CODEWITHSUNDEEP
javascriptbit-manipulation
Royi Namir
Royi Namir

Reputation: 148744

Bitwise operator in js is inconsistent?

I know that there is a trick to make a decimal number into an integer using x|0 :

for example :

3.14|0 -->3

There is no rounding involved here :

3.999 |0 -->3`

Question

However : why does

9.99999999999|0 yields 9

while

9.9999999999999999|0 yields 10?

Upvotes: 0

Views: 83

Answers (2)

mvw
mvw

Reputation: 5105

I would need to look in the source code of firefox, nonetheless it is interesting to try some samples:

>>> 9.99999999999999999.toString(2)
"1010"
>>> 9.9999999999999999.toString(2)
"1010"
>>> 9.999999999999999.toString(2)
"1001.1111111111111111111111111111111111111111111111111"
>>> 9.999999999999999.toString(2).length
54
>>> 9.99999999999999.toString(2)
"1001.111111111111111111111111111111111111111111111101"    
>>> 9.99999999999999.toString(2).length
53

Rough estimate: about 49 bits go into the mantissa which leaves 15 bits for sign and exponent.

Update: I found an interesting entry in the comp.lang.javascript FAQ, see http://www.jibbering.com/faq/#FAQ4_7 It limits the precision of a double type to 53 bits.

I think the precision is exhausted during the step when it converts the double into an 32-bit int prior to the bitwise logical or operation.

Upvotes: 0

Tibos
Tibos

Reputation: 27853

9.9999999999999999 has too many decimals and loses precision in Javascript representation, becoming 10. You can test this: 9.9999999999999999 === 10 will be true

Upvotes: 2

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