I am using Math.pow() function in my code but when I try to execute below condition, this function return infinity which is not required. I am looking for an alternate solutions.
math.pow(451939.27436410653, 299);
Please help if anyone any idea
Use logs
What result were you expecting? The result is extremely large and exceeds the ability of the conventional floating point format to represent. I would be surprised if this was really necessary in a real-world calculation. Can you provide some context?
If really necessary, perhaps you could resolve the difficulties by using logarithms. If y=Math.pow(451939,299) then
Math.log(y) = Math.log(451939) * 299.
You could do any multiplication/division by adding/subtracting logs, and then do a Math.exp at the end to generate your result.
This may be easier than using a special library like bignumber.js for arbitrary-precision arithmetic. For example, the code below returns "7.395117980030695 x 10^ 1690", which has 1691 digits before the decimal point.
let log10Y=Math.log(451939.27436410653)*299/Math.log(10);
let b = Math.floor(log10Y);
let a = log10Y-b;
console.log("Answer: ",10**a," x 10^",b);
If you want to store such enormous numbers, you can use a library like bignumber.js. It stores floating-point values with an exponent as large as necessary to store them.
Related
I am using toFixed but the method does not operate as expected
parseFloat(19373.315).toFixed(2);
//19373.31 Chrome
Expected Output : 19373.32
parseFloat(9373.315).toFixed(2);
// 9373.32 Working fine
Why does the first example round down, whereas the second example round up?
The problem is that binary floating point representation of most decimal fractions is not exact. The internal representation of 19373.315 may actually be something like 19373.314999999, so toFixed rounds down, while 19373.315 might be 19373.315000001, which rounds up.
Why does the first example round down, whereas the second example round up?
Look at the binary representation of the two values in memory.
const farr = new Float64Array(2);
farr[0] = 19373.315;
farr[1] = 9373.315;
const uarr = new Uint32Array(farr.buffer);
console.log(farr[0], uarr[1].toString(2).padStart(32, 0) + uarr[0].toString(2).padStart(32, 0));
console.log(farr[1], uarr[3].toString(2).padStart(32, 0) + uarr[2].toString(2).padStart(32, 0));
Without diving into the details, we can see that the second value has an additional '1' at the end, which is lost in the first larger value when it is fit into 64 bits.
Other answers have explained why, I would suggest using a library like numeral.js which will round things as you would expect.
Assuming toFixed casts to 32-bit float;
Check with this utility...
19373.315 is stored as 19373.314453125 (an error of -0.000546875) in 32-bit floating point format.
This is despite (19373.315).toFixed(4) coming out as 19373.3150.
Even if this is "expected" or "intended", I'd still report it as a bug.
It should use a double during the rounding check, and thus proper rounding during conversion to fixed string.
I think the spec even says so. :\
In the V8 javascript engine source, the Number.prototype.toFixed function invokes DoubleToFixedCString in this file ...
There's probably some inappropriate optimization in there... (Looking into it.)
I'd suggest submitting an additional test case for V8 with 19373.315 specifically.
(19373.3150).toFixed(39) yields 19373.314999999998690327629446983337402343750.
Rounding occurs once to bring it up to 19373.315 - which is correct - but not at the right digit when rounding to 2 digits.
I think this should have a second pass on rounding here to catch edge cases like this. I think it might have to round to n+1 digits, then again to n digits. Maybe there's some other clever way to fix it though.
function toFixedFixed(a,n) {
return (a|0) + parseFloat((a % 1).toFixed(n+1)).toFixed(n).substr(1);
}
console.log(toFixedFixed(19373.315,2)); // "19373.32"
console.log(toFixedFixed(19373.315,3)); // "19373.315"
console.log(toFixedFixed(19373.315,4)); // "19373.3150"
console.log(toFixedFixed(19373.315,37)); // "19373.3149999999986903276294469833374023438"
console.log(toFixedFixed(19373.315,38)); // "19373.31499999999869032762944698333740234375"
console.log(toFixedFixed(19373.315,39)); // "19373.314999999998690327629446983337402343750"
(Adopted from my comments on Vahid Rahmani's answer, who is correct.)
I'm trying to increase by one this number: 9223372036854775808:
var number = 9223372036854775808;
var plusOne = number + 1;
This should yield 9223372036854775809, but it instead yields 9223372036854776000.
Why? More important, how can I fix this?
The largest representable number in JavaScript is (2^53) - 1, or, written out, 9007199254740991. The number you have, 9223372036854775808, is more than 1024 times that quantity.
If you want to work with numbers larger than the one above, you should use a big integer library. JavaScript does not have one built in, however, so you'll need to grab it yourself. Personally, I use big-integer when I'm working on things that deal with really large numbers, e.g., Project Euler.
This is to do with JavaScript storing numbers internally as (double precision) floating point. As you go up the scale of floating point numbers, the numbers get more and more sparse until the point where you get incorrect results, because the next representable number is more that 1 away (As in your example). To properly handle large numbers, you will need to use a proper large number library such as javascript-bignum. If you only need integers, you can use BigInteger.js
how I should avoid decimal rounding in javascript
suppose I have input like
input : 99999999999.999999
Expected output : 99999999999.999999
but i am getting output like
100000000000
here, internally javascript rounding the decimal place.
please let me know how should I avoid this javascript rounding???.
For example:
var test = 99999999999.999999;
console.log(test); //This will print 100000000000 instead
As #Aliendroid suggested, you can use the BigDecimal.js library. In my perspective, this is the best way to handle double and floats in JavaScript
To answer to your question more specifically, you can actually do this:
var test_number = new BigDecimal('99999999999.999999');
console.log(test_number.setScale(6).toString());
and to your original question:
please let me know how should I avoid this javascript rounding???
Well you can't as there is not double number in JavaScript, only float. So for example if you do something like this:
console.log(99999999999.99); //This is a float number
This will actually output: 99999999999.99
It is NOT Rounding
It is floating point error
learn floating point number in programming !
use string if needed and some Big number library
Ref:
http://floating-point-gui.de/
https://github.com/dtrebbien/BigDecimal.js
Why is it if I do this in javascript, I get the following result:
1234.56 * 10 = 12345.599999999999
It should be 123456. How can I get around this problem?
Thanks.
Floating points are not exact, since there are ifinite numbers at their range [or in any range to be more exact], and only a finite number of bits to store this data.
Have a look at what every programmer should know about floating point arithmetics.
Another easy solution:
parseFloat((1234.56 * 10).toPrecision(12))
and the result will be: 12345.6, and YES... it works with decimal numbers.
As the others said, floating points and so on.
Easy solution would be to do something like this:
var answer = parseInt(1234.56 * 10);
Or just use Math.round?
All numbers in JS are internally defined by float and drop the less significant digits if needed.
(10000000000000000000000000000 + 1) == 10000000000000000000000000000
// this will return true
And javascript is well known for droping bits quite often in numbers. So handle with care
As you all know since it is one of the most asked topic on SO, I am having problems with rounding errors (it isn't actually errors, I am well aware).
Instead of explaining my point, I'll give an example of what possible numbers I have and which input I want to be able to obtain:
Let's say
var a = 15 * 1e-9;
alert(a)
outputs
1.5000000000000002e-8
I want to be able to obtain 1.5e-8 instead, but I cannot just multiply by 10e8, round and divide by 10e8 because I don't know if it will be e-8 or e-45 or anything else.
So basically I want to be able to obtain the 1.5000002 part, apply toFixed(3) and put back the exponent part.
I could convert into a string and parse but it just doesn't seem right...
Any idea ?
(I apologize in advance if you feel this is one of many duplicates, but I could not find a similar question, only related ones)
Gael
You can use the toPrecision method:
var a = 15 * 1e-9;
a.toPrecision(2); // "1.5e-8"
If you're doing scientific work and need to round with significant figures in mind: Rounding to an arbitrary number of significant digits
var a = 15 * 1e-9;
console.log(Number.parseFloat(a).toExponential(2));
//the above formula will display the result in the console as: "1.50e-8"