I have an annoying problem in JavaScript.
> parseInt(1 / 0, 19)
> 18
Why does the parseInt
function return 18
?
if
might help.
The result of 1/0
is Infinity
.
parseInt
treats its first argument as a string which means first of all Infinity.toString()
is called, producing the string "Infinity"
. So it works the same as if you asked it to convert "Infinity"
in base 19 to decimal.
Here are the digits in base 19 along with their decimal values:
Base 19 Base 10 (decimal)
---------------------------
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
a 10
b 11
c 12
d 13
e 14
f 15
g 16
h 17
i 18
What happens next is that parseInt
scans the input "Infinity"
to find which part of it can be parsed and stops after accepting the first I
(because n
is not a valid digit in base 19).
Therefore it behaves as if you called parseInt("I", 19)
, which converts to decimal 18 by the table above.
Here's the sequence of events:
1/0 evaluates to Infinity
parseInt reads Infinity and happily notes that I is 18 in base 19
parseInt ignores the remainder of the string, since it can't be converted.
Note that you'd get a result for any base >= 19
, but not for bases below that. For bases >= 24
, you'll get a larger result, as n
becomes a valid digit at that point.
parseInt
will accept is 36, since there are 26 letters in the English alphabet, and the convention is to use digits then letters as the set of valid digits in the given base.
Infinity
to "Infinity"
...
parseInt
expects a string as its first argument, so Infinity
is coerced to "Infinity"
; also that should be an ordered list, not an unordered one. But the accepted answer already explains everything…
To add to the above answers:
parseInt is intended to parse strings into numbers (the clue is in the name). In your situation, you don't want to do any parsing at all since 1/0 is already a number, so it's a strange choice of function. If you have a number (which you do) and want to convert it to a particular base, you should use toString with a radix instead.
var num = 1 / 0;
var numInBase19 = num.toString(19); // returns the string "Infinity"
To add to the above answers
parseInt(1/0,19)
is equivalent to parseInt("Infinity",19)
Within base 19 numbers 0-9
and A-I
(or a-i)
are a valid numbers. So, from the "Infinity" it takes I
of base 19 and converts to base 10 which becomes 18 Then it tries to take the next character i.e. n
which is not present in base 19 so discards next characters (as per javascript's behavior of converting string to number)
So, if you write parseInt("Infinity",19)
OR parseInt("I",19)
OR parseInt("i",19)
the result will be same i.e 18
.
Now, if you write parseInt("I0",19)
the result will be 342
as I X 19 (the base)^1 + 0 X 19^0
= 18 X 19^1 + 0 X 19^0
= 18 X 19 + 0 X 1
= 342
Similarly, parseInt("I11",19)
will result in 6518
i.e.
18 X 19^2 + 1 X 19^1 + 1 X 19^0
= 18 X 19^2 + 1 X 19^1 + 1 X 19^0
= 18 X 361 + 1 X 19 + 1 X 1
= 6498 + 19 + 1
= 6518
Success story sharing
parseInt('Infini',24)
.n
is also a valid digit, so it actually ends up doingparseInt("Infini", 24)
.