Enthuware

Not sure what is meant by "...implicit narrowing is permitted only among byte, char, short, and int."

Assigning an int to a short requires a cast.

Hi,
Implicit narrowing happens only for constants. A cast is not required for says byte b = 10; even though 10 is an int.

HTH,
Paul.

Are we supposed to know that 43e1 produces a double and 0x0123 produces a float?

You do need to know the ways in which you can write a double value (such as 1.0 or 43e1 ) and the ways in which you can write an integral value (such as 1 or 0x0123 ). You also need to know that an integer value can be assigned to a float variable without a cast.

I missed because I wasn't sure about what 0x0123 means, and I thought that 43e1 fits into a double. Sadly I was wrong...