Like the title says, I don't understand how Java's primitive data type long maximum and minimum values are smaller than a float's maximum and minimum. Despite the long being 64bit and float being 32bit.
What is going on?
Asked
Active
Viewed 216 times
-1
AlanSTACK
- 5,525
- 3
- 40
- 99
-
Take a look at [this](https://stackoverflow.com/questions/15004944/max-value-of-integer). It is not a direct answer, but it contains the exact explanation. – campovski Sep 04 '17 at 22:21
2 Answers
4
the reason is because a float using floating point precision. A Long can store a high number of precise digits, while I float can store a high value but without the same precision in its lower bits.
In a sense, a float stores values in scientific/exponential notation. thus a large value can be stored in a small number of bits. Think 2 x10^200, this is huge number but can be stored in a small number of bits.
adamM
- 1,116
- 10
- 29
3
A 32-bit float isn't just a simple number, like a long; it's broken up into fields. The fields are:
- 1 bit for the sign;
- 8 bits for the exponent, which is a power of 2;
- 23 bits for the mantissa.
The exponent field is encoded (usually, it's just an offset from the real exponent), but it represents exponents from -126 to 127. (That covers 254 of the possible 256 values; the other two are used for special purposes.) The mantissa represents a value that can be from 1 to a value just below 2 (actually 2 - 2-23). This means that the largest possible float is 2127 x (2 - 2-23), or almost 2128, which is larger than the largest possible long value which is 263-1.
More information:
ajb
- 31,309
- 3
- 58
- 84