Closed whichislovely closed 1 month ago
If the scanning variable (for instance dt) is being used to define edge id5 and then id5 is used in the evaluation of another edge id6 = id5 + 1. Then the id6 is not recognized as a scanned variable and the program does not work as expected.
It was resolved
If the scanning variable (for instance dt) is being used to define edge id5 and then id5 is used in the evaluation of another edge id6 = id5 + 1. Then the id6 is not recognized as a scanned variable and the program does not work as expected.