Why are these two numbers equivalent in javascript? [duplicate] - javascript

This question already has answers here:
What is JavaScript's highest integer value that a number can go to without losing precision?
(21 answers)
Closed 8 years ago.
Type this in the console of your browser:
9999999999999999==10000000000000000
It says they are equal, why?

JavaScript only supports 53 bit integers
All numbers in JavaScript are floating point which means that integers are always represented as
sign × mantissa × 2exponent
The mantissa has 53 bits. You can use the exponent to get higher integers, but then they won’t be contiguous, any more. For example, you generally need to multiply the mantissa by two (exponent 1) in order to reach the 54th bit. However, if you multiply by two, you will only be able to represent every second integer:

Related

Adding integer numbers javascript, double float [duplicate]

This question already has answers here:
Floating point representations seem to do integer arithmetic correctly - why?
(8 answers)
Closed 5 years ago.
So I have read this answer here:
Is floating point math broken?
That because every number in JS is double float 0.1+0.2 for example will NOT equal to 0.3.
But I don't understand why it never happens with integers? Why then 1+2 always equals 3 etc. It would seem that integers like 1 or 2 similarly to 0.1 and 0.2 don't have perfect representation in the binary64 so their math should also sometimes break but that never happens.
Why is that?
but I don't understand why it never happens with integers?
It does, the integer just has to be really big before they hit the limits of the IEEE-754 format:
var a = 9007199254740992;
console.log(a); // 9007199254740992
var b = a + 1;
console.log(b); // still 9007199254740992
console.log(a == b); // true
Floating point formats, such as IEEE-754 are essentially an expression that describes the value as the following:
value := sign * mantissa * 2 ^ exponent
The mantissa is an integer of various sizes. For four byte floating point, the mantissa is 24 bits and, for eight byte floating point, the mantissa is 48 bits. If the exponent is 0, the value of the expression is determined only by the sign and the mantissa. This is, in fact, how integers are represented by JavaScript.
What seems to take most people by surprise is due to the base 2 exponent instead of base 10. We accept, that, in base 10, the result of 1/3 or 2/3 cannot be exactly represented without an infinite number of digits or the acceptance of round-off error. Similarly, there are fractions in base 2 that have similar issues. Unfortunately for our base 10 mindset, these fractions most often involve negative powers of 10.

parseInt behaviour in Javascript [duplicate]

This question already has answers here:
What is JavaScript's highest integer value that a number can go to without losing precision?
(21 answers)
Closed 7 years ago.
I open terminal browser (Chrome for example).
I write this:
var y = "11000011010101011";
"11000011010101011"
parseInt(y)
11000011010101012
I expected 11000011010101011 but it returns me 11000011010101012.
Does anybody know why?
Every number in Javascript is represented as a double precision floating point. JavaScript can accurately represent integers only up to 9007199254740991 (2^53 - 1). Once you get over that limit, you will loose precision.
According to this page.
All number in Javascript are 64-bit floating point number, and integers are represented by the 53-bit mantisa.
Because of that, you can't store a integer larger than 2^53 -1 and smaller than -2^53 +1 without losing precission (Javascript rounds your number in order to be able to store it).
Your number is larger than 2^53 -1. Even though a String can store it, in order to store it in a "Number" variable, it has to be rounded, losing the precision and returning you a slightly different number.
Only 9007199254740991 is the safe integer in javascript.
This is case something like
9007199254740992 + 1 // 9007199254740992
9007199254740993 + 1 // 9007199254740992
9007199254740994 + 1 //9007199254740996
Please see the links for more details
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number
http://www.2ality.com/2012/04/number-encoding.html
Also see this one
how addition of Number works on max limit numbers?
This is the dublicate questions see its original

Convert number value to string [duplicate]

This question already has answers here:
What is JavaScript's highest integer value that a number can go to without losing precision?
(21 answers)
Closed 8 years ago.
Can anybody solve the following problem with javascript
var i = 10152233307863175;
alert(i.toString());
alert shows value 10152233307863176. Any solution. Problem is when I get json object on client and when string is converted to json it contains wrong values.
This is a limitation in the precision of the numeric data format that javascript uses (double precision floating point).
The best way of storing that value, assuming you don't need to do any mathematical operations, is storing it as a string in the first place.
MDN has this to say about numbers in JavaScript.
Numbers in JavaScript are "double-precision 64-bit format IEEE 754 values", according to the spec.
There is no real integers in JavaScript. According to this source:
ECMAScript numbers are represented in binary as IEEE-754 (IEC 559) Doubles, with a resolution of 53 bits, giving an accuracy of 15-16 decimal digits; integers up to just over 9e15 are precise, ...
Your number 10152233307863175 contains 17 digits. Since the number is represented as a floating point number, JavaScript tries to do it's best and set bits in a way that the resulting number is closest to the supplied number.

4.25.toFixed(1) == 4.35.toFixed(1) == 4.3 but 2.35.toFixed(1)==2.4 [duplicate]

This question already has answers here:
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&gt4:
5.35.toFixed(1) => 5.3
15.35.toFixed(1) => 15.3
131.35.toFixed(1) => 131.3
2.nr&lt4:
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.

Multiplication int and float(double) in JavaScript [duplicate]

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.

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