On Wed, Jan 11, 2012 at 10:38 PM, Luis Mochan <[email protected]> wrote:
> In the examples of Craig the argument of floor() is already a PDL, but the > confusion arises when floor is applied to a perl scalar. If I take another > common function, say cos() and apply it to a perl scalar I get a perl > scalar, and if I apply it to a PDL I get a PDL. Isn't that less surprising > than floor() making conversions? > Best regards, > Luis > But this happens because cos() is a built-in whereas floor is not. Probably, Chris should have use int(), which is a Perl built-in that is not overloaded by Perl. I would have mentioned that earlier, but it didn't occur to me. David
_______________________________________________ Perldl mailing list [email protected] http://mailman.jach.hawaii.edu/mailman/listinfo/perldl
