Initialize a long in Java

Question

1. You should add L: long i = 12345678910L;.
2. Yes.

BTW: it doesn't have to be an upper case L, but lower case is confused with 1 many times :).

1. You need to add the L character to the end of the number to make Java recognize it as a long.

   long i = 12345678910L;
   

2. Yes.

See Primitive Data Types which says "An integer literal is of type long if it ends with the letter L or l; otherwise it is of type int."

You need to add uppercase L at the end like so

long i = 12345678910L;

Same goes true for float with 3.0f

Which should answer both of your questions

To initialize long you need to append "L" to the end.
It can be either uppercase or lowercase.

All the numeric values are by default int. Even when you do any operation of byte with any integer, byte is first promoted to int and then any operations are performed.

Try this

byte a = 1; // declare a byte
a = a*2; //  you will get error here

You get error because 2 is by default int.
Hence you are trying to multiply byte with int. Hence result gets typecasted to int which can't be assigned back to byte.