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I do not understand about bitwise operator in Javascript

I am student studying programming.
As far as I know, Javascript saves number as float.
However, bitwise operator in Javascript run as a type of number is integer.
For instance,
> 1 | 2 // -> 3
> 2 << 4 // -> 32
How is it possible?
I find official documentation(mdn web docs), but I can not find the reason.

The bitwise operators in JavaScript convert numbers to 32-bit integers before performing the operation. This means that even though JavaScript saves numbers as floats, the bitwise operators treat them as 32-bit integers. This is why you are able to use the bitwise operators on numbers and get integer results.

As #rviretural_001 mentioned, JavaScript does some automatic conversions by spec. For example, in addition to IEEE 754 doubles to 32 signed integers:
0 == '0' // true
String to Integer (Shift Right)
'4' >> 1 // 2
Converting to Integer (Bitwise OR)
'1.3' | 0 // 1
This last one was used in asm.js for 32 signed integers

just to contribute to the answers above note that the conversion of a floating-point number to an integer is done by removing the fractional part of the number.
For example, if you have the number 4.5 and you use a bitwise operator on it, JavaScript will convert it to the integer 4 before performing the operation.
This tutorial might help you to find out more https://www.programiz.com/javascript/bitwise-operators
Also note that performing bitwise operations on floating-point numbers can lead to unexpected results. This is due to the way floating-point numbers are represented in memory. So, it's recommended to avoid using bitwise operators on floating-point numbers in JavaScript.

Advice on converting 8 byte (u64) unsigned integer into javascript

I have a u64 (unsigned integer) stored in 8 bytes of memory. Clearly the range is 0-2^64 integers.
I am converting it to a javascript number by turning each byte into hex and making a hex string:
let s = '0x'
s += buffer.slice(0,1).toString("hex")
s += buffer.slice(1,2).toString("hex")
...
n = parseInt(s)
Works great for everything I have done so far.
But when I look at how javascript stores numbers, I become unsure. Javascript uses 8 bytes for numbers, but treats all numbers the same. This internal javascript "number" representation can also hold floating point numbers.
Can a javascript number store all integers from 0 to 2^64? seems not.
At what point do I get into trouble?
What do people do to get round this?

An unsigned 64 bit integer has the range of a 0 to 18.446.744.073.709.551.615.
You could use the Number wrapper object with the .MAX_VALUE property, it represents the maximum numeric value representable in JavaScript.
The JavaScript Number type is a double-precision 64-bit binary format IEEE 754 value, like double in Java or C#.
General Info:
Integers in JS:
JavaScript has only floating-point numbers. Integers appear internally in two ways. First, most JavaScript engines store a small enough number without a decimal fraction as an integer (with, for example, 31 bits) and maintain that representation as long as possible. They have to switch back to a floating point representation if a number’s magnitude grows too large or if a decimal fraction appears.
Second, the ECMAScript specification has integer operators: namely, all of the bitwise operators. Those operators convert their operands to 32-bit integers and return 32-bit integers. For the specification, integer only means that the numbers don’t have a decimal fraction, and 32-bit means that they are within a certain range. For engines, 32-bit integer means that an actual integer (non-floating-point) representation can usually be introduced or maintained.
Ranges of integers
Internally, the following ranges of integers are important in JavaScript:
Safe integers [1], the largest practically usable range of integers that JavaScript supports:
53 bits plus a sign, range (−2^53, 2^53) which relates to (+/-) 9.007.199.254.740.992
Array indices [2]:
32 bits, unsigned
Maximum length: 2^32−1
Range of indices: [0, 2^32−1) (excluding the maximum length!)
Bitwise operands [3]:
Unsigned right shift operator (>>>): 32 bits, unsigned, range [0, 2^32)
All other bitwise operators: 32 bits, including a sign, range [−2^31, 2^31)
“Char codes”, UTF-16 code units as numbers:
Accepted by String.fromCharCode()
Returned by String.prototype.charCodeAt()
16 bit, unsigned
References:
[1] Safe integers in JavaScript
[2] Arrays in JavaScript
[3] Label bitwise_ops
Source: https://2ality.com/2014/02/javascript-integers.html

In javascript, whether Bitwise OR use iee-754 as storage format during calculation nor not ？

if Bitwise OR was used ieee-754 to calc the result in Javascript,i can't understand the result.
for example:
2|1 =>3
In ieee-754,
2 is stored as 0 10000000000 0000...0000,and
1 is stored as 0 01111111111 0000...0000
if exec bitwise or,i think the result is 0 11111111111 0000...0000,but why does it output 3?
As same as above,
example:
0.1|0 =>0
0 is stored as 0 00000000000 0000...0000,and
0.1 is stored as 0 01111111011 1001100110011001100110011001100110011001100110011010
if exec bitwise or,i think the result is 0 01111111011 1001100110011001100110011001100110011001100110011010,but why does it output 0,and lose decimal?
example:
2|-1 =>-1
2 is stored as 0 10000000000 0000...0000,and
-1 is stored as 1 01111111111 0000...0000
if exec bitwise or,i think the result is 1 11111111111 0000...0000,but why does it output -1?

JavaScript is an implementation of ECMAScript. The ECMAScript 2018 Language Specification (9th edition, June 2018) specifies the binary bitwise operators in clause 12.12. The evaluation steps in 12.12.3 include (using “#” to stand for one of the bitwise operations):
Let lnum be ? ToInt32(lval).
Let rnum be ? ToInt32(rval).
Return the result of applying the bitwise operator # to lnum and rnum. The result is a signed 32-bit integer.
Thus, the bitwise operation is not performed on the bytes that represent a Number. The value of the Number is converted to a 32-bit two’s complement representation of an integer, and the operation is performed on those bits. Then resulting bits are then interpreted as a 32-bit two’s complement representation, so the value they represent becomes the value produced by the operation.

Math.pow gives wrong result

I was trying to repeat a character N times, and came across the Math.pow function.
But when I use it in the console, the results don't make any sense to me:
Math.pow(10,15) - 1 provides the correct result 999999999999999
But why does Math.pow(10,16) - 1 provide 10000000000000000?

You are producing results which exceed the Number.MAX_SAFE_INTEGER value, and so they are not accurate any more up to the unit.
This is related to the fact that JavaScript uses 64-bit floating point representation for numbers, and so in practice you only have about 16 (decimal) digits of precision.
Since the introduction of BigInt in EcmaScript, you can get an accurate result with that data type, although it cannot be used in combination with Math.pow. Instead you can use the ** operator.
See how the use of number and bigint (with the n suffix) differ:
10 ** 16 - 1 // == 10000000000000000
10n ** 16n - 1n // == 9999999999999999n

Unlike many other programming languages, JavaScript does not define different types of numbers, like integers, short, long, floating-point etc.
JavaScript numbers are always stored as double-precision floating-point numbers, following the international IEEE 754 standard.
This format stores numbers in 64 bits, where the number (the fraction) is stored in bits 0 to 51, the exponent in bits 52 to 62, and the sign-in bit 63:
Integers (numbers without a period or exponent notation) are accurate up to 15 digits.

Negative numbers to binary string in JavaScript

Anyone knows why javascript Number.toString function does not represents negative numbers correctly?
//If you try
(-3).toString(2); //shows "-11"
// but if you fake a bit shift operation it works as expected
(-3 >>> 0).toString(2); // print "11111111111111111111111111111101"
I am really curious why it doesn't work properly or what is the reason it works this way?
I've searched it but didn't find anything that helps.

Short answer:
The toString() function takes the decimal, converts it
to binary and adds a "-" sign.
A zero fill right shift converts it's operands to signed 32-bit
integers in two complements format.
A more detailed answer:
Question 1:
//If you try
(-3).toString(2); //show "-11"
It's in the function .toString(). When you output a number via .toString():
Syntax
numObj.toString([radix])
If the numObj is negative, the sign is preserved. This is the case
even if the radix is 2; the string returned is the positive binary
representation of the numObj preceded by a - sign, not the two's
complement of the numObj.
It takes the decimal, converts it to binary and adds a "-" sign.
Base 10 "3" converted to base 2 is "11"
Add a sign gives us "-11"
Question 2:
// but if you fake a bit shift operation it works as expected
(-3 >>> 0).toString(2); // print "11111111111111111111111111111101"
A zero fill right shift converts it's operands to signed 32-bit integers. The result of that operation is always an unsigned 32-bit integer.
The operands of all bitwise operators are converted to signed 32-bit
integers in two's complement format.

-3 >>> 0 (right logical shift) coerces its arguments to unsigned integers, which is why you get the 32-bit two's complement representation of -3.
http://en.wikipedia.org/wiki/Two%27s_complement
http://en.wikipedia.org/wiki/Logical_shift

var binary = (-3 >>> 0).toString(2); // coerced to uint32
console.log(binary);
console.log(parseInt(binary, 2) >> 0); // to int32
on jsfiddle
output is
11111111111111111111111111111101
-3

.toString() is designed to return the sign of the number in the string representation. See EcmaScript 2015, section 7.1.12.1:
If m is less than zero, return the String concatenation of the String "-" and ToString(−m).
This rule is no different for when a radix is passed as argument, as can be concluded from section 20.1.3.6:
Return the String representation of this Number value using the radix specified by radixNumber. [...] the algorithm should be a generalization of that specified in 7.1.12.1.
Once that is understood, the surprising thing is more as to why it does not do the same with -3 >>> 0.
But that behaviour has actually nothing to do with .toString(2), as the value is already different before calling it:
console.log (-3 >>> 0); // 4294967293
It is the consequence of how the >>> operator behaves.
It does not help either that (at the time of writing) the information on mdn is not entirely correct. It says:
The operands of all bitwise operators are converted to signed 32-bit integers in two's complement format.
But this is not true for all bitwise operators. The >>> operator is an exception to the rule. This is clear from the evaluation process specified in EcmaScript 2015, section 12.5.8.1:
Let lnum be ToUint32(lval).
The ToUint32 operation has a step where the operand is mapped into the unsigned 32 bit range:
Let int32bit be int modulo 232.
When you apply the above mentioned modulo operation (not to be confused with JavaScript's % operator) to the example value of -3, you get indeed 4294967293.
As -3 and 4294967293 are evidently not the same number, it is no surprise that (-3).toString(2) is not the same as (4294967293).toString(2).

Just to summarize a few points here, if the other answers are a little confusing:
what we want to obtain is the string representation of a negative number in binary representation; this means the string should show a signed binary number (using 2's complement)
the expression (-3 >>> 0).toString(2), let's call it A, does the job; but we want to know why and how it works
had we used var num = -3; num.toString(-3) we would have gotten -11, which is simply the unsigned binary representation of the number 3 with a negative sign in front, which is not what we want
expression A works like this:
1) (-3 >>> 0)
The >>> operation takes the left operand (-3), which is a signed integer, and simply shifts the bits 0 positions to the left (so the bits are unchanged), and the unsigned number corresponding to these unchanged bits.
The bit sequence of the signed number -3 is the same bit sequence as the unsigned number 4294967293, which is what node gives us if we simply type -3 >>> 0 into the REPL.
2) (-3 >>> 0).toString
Now, if we call toString on this unsigned number, we will just get the string representation of the bits of the number, which is the same sequence of bits as -3.
What we effectively did was say "hey toString, you have normal behavior when I tell you to print out the bits of an unsigned integer, so since I want to print out a signed integer, I'll just convert it to an unsigned integer, and you print the bits out for me."

Daan's answer explains it well.
toString(2) does not really convert the number to two's complement, instead it just do simple translation of the number to its positive binary form, while preserve the sign of it.
Example
Assume the given input is -15,
1. negative sign will be preserved
2. `15` in binary is 1111, therefore (-15).toString(2) gives output
-1111 (this is not in 2's complement!)
We know that in 2's complement of -15 in 32 bits is
11111111 11111111 11111111 11110001
Therefore in order to get the binary form of (-15), we can actually convert it to unsigned 32 bits integer using the unsigned right shift >>>, before passing it to toString(2) to print out the binary form. This is the reason we do (-15 >>> 0).toString(2) which will give us 11111111111111111111111111110001, the correct binary representation of -15 in 2's complement.