How does JS convert large numbers (above 2^53) to string - javascript

Since integers above 2^53 can't be accurately represented in doubles, how does JS decide on their decimal representation when they are printed as strings?
For example, 2^55 is 36028797018963968, and printf("%lf",(double)(1LL<<55)) in C will print that number correctly, since it has trailing zeroes in its binary representation that do not cause precision loss when truncated.
However, in Javascript, we get 36028797018963970 instead. It seems to try to round numbers to get a 0 at the end, but not always - for instance, 2^55-4 is represented correctly with 4 at the end.
Is there some place in the spec that defines this weird behavior?

Question- Since integers above 2^53 can't be accurately represented in doubles, how does JS decide on their decimal representation when they are printed as strings?
1. Way of Printing decimal numbers in JS
JavaScript numbers are internally stored in binary floating point and usually displayed in the decimal system.
There are two decimal notations used by JavaScript:
Fixed notation
[ "+" | "-" ] digit+ [ "." digit+ ]
Exponential notation
[ "+" | "-" ] digit [ "." digit+ ] "e" [ "+" | "-" ] digit+
An example of exponential notation is 1.2345678901234568e+21.
Rules for Displaying decimal numbers:
A. Use exponential notation if there are more than 21 digits before the decimal point.
B. Use exponential notation if the number starts with “0.” followed by more than five zeros.
2. The ECMAScript 5.1 display algorithm
Here is a details of Sect. 9.8.1 of the ECMAScript 5.1 specification describes the algorithm for displaying a decimal number
Given a number
mantissa × 10^pointPos−digitCount
The mantissa of a floating point number is an integer – the significant digits plus a sign. Leading and trailing zeros are discarded. Examples:
The mantissa of 12.34 is 1234.
Case-1. No decimal point: digitCount ≤ pointPos ≤ 21
Print the digits (without leading zeros), followed by pointPos−digitCount zeros.
Case-2. Decimal point inside the mantissa: 0 < pointPos ≤ 21, pointPos < digitCount
Display the pointPos first digits of the mantissa, a point and then the remaining digitCount−pointPos digits.
Case-3. Decimal point comes before the mantissa: −6 < pointPos ≤ 0
Display a 0 followed by a point, −pointPos zeros and the mantissa.
Case-4. Exponential notation: pointPos ≤ -6 or pointPos > 21
Display the first digit of the mantissa. If there are more digits then display a point and the remaining digits. Next, display the character e and a plus or minus sign (depending on the sign of pointPos−1), followed by the absolute value of pointPos−1. Therefore, the result looks as follows.
mantissa0 [ "." mantissa1..digitCount ]
"e" signChar(pointPos−1) abs(pointPos−1)
Question-
However, in Javascript, we get 36028797018963970 instead. It seems to try to round numbers to get a 0 at the end, but not always - for instance, 2^55-4 is represented correctly with 4 at the end.
Is there some place in the spec that defines this weird behavior?
Check: How numbers are encoded in JavaScript specially ==>5. The maximum integer
Additional Reference: https://medium.com/dailyjs/javascripts-number-type-8d59199db1b6

Related

What 17 decimal places of precision means in number js mdn doc?

const a = 10;
const b = 0.123456789123456789;
console.log((a + b).toFixed(17));
// 10.12345678912345726
As you can see from example above, .12345678912345 , only this part are shown correctly , as I understand Javascript only consider 15 places precision ( including .). If I will change 10 to 100 , it will be same amount , but I was thinking it should be 17 places precision by MDN doc. What doesn't this phrase exactly mean 17 decimal places of precision ?
If I will show it without .toFixed() method , it will show same 15 precision 10.123456789123457 - response of a + b
Url: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number
According to JS/ECMAScript specification, the Number type uses double-precision floating point which has 64-bit format (binary64), consists of a sign bit (determines positive or negative value), 11 exponent bits and 52 fraction bits (each digit represents 4-bits, hence 64-bit has 16 digits):
The Number type representing the double-precision 64-bit format IEEE
754-2008 values as specified in the IEEE Standard for Binary
Floating-Point Arithmetic.
The maximum positive number which can be represented properly using double precision is 9007199254740992, which can be achieved by using Math.pow(2, 53). If the number range is between Math.pow(2, 53) and Math.pow(2, 54) (or between Math.pow(2, -53) and Math.pow(2, -54)), only even numbers can be represented properly because the exponent bits will affect LSB (least-significant bit) on the fraction bits.
Let's review the large number part:
var x = 12345678912345.6789
var x = new Number(12345678912345.6789)
This number contains more than 52 fractional bits (72 bits in total), hence the rounding used to keep the fractional bits to 52.
Also with this decimal number:
var x = new Number(.12345678912367890)
This number contains 68 fractional bits, hence the last zero is chopped off to keep 64-bit length.
Usually numeric representation larger than 9007199254740992 or smaller than 1.1102230246251565E-16 are stored as literal strings instead of Number. If you need to compute very large numbers, there are certain external libraries available to perform arbitrary precision arithmetic.
If you want to cast more then 16 points after the decimal point you can either:
Use literal string to represent your number
Use external libraries like math.js, BigInteger.js or strint library.

Javascript - What is the maximum number in scientific notation restricted by?

In JavaScript, I was curious to find out what was the maximum possible number representable in scientific notation without getting "Infinity" as a result, so I wrote a little program and found out it's this one:
17976931348623158079372897140530341507993413271003782693617377898044496829276475094664901797758720709633028641669288791094655554785194040263065748867150582068190890200070838367627385484581771153176447573027006985557136695962284291481986083493647529271907416844436551070434271155969950809304288017790417449779
which can be abbreviated to 1.7976931348623157e+308.
My question is, what makes this specific number the maximum possible in JavaScript? Is it hardware-dependent (maybe maximum one on 64 bit?) or language-specific? Why exactly is 308 the maximum usable power of 10?
And also, how different is it in other languages?
Short answer:
Double precision float. Due to how the double data-type is defined.
Long answer:
All floating point numbers (double is a double-precision float) are written as a product of two values, the mantissa and the exponent. In principle, this works similar to how numbers are written in scientific notation: for the number 1.34 * 10^24, the mantissa is 1.34 and the exponent is 24.
https://en.wikipedia.org/wiki/Double-precision_floating-point_format
Number.MAX_VALUE
The value of Number.MAX_VALUE is the largest positive finite value of the Number type, which is approximately 1.7976931348623157e+308.
This property has the attributes { [[Writable]]: false, [[Enumerable]]: fafalselse, [[Configurable]]: false }.
http://ecma262-5.com/ELS5_HTML.htm#Section_8.5
What differs for floats (and doubles) is that you split the total bytes that hold the number into two parts, one for the mantissa and one for the exponent.
That gives you an exponent of 10 bits, and one sign bit for the exponent, so that would give you a number from -1023 to +1024.
However, the base of the exponent is not 10, but 2. The way the floating point number exponent is stored uses 8 bits (for floats) or 11 bits (for doubles), meaning you get exponent values of -127 to +128 (float) or -1023 to +1024 (double).
And 2^1024 gives us a value of 1.797693134862315907729305190789 * 10^308, which is the largest exponent of a double precision float.

JavaScript number automatically "rounded" [duplicate]

See this code:
var jsonString = '{"id":714341252076979033,"type":"FUZZY"}';
var jsonParsed = JSON.parse(jsonString);
console.log(jsonString, jsonParsed);
When I see my console in Firefox 3.5, the value of jsonParsed is the number rounded:
Object id=714341252076979100 type=FUZZY
Tried different values, the same outcome (number rounded).
I also don't get its rounding rules. 714341252076979136 is rounded to 714341252076979200, whereas 714341252076979135 is rounded to 714341252076979100.
Why is this happening?
You're overflowing the capacity of JavaScript's number type, see §8.5 of the spec for details. Those IDs will need to be strings.
IEEE-754 double-precision floating point (the kind of number JavaScript uses) can't precisely represent all numbers (of course). Famously, 0.1 + 0.2 == 0.3 is false. That can affect whole numbers just like it affects fractional numbers; it starts once you get above 9,007,199,254,740,991 (Number.MAX_SAFE_INTEGER).
Beyond Number.MAX_SAFE_INTEGER + 1 (9007199254740992), the IEEE-754 floating-point format can no longer represent every consecutive integer. 9007199254740991 + 1 is 9007199254740992, but 9007199254740992 + 1 is also 9007199254740992 because 9007199254740993 cannot be represented in the format. The next that can be is 9007199254740994. Then 9007199254740995 can't be, but 9007199254740996 can.
The reason is we've run out of bits, so we no longer have a 1s bit; the lowest-order bit now represents multiples of 2. Eventually, if we keep going, we lose that bit and only work in multiples of 4. And so on.
Your values are well above that threshold, and so they get rounded to the nearest representable value.
As of ES2020, you can use BigInt for integers that are arbitrarily large, but there is no JSON representation for them. You could use strings and a reviver function:
const jsonString = '{"id":"714341252076979033","type":"FUZZY"}';
// Note it's a string −−−−^−−−−−−−−−−−−−−−−−−^
const obj = JSON.parse(jsonString, (key, value) => {
if (key === "id" && typeof value === "string" && value.match(/^\d+$/)) {
return BigInt(value);
}
return value;
});
console.log(obj);
(Look in the real console, the snippets console doesn't understand BigInt.)
If you're curious about the bits, here's what happens: An IEEE-754 binary double-precision floating-point number has a sign bit, 11 bits of exponent (which defines the overall scale of the number, as a power of 2 [because this is a binary format]), and 52 bits of significand (but the format is so clever it gets 53 bits of precision out of those 52 bits). How the exponent is used is complicated (described here), but in very vague terms, if we add one to the exponent, the value of the significand is doubled, since the exponent is used for powers of 2 (again, caveat there, it's not direct, there's cleverness in there).
So let's look at the value 9007199254740991 (aka, Number.MAX_SAFE_INTEGER):
+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− sign bit
/ +−−−−−−−+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− exponent
/ / | +−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−+− significand
/ / | / |
0 10000110011 1111111111111111111111111111111111111111111111111111
= 9007199254740991 (Number.MAX_SAFE_INTEGER)
That exponent value, 10000110011, means that every time we add one to the significand, the number represented goes up by 1 (the whole number 1, we lost the ability to represent fractional numbers much earlier).
But now that significand is full. To go past that number, we have to increase the exponent, which means that if we add one to the significand, the value of the number represented goes up by 2, not 1 (because the exponent is applied to 2, the base of this binary floating point number):
+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− sign bit
/ +−−−−−−−+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− exponent
/ / | +−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−+− significand
/ / | / |
0 10000110100 0000000000000000000000000000000000000000000000000000
= 9007199254740992 (Number.MAX_SAFE_INTEGER + 1)
Well, that's okay, because 9007199254740991 + 1 is 9007199254740992 anyway. But! We can't represent 9007199254740993. We've run out of bits. If we add just 1 to the significand, it adds 2 to the value:
+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− sign bit
/ +−−−−−−−+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− exponent
/ / | +−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−+− significand
/ / | / |
0 10000110100 0000000000000000000000000000000000000000000000000001
= 9007199254740994 (Number.MAX_SAFE_INTEGER + 3)
The format just cannot represent odd numbers anymore as we increase the value, the exponent is too big.
Eventually, we run out of significand bits again and have to increase the exponent, so we end up only being able to represent multiples of 4. Then multiples of 8. Then multiples of 16. And so on.
What you're seeing here is actually the effect of two roundings. Numbers in ECMAScript are internally represented double-precision floating-point. When id is set to 714341252076979033 (0x9e9d9958274c359 in hex), it actually is assigned the nearest representable double-precision value, which is 714341252076979072 (0x9e9d9958274c380). When you print out the value, it is being rounded to 15 significant decimal digits, which gives 14341252076979100.
It is not caused by this json parser. Just try to enter 714341252076979033 to fbug's console. You'll see the same 714341252076979100.
See this blog post for details:
http://www.exploringbinary.com/print-precision-of-floating-point-integers-varies-too
JavaScript uses double precision floating point values, ie a total precision of 53 bits, but you need
ceil(lb 714341252076979033) = 60
bits to exactly represent the value.
The nearest exactly representable number is 714341252076979072 (write the original number in binary, replace the last 7 digits with 0 and round up because the highest replaced digit was 1).
You'll get 714341252076979100 instead of this number because ToString() as described by ECMA-262, §9.8.1 works with powers of ten and in 53 bit precision all these numbers are equal.
The problem is that your number requires a greater precision than JavaScript has.
Can you send the number as a string? Separated in two parts?
JavaScript can only handle exact whole numbers up to about 9000 million million (that's 9 with 15 zeros). Higher than that and you get garbage. Work around this by using strings to hold the numbers. If you need to do math with these numbers, write your own functions or see if you can find a library for them: I suggest the former as I don't like the libraries I've seen. To get you started, see two of my functions at another answer.

Javascript number conversion [duplicate]

See this code:
var jsonString = '{"id":714341252076979033,"type":"FUZZY"}';
var jsonParsed = JSON.parse(jsonString);
console.log(jsonString, jsonParsed);
When I see my console in Firefox 3.5, the value of jsonParsed is the number rounded:
Object id=714341252076979100 type=FUZZY
Tried different values, the same outcome (number rounded).
I also don't get its rounding rules. 714341252076979136 is rounded to 714341252076979200, whereas 714341252076979135 is rounded to 714341252076979100.
Why is this happening?
You're overflowing the capacity of JavaScript's number type, see §8.5 of the spec for details. Those IDs will need to be strings.
IEEE-754 double-precision floating point (the kind of number JavaScript uses) can't precisely represent all numbers (of course). Famously, 0.1 + 0.2 == 0.3 is false. That can affect whole numbers just like it affects fractional numbers; it starts once you get above 9,007,199,254,740,991 (Number.MAX_SAFE_INTEGER).
Beyond Number.MAX_SAFE_INTEGER + 1 (9007199254740992), the IEEE-754 floating-point format can no longer represent every consecutive integer. 9007199254740991 + 1 is 9007199254740992, but 9007199254740992 + 1 is also 9007199254740992 because 9007199254740993 cannot be represented in the format. The next that can be is 9007199254740994. Then 9007199254740995 can't be, but 9007199254740996 can.
The reason is we've run out of bits, so we no longer have a 1s bit; the lowest-order bit now represents multiples of 2. Eventually, if we keep going, we lose that bit and only work in multiples of 4. And so on.
Your values are well above that threshold, and so they get rounded to the nearest representable value.
As of ES2020, you can use BigInt for integers that are arbitrarily large, but there is no JSON representation for them. You could use strings and a reviver function:
const jsonString = '{"id":"714341252076979033","type":"FUZZY"}';
// Note it's a string −−−−^−−−−−−−−−−−−−−−−−−^
const obj = JSON.parse(jsonString, (key, value) => {
if (key === "id" && typeof value === "string" && value.match(/^\d+$/)) {
return BigInt(value);
}
return value;
});
console.log(obj);
(Look in the real console, the snippets console doesn't understand BigInt.)
If you're curious about the bits, here's what happens: An IEEE-754 binary double-precision floating-point number has a sign bit, 11 bits of exponent (which defines the overall scale of the number, as a power of 2 [because this is a binary format]), and 52 bits of significand (but the format is so clever it gets 53 bits of precision out of those 52 bits). How the exponent is used is complicated (described here), but in very vague terms, if we add one to the exponent, the value of the significand is doubled, since the exponent is used for powers of 2 (again, caveat there, it's not direct, there's cleverness in there).
So let's look at the value 9007199254740991 (aka, Number.MAX_SAFE_INTEGER):
+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− sign bit
/ +−−−−−−−+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− exponent
/ / | +−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−+− significand
/ / | / |
0 10000110011 1111111111111111111111111111111111111111111111111111
= 9007199254740991 (Number.MAX_SAFE_INTEGER)
That exponent value, 10000110011, means that every time we add one to the significand, the number represented goes up by 1 (the whole number 1, we lost the ability to represent fractional numbers much earlier).
But now that significand is full. To go past that number, we have to increase the exponent, which means that if we add one to the significand, the value of the number represented goes up by 2, not 1 (because the exponent is applied to 2, the base of this binary floating point number):
+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− sign bit
/ +−−−−−−−+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− exponent
/ / | +−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−+− significand
/ / | / |
0 10000110100 0000000000000000000000000000000000000000000000000000
= 9007199254740992 (Number.MAX_SAFE_INTEGER + 1)
Well, that's okay, because 9007199254740991 + 1 is 9007199254740992 anyway. But! We can't represent 9007199254740993. We've run out of bits. If we add just 1 to the significand, it adds 2 to the value:
+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− sign bit
/ +−−−−−−−+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− exponent
/ / | +−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−+− significand
/ / | / |
0 10000110100 0000000000000000000000000000000000000000000000000001
= 9007199254740994 (Number.MAX_SAFE_INTEGER + 3)
The format just cannot represent odd numbers anymore as we increase the value, the exponent is too big.
Eventually, we run out of significand bits again and have to increase the exponent, so we end up only being able to represent multiples of 4. Then multiples of 8. Then multiples of 16. And so on.
What you're seeing here is actually the effect of two roundings. Numbers in ECMAScript are internally represented double-precision floating-point. When id is set to 714341252076979033 (0x9e9d9958274c359 in hex), it actually is assigned the nearest representable double-precision value, which is 714341252076979072 (0x9e9d9958274c380). When you print out the value, it is being rounded to 15 significant decimal digits, which gives 14341252076979100.
It is not caused by this json parser. Just try to enter 714341252076979033 to fbug's console. You'll see the same 714341252076979100.
See this blog post for details:
http://www.exploringbinary.com/print-precision-of-floating-point-integers-varies-too
JavaScript uses double precision floating point values, ie a total precision of 53 bits, but you need
ceil(lb 714341252076979033) = 60
bits to exactly represent the value.
The nearest exactly representable number is 714341252076979072 (write the original number in binary, replace the last 7 digits with 0 and round up because the highest replaced digit was 1).
You'll get 714341252076979100 instead of this number because ToString() as described by ECMA-262, §9.8.1 works with powers of ten and in 53 bit precision all these numbers are equal.
The problem is that your number requires a greater precision than JavaScript has.
Can you send the number as a string? Separated in two parts?
JavaScript can only handle exact whole numbers up to about 9000 million million (that's 9 with 15 zeros). Higher than that and you get garbage. Work around this by using strings to hold the numbers. If you need to do math with these numbers, write your own functions or see if you can find a library for them: I suggest the former as I don't like the libraries I've seen. To get you started, see two of my functions at another answer.

Large numbers erroneously rounded in JavaScript

See this code:
var jsonString = '{"id":714341252076979033,"type":"FUZZY"}';
var jsonParsed = JSON.parse(jsonString);
console.log(jsonString, jsonParsed);
When I see my console in Firefox 3.5, the value of jsonParsed is the number rounded:
Object id=714341252076979100 type=FUZZY
Tried different values, the same outcome (number rounded).
I also don't get its rounding rules. 714341252076979136 is rounded to 714341252076979200, whereas 714341252076979135 is rounded to 714341252076979100.
Why is this happening?
You're overflowing the capacity of JavaScript's number type, see §8.5 of the spec for details. Those IDs will need to be strings.
IEEE-754 double-precision floating point (the kind of number JavaScript uses) can't precisely represent all numbers (of course). Famously, 0.1 + 0.2 == 0.3 is false. That can affect whole numbers just like it affects fractional numbers; it starts once you get above 9,007,199,254,740,991 (Number.MAX_SAFE_INTEGER).
Beyond Number.MAX_SAFE_INTEGER + 1 (9007199254740992), the IEEE-754 floating-point format can no longer represent every consecutive integer. 9007199254740991 + 1 is 9007199254740992, but 9007199254740992 + 1 is also 9007199254740992 because 9007199254740993 cannot be represented in the format. The next that can be is 9007199254740994. Then 9007199254740995 can't be, but 9007199254740996 can.
The reason is we've run out of bits, so we no longer have a 1s bit; the lowest-order bit now represents multiples of 2. Eventually, if we keep going, we lose that bit and only work in multiples of 4. And so on.
Your values are well above that threshold, and so they get rounded to the nearest representable value.
As of ES2020, you can use BigInt for integers that are arbitrarily large, but there is no JSON representation for them. You could use strings and a reviver function:
const jsonString = '{"id":"714341252076979033","type":"FUZZY"}';
// Note it's a string −−−−^−−−−−−−−−−−−−−−−−−^
const obj = JSON.parse(jsonString, (key, value) => {
if (key === "id" && typeof value === "string" && value.match(/^\d+$/)) {
return BigInt(value);
}
return value;
});
console.log(obj);
(Look in the real console, the snippets console doesn't understand BigInt.)
If you're curious about the bits, here's what happens: An IEEE-754 binary double-precision floating-point number has a sign bit, 11 bits of exponent (which defines the overall scale of the number, as a power of 2 [because this is a binary format]), and 52 bits of significand (but the format is so clever it gets 53 bits of precision out of those 52 bits). How the exponent is used is complicated (described here), but in very vague terms, if we add one to the exponent, the value of the significand is doubled, since the exponent is used for powers of 2 (again, caveat there, it's not direct, there's cleverness in there).
So let's look at the value 9007199254740991 (aka, Number.MAX_SAFE_INTEGER):
+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− sign bit
/ +−−−−−−−+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− exponent
/ / | +−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−+− significand
/ / | / |
0 10000110011 1111111111111111111111111111111111111111111111111111
= 9007199254740991 (Number.MAX_SAFE_INTEGER)
That exponent value, 10000110011, means that every time we add one to the significand, the number represented goes up by 1 (the whole number 1, we lost the ability to represent fractional numbers much earlier).
But now that significand is full. To go past that number, we have to increase the exponent, which means that if we add one to the significand, the value of the number represented goes up by 2, not 1 (because the exponent is applied to 2, the base of this binary floating point number):
+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− sign bit
/ +−−−−−−−+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− exponent
/ / | +−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−+− significand
/ / | / |
0 10000110100 0000000000000000000000000000000000000000000000000000
= 9007199254740992 (Number.MAX_SAFE_INTEGER + 1)
Well, that's okay, because 9007199254740991 + 1 is 9007199254740992 anyway. But! We can't represent 9007199254740993. We've run out of bits. If we add just 1 to the significand, it adds 2 to the value:
+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− sign bit
/ +−−−−−−−+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− exponent
/ / | +−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−+− significand
/ / | / |
0 10000110100 0000000000000000000000000000000000000000000000000001
= 9007199254740994 (Number.MAX_SAFE_INTEGER + 3)
The format just cannot represent odd numbers anymore as we increase the value, the exponent is too big.
Eventually, we run out of significand bits again and have to increase the exponent, so we end up only being able to represent multiples of 4. Then multiples of 8. Then multiples of 16. And so on.
What you're seeing here is actually the effect of two roundings. Numbers in ECMAScript are internally represented double-precision floating-point. When id is set to 714341252076979033 (0x9e9d9958274c359 in hex), it actually is assigned the nearest representable double-precision value, which is 714341252076979072 (0x9e9d9958274c380). When you print out the value, it is being rounded to 15 significant decimal digits, which gives 14341252076979100.
It is not caused by this json parser. Just try to enter 714341252076979033 to fbug's console. You'll see the same 714341252076979100.
See this blog post for details:
http://www.exploringbinary.com/print-precision-of-floating-point-integers-varies-too
JavaScript uses double precision floating point values, ie a total precision of 53 bits, but you need
ceil(lb 714341252076979033) = 60
bits to exactly represent the value.
The nearest exactly representable number is 714341252076979072 (write the original number in binary, replace the last 7 digits with 0 and round up because the highest replaced digit was 1).
You'll get 714341252076979100 instead of this number because ToString() as described by ECMA-262, §9.8.1 works with powers of ten and in 53 bit precision all these numbers are equal.
The problem is that your number requires a greater precision than JavaScript has.
Can you send the number as a string? Separated in two parts?
JavaScript can only handle exact whole numbers up to about 9000 million million (that's 9 with 15 zeros). Higher than that and you get garbage. Work around this by using strings to hold the numbers. If you need to do math with these numbers, write your own functions or see if you can find a library for them: I suggest the former as I don't like the libraries I've seen. To get you started, see two of my functions at another answer.

Categories