@ValentinTihomirov it is not mistake 16#E#E1 it is base 16 value, E (which means 14) and E1 is shift by one value to the left. 2E1 is equal to 2*10^1, but in base 16 it is not multiply by 10^1 but 16^1! So given expression evaluates to E0 in hexadecimal, which is 224.
To me this is weird notation, I'd put 0x in front so 0xE#E1, or from context and existence of E, implicitly E#E1. Why # is needed? Because E#E1 is E0 and EE1 would be ambigous. In decimal base we also ommit #, because E is not valid digit.
