 0, 1.) Andrew --- This email has been checked for viruses by Avast antivirus software. https://www.avast.com/antivirus ```