Java long byte length

Question

If Java's long primitive data type is a 64-bit signed integer, then why do the Java docs state that the static long maximum value is 2^63-1? Is there a larger reason that affects other primitive data types similarly in Java?

I'm learning Java and am just genuinely curious about this disparity. Thank you.

A signed value needs to be able to distuinguish between `x` and `-x`. That's why the maximum value is not the size. — Ben, Jun 04 '18 at 11:22
What he means is that there are `long` values less than zero. In fact, there are **more** `long` values less than zero than greater than zero! — Stephen C, Jun 04 '18 at 11:24
in any programming language, the Most Significant Bit will be the sign bit, which denotes the sign of the number. So in your question, for a 64-bit signed integer, leaving the MSB, the remaining 63 bits are used to denote the value of the integer. Which is why 2^63-1 — Ashwin K Kumar, Jun 04 '18 at 11:26
@AshwinKKumar that is wrong. Java uses the "Two`s Complement" and does not just use the most significant bit for the sign (the fact that the msb is 1 for negative numbers is true but the remaining bits are not the same as for the positive number) — Ben, Jun 04 '18 at 11:28
*"in any programming language, the Most Significant Bit will be the sign bit"* - Incorrect. 1) In a programming language with unsigned integers, the top bit is not the sign bit. 2) You are assuming the programming language uses the hardware-provided native (signed) integer representation. In some languages, that is not always true ... or never true. (Example of "never true" - the CLU programming language, which "stole" a bit for use by the GC to distinguish scalars from pointers.) — Stephen C, Jun 04 '18 at 11:32
Read this https://stackoverflow.com/questions/1049722/what-is-2s-complement — jschnasse, Jun 04 '18 at 11:35
Ahh. My bad, I went through the Java Documentation to clear things up. Shouldn't have assumed things. — Ashwin K Kumar, Jun 04 '18 at 11:39
I don't get the question - what disparity? *64-bit **signed** integer* ranges from `-2^63` to `+2^63-1` (-1 to count for `0`) — user85421, Jun 04 '18 at 11:41

score 4 · Accepted Answer · answered Jun 04 '18 at 11:35

Java's long type is indeed a 64-bit integer, but it is also signed.

With 64-bit, you can represent 2^64 different numbers. If you ignore all the negative numbers, then the maximum value would be 2^64-1, and minimum will be 0. 0 plus all 2^64-1 positive numbers is 2^64 numbers in total.

However, if you consider all the negative numbers as well as the positives, about half of the 2^64 different numbers will be negative. Java just chose to represent 2^63 negative numbers, the number 0, and 2^63-1 positive numbers. If you add all these up, you get 2^64 total numbers.

2^63 + 1 + 2^63 - 1 = 2^64
       ^
       |
  this is the number zero

score 0 · Answer 2 · edited Jun 04 '18 at 11:45

0

Its just a range formula just like range for octal and hexadecimal numbers 2^n -1 different for signed and unsigned numbers. 2^n and 2^n-1(Range)

edited Jun 04 '18 at 11:45

Sport

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answered Jun 04 '18 at 11:38

Talha Shan

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Java long byte length

2 Answers2