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Closed 10 years ago.
Possible Duplicate:
Is JavaScript’s Floating-Point Math Broken?
Note - I have read other posts on parseFloat(), but I have not found a good explanation as to why the problem occurs. (Or I just didn't understand it).
This code ...
var sum = parseFloat("1.001") + parseFloat(".001");
alert(parseFloat(sum));
outputs ...
1.0019999999999998
I've read that adding sum.toFixed(2) will include only 2 decimal points.
However, I do not 100% understand why this long decimal occurs.
Does parseFloat(sum) represent sum in binary? If so, then 1.001 cannot be represented in binary since 1/2^x + ... can never equal .001 or 1/1000 exactly?
This isn't specific to Javascript, but rather how IEEE Floating Point Numbers are represented internally that cause precision errors.
I won't reproduce the content here, but there are a bunch of resources available to help explain what is going on in your example.
Here is one: http://www.petebecker.com/js/js200006.html
Related
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What is JavaScript's highest integer value that a number can go to without losing precision?
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Large numbers erroneously rounded in JavaScript
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javascript large integer round because precision? (why?)
(2 answers)
Closed 1 year ago.
So, I found something I couldn't understand and can't find any internet resource that explains it.
Please see the code below:
var num = 35422484817926290
// subtract 5 from num
console.log(num-5)
Output (Wrong) : 35422484817926284
I checked it in Node, Opera, and Chrome, all of them give the wrong answers.
I do understand the fact that arithmetic with unsafe Integers in JS is faulty, for example:
console.log(100000000000000005-1)
Output (Wrong) : 100000000000000000
So what's the deal with big number arithmetic in JS?
When I run this code:
var num = 35422484817926290
// subtract 5 from num
console.log(num-5)
in Visual Studio Code, i get the following warning:
"Numeric literals with absolute values equal to 2^53 or greater are too large to be represented accurately as integers."
So the correct way to make this calculation would be like this:
var num = 35422484817926290n
// subtract 5 from num
console.log(num-5n)
JavaScript is not faulty, this is how the Floating point arithmetic works. Looks duplicate to this post. For better calculation involving floating-point numbers you should use BigNumber API.
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Javascript toFixed function
(6 answers)
Closed 8 years ago.
When the non fractional part is bigger than 4 the fractional part it is truncated to .3 but when it is smaller than 4 it is rounded to .4.
Examples:
1.nr>4:
5.35.toFixed(1) => 5.3
15.35.toFixed(1) => 15.3
131.35.toFixed(1) => 131.3
2.nr<4:
2.35.toFixed(1) =>2.4
1.35.toFixed(1) =>1.4
Is this kind of behaviour normal?
The problem is that the exact values you're calling toFixed on aren't 1.35 etc... they're the nearest IEEE-754 64-bit representation. In this case, the exact values are:
1.350000000000000088817841970012523233890533447265625
2.350000000000000088817841970012523233890533447265625
5.3499999999999996447286321199499070644378662109375
15.3499999999999996447286321199499070644378662109375
Now look at those values and work out what you'd do in terms of rounding to 1 decimal place.
Basically you're falling foul of the fact that these are floating binary point values, so the value you express in decimal isn't always the value that's actually used. It's just an approximation. In other languages the preferred alternative is to use a type which represents floating decimal point values (e.g. BigDecimal in Java or decimal in C#) but I don't know of anything similar within standard Javascript. You may find some third party libraries though.
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 9 years ago.
I tried this expression in both my FireFox and Chrome console:
17.99 * 100;
Expected Result: 1799
Actual Result: 1798.9999999999998
I also tried:
parseInt(17.99 * 100);
Expected Result: 1799
Actual Result: 1798
Why is this happening, and how do I get the expected result?
Floating point arithmetic isn't an exact science. The reason is that in memory, your precision is stored in binary (all powers of two). If it can't be represented by an exact power of two you can get some lost precision.
Your number, 1798.9999999999998 had enough lost precision that it didn't round up in the multiplication.
http://en.wikipedia.org/wiki/IEEE_floating_point
Try this:
Math.round(17.99*100)
As the previous answer explained, multiplying a float number is not exact science; what you can do is to expect a result in a certain precision range. Take a look at Number.prototype.toPrecision().
(17.99*100).toPrecision(4)
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 9 years ago.
I want to make multiplication using JavaScript.
Multiplication of 2 and 0.15 is 0.3 but 3 and 0.15 is 0.44999999999999996. I want to get 0.45 such as result. How can I do it with JavaScript?
It's a rounding error. You may want to round your number to a fixed amount of digits, like this:
parseFloat((your_number).toFixed(2));
Unfortunately this happens in any language using floating point arithmetic. It's an artifact arising when floating point operations are encoded into binary, operations performed, and decoded from binary to report the answer in a form you'd expect.
Depending on what you want to do with the output, you can call a round()-like function to round to a number of decimal places or compare the output to a value using a (small) number called a tolerance. E.g. two numbers are considered equal if their absolute difference is less than this tolerance.
This question already has answers here:
Closed 12 years ago.
Possible Duplicate:
Is JavaScript’s Math broken?
Why does JS screw up this simple math?
console.log(.1 + .2) // 0.3000000000000004
console.log(.3 + .6) // 0.8999999999999999
The first example is greater than the correct result, while the second is less. ???!! How do you fix this? Do you have to always convert decimals into integers before performing operations? Do I only have to worry about adding (* and / don't appear to have the same problem in my tests)?
I've looked in a lot of places for answers. Some tutorials (like shopping cart forms) pretend the problem doesn't exist and just add values together. Gurus provide complex routines for various math functions or mention JS "does a poor job" in passing, but I have yet to see an explanation.
It's not a JS problem but a more general computer one. Floating number can't store properly all decimal numbers, because they store stuff in binary
For example:
0.5 is store as b0.1
but 0.1 = 1/10 so it's 1/16 + (1/10-1/16) = 1/16 + 0.0375
0.0375 = 1/32 + (0.0375-1/32) = 1/32 + 00625 ... etc
so in binary 0.1 is 0.00011...
but that's endless.
Except the computer has to stop at some point. So if in our example we stop at 0.00011 we have 0.09375 instead of 0.1.
Anyway the point is, that doesn't depend on the language but on the computer. What depends on the language is how you display numbers. Usually, the language rounds numbers to an acceptable representation. Apparently JS doesn't.
So what you have to do (the number in memory is accurate enough) is just to tell somehow to JS to round "nicely" number when converting them to text.
You may try the sprintf function which give you a fine control of how to display a number.
From The Floating-Point Guide:
Why don’t my numbers, like 0.1 + 0.2
add up to a nice round 0.3, and
instead I get a weird result like
0.30000000000000004?
Because internally, computers use a
format (binary floating-point) that
cannot accurately represent a number
like 0.1, 0.2 or 0.3 at all.
When the code is compiled or
interpreted, your “0.1” is already
rounded to the nearest number in that
format, which results in a small
rounding error even before the
calculation happens.
The site has detailed explanations as well as information on how to fix the problem (and how to decide whether it is a problem at all in your case).
This is not a javascript only limitation, it applies to all floating point calculations. The problem is that 0.1 and 0.2 and 0.3 are not exactly representable as javascript (or C or Java etc) floats. Thus the output you are seeing is due to that inaccuracy.
In particular only certain sums of powers of two are exactly representable. 0.5 = =0.1b = 2^(-1), 0.25=0.01b=(2^-2), 0.75=0.11b = (2^-1 + 2^-2) are all OK. But 1/10 = 0.000110001100011..b can only be expressed as an infinite sum of powers of 2, which the language chops off at some point. Its this chopping that is causing these slight errors.
This is normal for all programming languages because not all decimal values can be represented exactly in binary. See What Every Computer Scientist Should Know About Floating-Point Arithmetic
It has to do with how computers handle floating numbers. You can read more about it here: http://docs.sun.com/source/806-3568/ncg_goldberg.html