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Why does this happen to numbers in JavaScript

Found this here:
How does this even work? What is happening here? Why does the number change in the first line?

JavaScript uses double-precision floating-point format numbers as specified in IEEE 754 and can only safely represent numbers between -(253 - 1) and 253 - 1.
The number 111111111111111111 (18 digits) is above that range.
Reference: Number.MAX_SAFE_INTEGER

As mentioned above, JavaScript uses the double-precision 64-bit floating point format for the numbers. 52 bits are reserved for the values, 11 bits for the exponent and 1 bit for the plus/minus sign.
The whole deal with the numbers is beautifully explained in this video. Essentially, JavaScript uses a pointer that moves along the 52 bits to mark the floating point. Naturally, you need more bits to express larger numbers such as your 111111111111111111.
To convert your number into the binary, it would be
sign - 0
exponent - 10000110111
mantissa - 1000101010111110111101111000010001100000011100011100
The more space is taken by the value, the less is available for the decimal digits.
Eventually, simple calculations such as the increment by 1 will become inaccurate due to the lack of bits on the far right and the lowest possible increment will depend on the position of your pointer.

javascript: why is returning so much decimals? After a multiply [duplicate]

I know a little bit about how floating-point numbers are represented, but not enough, I'm afraid.
The general question is:
For a given precision (for my purposes, the number of accurate decimal places in base 10), what range of numbers can be represented for 16-, 32- and 64-bit IEEE-754 systems?
Specifically, I'm only interested in the range of 16-bit and 32-bit numbers accurate to +/-0.5 (the ones place) or +/- 0.0005 (the thousandths place).

For a given IEEE-754 floating point number X, if
2^E <= abs(X) < 2^(E+1)
then the distance from X to the next largest representable floating point number (epsilon) is:
epsilon = 2^(E-52) % For a 64-bit float (double precision)
epsilon = 2^(E-23) % For a 32-bit float (single precision)
epsilon = 2^(E-10) % For a 16-bit float (half precision)
The above equations allow us to compute the following:
For half precision...
If you want an accuracy of +/-0.5 (or 2^-1), the maximum size that the number can be is 2^10. Any X larger than this limit leads to the distance between floating point numbers greater than 0.5.
If you want an accuracy of +/-0.0005 (about 2^-11), the maximum size that the number can be is 1. Any X larger than this maximum limit leads to the distance between floating point numbers greater than 0.0005.
For single precision...
If you want an accuracy of +/-0.5 (or 2^-1), the maximum size that the number can be is 2^23. Any X larger than this limit leads to the distance between floating point numbers being greater than 0.5.
If you want an accuracy of +/-0.0005 (about 2^-11), the maximum size that the number can be is 2^13. Any X larger than this lmit leads to the distance between floating point numbers being greater than 0.0005.
For double precision...
If you want an accuracy of +/-0.5 (or 2^-1), the maximum size that the number can be is 2^52. Any X larger than this limit leads to the distance between floating point numbers being greater than 0.5.
If you want an accuracy of +/-0.0005 (about 2^-11), the maximum size that the number can be is 2^42. Any X larger than this limit leads to the distance between floating point numbers being greater than 0.0005.

For floating-point integers (I'll give my answer in terms of IEEE double-precision), every integer between 1 and 2^53 is exactly representable. Beyond 2^53, integers that are exactly representable are spaced apart by increasing powers of two. For example:
Every 2nd integer between 2^53 + 2 and 2^54 can be represented exactly.
Every 4th integer between 2^54 + 4 and 2^55 can be represented exactly.
Every 8th integer between 2^55 + 8 and 2^56 can be represented exactly.
Every 16th integer between 2^56 + 16 and 2^57 can be represented exactly.
Every 32nd integer between 2^57 + 32 and 2^58 can be represented exactly.
Every 64th integer between 2^58 + 64 and 2^59 can be represented exactly.
Every 128th integer between 2^59 + 128 and 2^60 can be represented exactly.
Every 256th integer between 2^60 + 256 and 2^61 can be represented exactly.
Every 512th integer between 2^61 + 512 and 2^62 can be represented exactly.
.
.
.
Integers that are not exactly representable are rounded to the nearest representable integer, so the worst case rounding is 1/2 the spacing between representable integers.

The precision quoted form Peter R's link to the MSDN ref is probably a good rule of thumb, but of course reality is more complicated.
The fact that the "point" in "floating point" is a binary point and not decimal point has a way of defeating our intuitions. The classic example is 0.1, which needs a precision of only one digit in decimal but isn't representable exactly in binary at all.
If you have a weekend to kill, have a look at What Every Computer Scientist Should Know About Floating-Point Arithmetic. You'll probably be particularly interested in the sections on Precision and Binary to Decimal Conversion.

First off, neither IEEE-754-2008 nor -1985 have 16-bit floats; but it is a proposed addition with a 5-bit exponent and 10-bit fraction. IEE-754 uses a dedicated sign bit, so the positive and negative range is the same. Also, the fraction has an implied 1 in front, so you get an extra bit.
If you want accuracy to the ones place, as in you can represent each integer, the answer is fairly simple: The exponent shifts the decimal point to the right-end of the fraction. So, a 10-bit fraction gets you ±211.
If you want one bit after the decimal point, you give up one bit before it, so you have ±210.
Single-precision has a 23-bit fraction, so you'd have ±224 integers.
How many bits of precision you need after the decimal point depends entirely on the calculations you're doing, and how many you're doing.
210 = 1,024
211 = 2,048
223 = 8,388,608
224 = 16,777,216
253 = 9,007,199,254,740,992 (double-precision)
2113 = 10,384,593,717,069,655,257,060,992,658,440,192 (quad-precision)
See also
Double-precision
Half-precision

See IEEE 754-1985:
Note (1 + fraction). As #bendin point out, using binary floating point, you cannot express simple decimal values such as 0.1. The implication is that you can introduce rounding errors by doing simple additions many many times or calling things like truncation. If you are interested in any sort of precision whatsoever, the only way to achieve it is to use a fixed-point decimal, which basically is a scaled integer.

If I understand your question correctly, it depends on your language.
For C#, check out the MSDN ref. Float has a 7 digit precision and double 15-16 digit precision.

It took me quite a while to figure out that when using doubles in Java, I wasn't losing significant precision in calculations. floating point actually has a very good ability to represent numbers to quite reasonable precision. The precision I was losing was immediately upon converting decimal numbers typed by users to the binary floating point representation that is natively supported. I've recently started converting all my numbers to BigDecimal. BigDecimal is much more work to deal with in the code than floats or doubles, since it's not one of the primitive types. But on the other hand, I'll be able to exactly represent the numbers that users type in.

Why does 230/100*100 not return 230? [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Is JavaScript’s Math broken?
In Javascript, I cannot figure out why 230/100*100 returns 229.99999999999997, while 240/100*100 returns 240.
This also applies to 460, 920 and so on...
Is there any solution?

The Issue:
230/100*100
= (230 / 100) * 100
= 2.3 * 100 in binary
2.3 in binary is the recurring decimal: 10.01001100110011001100110011001100...
This recurring decimal, cannot be accurately represented, due to limited precision, we get something like 2.29999999981373548507....
Interestingly, if you chose a division operation like such that it was accurately representable (not a recurring decimal and all digits lying within the maximum significant digits accommodated by the FP standard) in binary, you wouldn't see any such discrepancy.
E.g. 225/100*100 = 225
2.25 in binary is 10.01
Test Conversion: Binary to/from Decimal
Dealing with it:
Always be wary of precision when checking for equality between floating point values. Rounding up/down to a certain number of significant digits is good practice.

In JavaScript all numeric values are stored as IEEE 754 64-bit floating-point values (also known as double in many languages). This representation has only finite precision (so not all numbers can be accurately represented) and it is binary, so values that seem to be easy to represent in decimal can turn out to be problematic to handle.
There is no fire-and-forget solution suitable for everyone. If you need an integer then simply round using Math.round.

This problem relates to floating point inaccuracy. See this question for more details:
Is floating point math broken?

For the same reason that if you were to be forced to stay to a certain precision, and to take every step, you'd give 10/3*3 as 9.99999....
Say the precision you had to keep to was 10 digits. After 10/3 you'd have 3.333333333. Then when you multiplied that by 3, you'd have 9.999999999.
Now, since we know that the 3s will go on forever, we know that the 9s will go on forever, and so we know that the answer is really 10. But that's not the deal here, the deal is you apply each step as best you can, and then go on to the next.
As well as numbers that will result in recurring representations, there could be those that could be represented precisely, but not with the number of digits you are using.
Just as 10/3 cannot be represented perfectly in decimal, so 230/100 cannot be represented perfectly in binary.

The division in JavaScript is not integer division, but floating point.
2.3 or 2.4 can't be exactly represented in floating points. The difference is that the closest fp for 2.4 is 2.4000000953, while 2.3 is about 2.2999999523.
One can use Math.round(x) or one can use a JavaScript trick:
(x|0) converts x to integer, as the '|' operator forces the operands to integers.
Even in this case 299.9943 is not rounded but truncated.

Unexpected value on multiplication in javascript [duplicate]

There is some problem, i can't understand anyway.
look at this code please
<script type="text/javascript">
function math(x)
{
var y;
y = x*10;
alert(y);
}
</script>
<input type="button" onclick="math(0.011)">
What must be alerted after i click on button?
i think 0.11, but no, it alerts
0.10999999999999999
explain please this behavior.
thanks in advance

Ok. Let me try to explain it.
The basic thing to remember with floating point numbers is this: They occupy a limited amount of bits and try to represent the original number using base-2 arithmetic.
As you know, in base-2 arithmetic integers are represented by the powers of 2 that they contain. Thus, 6 would be represented as 4 + 2, ie. in binary as 110.
In order to understand how fractional numbers are represented, you have to think about how we represent fractional numbers in our decimal system. The fractional part of numbers (for example 0.11) is represented as multiples of inverse powers of 10 (since the base is 10). Thus 0.11 is actually 1/10 + 1/100. As you can appreciate, this is not powerful enough to represent all fractional numbers in a limited number of digits. For example, 1/3 would be 0.333333.... in a never ending fashion. If we had only 32 digits of space to write the number down, we would end up having only an approximation to the original number, 0.33333333333333333333333333333333. This number, for example, would give 0.99999999999999999999999999999999 if it was multiplied by 3 and not 1 as you would have expected.
The situation is similar in base-2. Each fractional number would be represented as multiples of inverse powers of 2. Thus 0.75 (in decimal) (ie 3/4) would be represented as 1/2 + 1/4, which would mean 0.11 (in base-2). Just as base 10 is not capable enough to represent every fractional number in a finite manner, base-2 cannot represent all fractional numbers given a limited amount of space.
Now, try to represent 0.11 in base-2; you start with 11/100 and try to find an inverse power of 2 that is just less than this number. 1/2 doesn't work, 1/4 neither, nor does 1/8. 1/16 fits the bill, so you mark a 1 in the 4th place after the decimal point and subtract 1/16 from 11/100. You are left with 19/400. Now try to find the next power of 2 that fits the description. 1/32 seems to be that one, mark the 5th place after the point and subtract 1/32 from 19/400, you get 13/800. Next one is 1/64 and you are left with 1/1600 thus the next one is all the way up at 1/2048, etc. etc. Thus we got as far as 0.00011100001 but it goes on and on; and you will see that there always is a fraction remaining. Now, I didn't go through the whole calculation, but after you have put in 32 binary digits after the dot you will still probably have some fraction left (and this is assuming that all of the 32 bits of space is spent representing the decimal part, which it is not). Thus, I am sure you can appreciate that the resulting number might differ from its actual value by some amount.
In your case, the difference is 0.00000000000000001 which is 1/100000000000000000 = 1/10^17 and I am sure that you can see why you might have that.

this is because you are dealing with floating point, and this is the expected behavior of floating point math.
what you need to do is format that number.
see this java explanation which also applies here if you want to know why this is happening.
in javascript all numbers are represented as 64bit floats, so you will run into this sort of thing often.
the quick overview of that article is that floating point tries to represent a range of values larger then would fit in 64bits, therefor there is going to be some imprecise representation, and this is what you are seeing.

With floating point number you get a representation of the number you try to encode. Mostly it is a number that is very close the the original number. More information on encoding/storing floating point numbers can be found here.
Note:
If you show the value of x, it still shows 0.011 because JavaScript has not yet decided what variable type x has. But after multiplying it with 10 the type got set to floating point (it is the only possibility) and the round error shows.

You can try to fix the nr of decimals with this one:
// fl is a float number with some nr of decimals
// d is how many decimals you want
function dec(fl, d) {
var p = Math.pow(10, d);
return Math.round(fl*p)/p;
}
Ex:
var n = 0.0012345;
console.log(dec(n,6)); // 0.001235
console.log(dec(n,5)); // 0.00123
console.log(dec(n,4)); // 0.0012
console.log(dec(n,3)); // 0.001
It works by first multiplying the float with 10^3 (1000) for three decimals, or 10^2 (100) for two decimals. Then do round on that and divide it back to original size.
Math.pow(10, d) makes 10^d (means that d will give us 1000).
In your case, do alert(dec(y,2));, it should work.

understanding floating point variables

There is some problem, i can't understand anyway.
look at this code please
<script type="text/javascript">
function math(x)
{
var y;
y = x*10;
alert(y);
}
</script>
<input type="button" onclick="math(0.011)">
What must be alerted after i click on button?
i think 0.11, but no, it alerts
0.10999999999999999
explain please this behavior.
thanks in advance

Ok. Let me try to explain it.
The basic thing to remember with floating point numbers is this: They occupy a limited amount of bits and try to represent the original number using base-2 arithmetic.
As you know, in base-2 arithmetic integers are represented by the powers of 2 that they contain. Thus, 6 would be represented as 4 + 2, ie. in binary as 110.
In order to understand how fractional numbers are represented, you have to think about how we represent fractional numbers in our decimal system. The fractional part of numbers (for example 0.11) is represented as multiples of inverse powers of 10 (since the base is 10). Thus 0.11 is actually 1/10 + 1/100. As you can appreciate, this is not powerful enough to represent all fractional numbers in a limited number of digits. For example, 1/3 would be 0.333333.... in a never ending fashion. If we had only 32 digits of space to write the number down, we would end up having only an approximation to the original number, 0.33333333333333333333333333333333. This number, for example, would give 0.99999999999999999999999999999999 if it was multiplied by 3 and not 1 as you would have expected.
The situation is similar in base-2. Each fractional number would be represented as multiples of inverse powers of 2. Thus 0.75 (in decimal) (ie 3/4) would be represented as 1/2 + 1/4, which would mean 0.11 (in base-2). Just as base 10 is not capable enough to represent every fractional number in a finite manner, base-2 cannot represent all fractional numbers given a limited amount of space.
Now, try to represent 0.11 in base-2; you start with 11/100 and try to find an inverse power of 2 that is just less than this number. 1/2 doesn't work, 1/4 neither, nor does 1/8. 1/16 fits the bill, so you mark a 1 in the 4th place after the decimal point and subtract 1/16 from 11/100. You are left with 19/400. Now try to find the next power of 2 that fits the description. 1/32 seems to be that one, mark the 5th place after the point and subtract 1/32 from 19/400, you get 13/800. Next one is 1/64 and you are left with 1/1600 thus the next one is all the way up at 1/2048, etc. etc. Thus we got as far as 0.00011100001 but it goes on and on; and you will see that there always is a fraction remaining. Now, I didn't go through the whole calculation, but after you have put in 32 binary digits after the dot you will still probably have some fraction left (and this is assuming that all of the 32 bits of space is spent representing the decimal part, which it is not). Thus, I am sure you can appreciate that the resulting number might differ from its actual value by some amount.
In your case, the difference is 0.00000000000000001 which is 1/100000000000000000 = 1/10^17 and I am sure that you can see why you might have that.

this is because you are dealing with floating point, and this is the expected behavior of floating point math.
what you need to do is format that number.
see this java explanation which also applies here if you want to know why this is happening.
in javascript all numbers are represented as 64bit floats, so you will run into this sort of thing often.
the quick overview of that article is that floating point tries to represent a range of values larger then would fit in 64bits, therefor there is going to be some imprecise representation, and this is what you are seeing.

With floating point number you get a representation of the number you try to encode. Mostly it is a number that is very close the the original number. More information on encoding/storing floating point numbers can be found here.
Note:
If you show the value of x, it still shows 0.011 because JavaScript has not yet decided what variable type x has. But after multiplying it with 10 the type got set to floating point (it is the only possibility) and the round error shows.

You can try to fix the nr of decimals with this one:
// fl is a float number with some nr of decimals
// d is how many decimals you want
function dec(fl, d) {
var p = Math.pow(10, d);
return Math.round(fl*p)/p;
}
Ex:
var n = 0.0012345;
console.log(dec(n,6)); // 0.001235
console.log(dec(n,5)); // 0.00123
console.log(dec(n,4)); // 0.0012
console.log(dec(n,3)); // 0.001
It works by first multiplying the float with 10^3 (1000) for three decimals, or 10^2 (100) for two decimals. Then do round on that and divide it back to original size.
Math.pow(10, d) makes 10^d (means that d will give us 1000).
In your case, do alert(dec(y,2));, it should work.