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)
Related
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 3 years ago.
Currently have a formula which essentially looks like this:
Math.round((20.49-14.99)*5);
The resulting number is 27.5 and AFAIK, Math.round() should round UP if the number is 0.5 and above and round down if it is below. Unless I've misunderstood, I was assuming the result would be rounded up to 28 (as it does in Microsoft Excel, for example), but instead, it rounds down to 27.
At first I assumed it had something to do with my code, but when I console logged it, the result was exactly the same.
Am I missing something and if I have, how would I round-up from 0.5 onwards?
Due to the way floating points work, the result of (20.49-14.99)*5 is actually 27.499999999999993, which is closer to 27 than 28
(20.49-14.99)*5 is 27.499999999999993, so JavaScript will round this to closer value. You should use Math.ceil to round above value
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 7 years ago.
Today I have entered random float number and multipled by hundred firstly using normal code and then in console as was giving me wrong number, console is returning me the same.
The given float number is: 1050.6
Therefore: 1050.6 * 100 should be 105060, but javascript is returning me 105059.99999999999
Anyone knows why?
JavaScript uses 64-bit floating point representation (double precision). Numbers are represented in this format as a whole number multiplied by a power of two.
This is based on the IEEE 754 standard
Rational numbers with a denominator that isn't a power of 2 can't be exactly represented. This is why floating point multiplication gives this result.
Source: https://en.wikipedia.org/wiki/IEEE_floating_point
If you want to get the real value, there are two methods you can use
Rounding with Math.round
Math.round(1050.6 * 100)
Or toFixed
(1050.6 * 1000).toFixed(0)
It's a feature of how computers handle floating point numbers. It turns out that decimal numbers can't always be perfectly represented in binary so the computer gives the closest approximation it can.
https://en.wikipedia.org/wiki/IEEE_floating_point
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:
Why does floating-point arithmetic not give exact results when adding decimal fractions?
(31 answers)
Closed 9 years ago.
Why this will not give the results as expected?
console.log(0.2+0.1); // gives 0.30000000000000004
console.log(0.2+0.3); // gives 0.5
console.log(0.2+0.5); // gives 0.7
console.log(0.2+0.4); // gives 0.6000000000000001
Why the first and last will not give the results as expected?
It’s because JavaScript used the IEEE Standard for Binary Floating-Point Arithmetic.
All floating point math is like this and is based on the IEEE 754 standard. JavaScript uses 64-bit floating point representation, which is the same as Java's double.
Can be possible duplicate of Is floating point math broken?
Try .toFixed() .This method formats a number with a specific number of digits to the right of the decimal.
console.log((0.2 + 0.1).toFixed(1)); // gives 0.3
console.log((0.2 + 0.3).toFixed(1)); // gives 0.5
console.log((0.2 + 0.5).toFixed(1)); // gives 0.7
console.log((0.2 + 0.4).toFixed(1)); // gives 0.6
This question already has answers here:
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