I was writing some code to generate sequence vectors of the appropriate dimensionality and I ran the following sequence:
pdl> $nmax = 1; pdl> $nvec = sequence(floor($nmax+1)); pdl> p $nvec 0 which surprised me since I expected the result to be [0,1] since floor($nmax+1) is 2. It turns out the result is correct according to the documentation _since_ the floor() routine generates a pdl output value even if the input is a perl scalar. If you use POSIX::floor instead, I get the expected result: pdl> $nvec = sequence(POSIX::floor($nmax+1)); pdl> p $nvec [0 1] Maybe the PDL::floor routine (and similar routines) should return perl scalars when passed perl scalars as input args instead of piddles. Goes to show you, PDL can always surprise you! :-) --Chris _______________________________________________ Perldl mailing list [email protected] http://mailman.jach.hawaii.edu/mailman/listinfo/perldl
