As I understand the difference between two integer from 32bit & 64bit are the following: 32bit range −2,147,483,648 to 2,147,483,647 64bit range: −9,223,372,036,854,775,808 to +9,223,372,036,854,775,807
I am using a 64bit jdk, I validate it by printing the following: System.out.println("JVM Bit size: " + System.getProperty("sun.arch.data.model"));
JVM Bit size: 64
when I try to init a new Integer variable with number bigger ther 10 letters I get a compilation error. why is that? it looks like the 64bit is larger
example (ran on netbeans): int x = 12345678910; => Error: integer is too large
3 Answers
The size of an int in Java is completely independent of the 32-bitness or 64-bitness of a JDK. It is always 4 bytes = 32 bits = −2,147,483,648 to 2,147,483,647.
If you want a 64-bit integer, use a long, which is always 64 bits = 8 bytes.
Unlike other languages, Java's numeric primitive types are always the same size, whatever the platform (32bit or 64bit, LE or BE); they are all big endian and are 1 byte long for byte, 2 bytes long for short and char, 4 bytes long for int and 8 bytes long for long.
If it were not the case, jars would not be portable across platforms...
Your best resource is JLS:
The integral types are byte, short, int, and long, whose values are 8-bit, 16-bit, 32-bit and 64-bit signed two's-complement integers, respectively, and char, whose values are 16-bit unsigned integers representing UTF-16 code units
4.2.1. Integral Types and Values
The values of the integral types are integers in the following ranges:
For byte, from -128 to 127, inclusive
For short, from -32768 to 32767, inclusive
For int, from -2147483648 to 2147483647, inclusive
For long, from -9223372036854775808 to 9223372036854775807, inclusive
For char, from '\u0000' to '\uffff' inclusive, that is, from 0 to 65535
