Initialize a long in Java

Question

Primitive Data Types - oracle doc says the range of long in Java is -9 223 372 036 854 775 808 to 9 223 372 036 854 775 807  But when I do something like this in my eclipse  long i   12345678910    it shows me  The literal 12345678910 of type int is out of range  error   There are 2 questions   1  How do I initialize the long with the value 12345678910   2  Are all numeric literals by default of type int

User · Answer

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

User · Answer

You should add L  long i   12345678910L   Yes    BTW  it doesn t have to be an upper case L  but lower case is confused with 1 many times

User · Answer

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

User · Answer

You need to add the L character to the end of the number to make Java recognize it as a long   long i   12345678910L   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

[java] Initialize a long in Java

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