Closed 17IMMANUEL17 closed 5 months ago
The result is correct, no bug. The order 3 in t2 is a "local" maximum order limiting size of memory and calculation as an optimisation, not a structural order that makes these tpsas incompatibles. When you print t1 and t2, they both have mo=4 and since they are empty, they have same values.
Discovered a bug for the equality operator: as for the print function eq checks the descriptor and not the tpsa. In this way mo = 4 d = gtpsad(nil,mo) t1 = tpsa(d) t2 = tpsa(d,3) t1==t2 -->true This should generate an error since the teo tpsas have a different order. This is due to the fact that the two different tpsas are created using the same descriptor and the check that eq does to ensure that the comparison is valid is based on the descriptor.