Open wouterbeek opened 7 years ago
Dear Wouter,
many thanks for reporting. Will have a look the next few days.
In the meantime you can use terms instead of singletons:
?- findall( row(N), between(1, 10000, N), L), x <- L. L = [row(1), row(2), row(3), row(4), row(5), row(6), row(7), row(8), row(...)|...].
?- <- x. [,1] [1,] 1 [2,] 2 [3,] 3
.....
Reproducible case: