Reputation: 7624
I found the following weird behavior while working on JavaScript numbers.
var baseNum = Math.pow(2, 53);
console.log(baseNum); //prints 9007199254740992
console.log(baseNum + 1); //prints 9007199254740992 again!
console.log(baseNum + 2); //prints 9007199254740994, 2 more than +1
console.log(baseNum + 3) // prints 9007199254740996, 2 more than +2
console.log(baseNum + 4) // prints 9007199254740996, same as +3
What is happening here? I understand that JavaScript can only represent numbers upto 2^53
(they are internally 'double'?), but why this behavior?
If 2^53
is the practical max, then why do we have Number.MAX_VALUE
(1.7976931348623157e+308
)?.
Upvotes: 4
Views: 843
Reputation: 28302
The number is really a double. The mantissa has 52-bits (source and extra information on doubles). Therefore, storing 2^53 chops off the one bit.
The number is stored using 3 pieces a sign bit (fairly straight forward) and two other pieces, the mantissa M and the exponent E. The number is calculated as:
(1 + M/2^53) * 2^(E-1023)
I might have some of the specifics there a little off, but the basic idea is there. So when the number is 2^53, 2^(E-1023) = 2^53 and since there are only 52 bits in M, you can no longer represent the lowest bit.
Upvotes: 8
Reputation: 33189
The answer that @CrazyCasta gave is good.
The only thing to add is to your second question:
If
2^53
is the practical max, then why do we haveNumber.MAX_VALUE
(1.7976931348623157e+308
)?
As you've demonstrated, it can store numbers larger than 2^53
, but with precision worse than 2^0
. As the numbers grow ever larger, they lose more and more precision.
So max value in Number.MAX_VALUE
means the "largest" value it can represent; but it doesn't mean that accuracy is the same as a value near 2^1
or 2^53
.
A corollary to this is that Number.MIN_VALUE
is the smallest value that Number can contain; not the most negative. That is, it is the closest non-zero number to zero: 5.00E-324
(notice that's a positive number!).
Upvotes: 3
Reputation: 413682
The maximum value storable in a long is much larger than the maximum value storable with exact precision in a long. Floating-point numbers have a fixed maximum number of significant digits, but the magnitude of the number can get much larger.
There are a certain number of bits allocated for an exponent (power of two), and that's implicitly multiplied by the mantissa stored in the rest of the bits. Beyond a certain point, you lose precision, but you can keep incrementing the exponent to represent larger and larger magnitudes.
Upvotes: 1