Of which datatype is the variable j?
Thanks for your reply Berndfi. I didn't define any datatype explicitly for "j " as I thought it takes automatically j a integer. Whether I need to define j as "int"explicitly? If so then can you give me the syntax for it?
Your for loop assigns integers to j, but your case is checking for j as string values (they are in quotation marks), so none of the branches of the case match...
Your case should be:
case(j (1 ...) (2 ...))
Now it works Andrew. You are right that it was taking it as a string. Thanks a lot.
I thought my question points out the error, but seemed not to be the case.Even if you do not have to declare datatypes explicitly in SKILL doesn’t mean that they don’t exist.