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How to store a 32-bit integer in an arraybuffer?

I seem to not be understanding the Uint32Array. According to what I've read about the Uint8Array I could just feed it a number inside an array (Uint8Array([16]) or Uint8Array([96,56])) and the results are exactly that. However, when I try the same thing for a larger number like Uint32Array([21640]), it seems to truncate it. Where 21640 should equal 5488 in hex, I only get 88. How does this actually work?
Edit: Elaborations
I am also attempting to concatenate several ArrayBuffers together. If I'm not mistaken readAsArrayBuffer produces an Uint8Array, and I am trying to append to that some 32-bit numbers using https://gist.github.com/72lions/4528834
There is so much information and examples on Uint8Array and what little there was on Uint32Array makes me think that one of these 32 would store a value as if it was 4 of the 8.

The largest value of an unsigned 8 bit number is 255. Larger numbers will be truncated or rolled over depending on the os/cpu. If you want to convert a 32 bit numbers in an 8 bit array try something like this.
var number = 21640;
var byte1 = 0xff & number;
var byte2 = 0xff & (number >> 8);
var byte3 = 0xff & (number >> 16);
var byte4 = 0xff & (number >> 24);
var arr1 = Uint8Array([byte1,byte2,byte3,byte4]);
Just reverse the order of the bytes when you create the array depending on if you want little or big endian.

Here is a working example showing 5488 in console
var bigNumber = new Uint32Array([21640]);
console.log(bigNumber[0].toString(16));
Since you've added more to the question. If you wanted to convert
var byte1 = 0x88;
var byte2 = 0x54;
var byte3 = 0;
var byte4 = 0;
var bigValue = (byte4 << 24) | (byte3 << 16) | (byte2 << 8) | (byte1);
console.log(bigValue);
Although you will need to factor in Endianness

Convert Bits to Bytes in Node-Red

I'm new to JavaScript and I use Node-Red to read an write from a Database.
I receive from the database an object that contains the status of 8 digital inputs.
Each inputs is represented as a bit.
I'm looking for a method to combine each bits into a byte.
This is the object that I receive from the database:
array[1]
0: object
idx: 10
ts: "2018-11-21T06:12:45.000Z"
in_0: 1
in_1: 1
in_2: 1
in_3: 1
in_4: 1
in_5: 1
in_6: 1
in_7: 1
in_x represent the input position.
As out I would like to receive a byte that represent the combination of each single byte.
For example:
in0: 0,
in1: 1,
in2: 0,
in3: 0,
in4: 0,
in5: 1,
in6: 0,
in7: 0,
The output byte will be: 00100001 in binary that converted to byte is 33
Any suggestions?
Thanks in advance.

The following code works as you requested*:
var output =
arr[0].in_0 +
(arr[0].in_1 << 1) +
(arr[0].in_2 << 2) +
(arr[0].in_3 << 3) +
(arr[0].in_4 << 4) +
(arr[0].in_5 << 5) +
(arr[0].in_6 << 6) +
(arr[0].in_7 << 7);
This code assumes that each variable can only be a 1 or a 0. Anything else will result in nonsense.
I have used the Left Bit Shift operator (<<) to obtain the power of two for each on bit.
You have specified that in_7 is the Most Significant Bit. If it is actually the Least Significant Bit, reverse the order of the in_x variables.
*The result is not a byte, but it does contain the number that I think you're expecting.

Is it possible to pack two integers to a hexadecimal and get them back

I have a weird requirement,
My destination only supports one integer, But I want to send two integers to it and later I want to get them back from a response.
for example,
allowed input:
{
'task': 2
}
I have subtask kind of a logic in my side, But my target is not aware of this. So, without letting know the target, can I somehow pack two integers and get decode them back in future?
Can this be achieved with hexadecimal?

You can combine any two numbers and get both numbers back using their product (a * b) as long as a * (a * b) + b < Number.MAX_SAFE_INTEGER
Here's a demo snippet:
(() => {
document.addEventListener("click", handleStuff);
// formula: c = (a * (a * b)) + b
// as long as c < 9007199254740991
const combine = (a, b) => ({
a: a,
b: b,
get c() { return this.a * this.b; },
get combined() { return this.a * this.c + this.b; },
get unraveled() { return [
Math.floor(this.combined / this.c),
this.combined % this.c ]; }
});
const log = txt => document.querySelector("pre").textContent = txt;
let numbers = combine(
+document.querySelector("#num1").value,
+document.querySelector("#num2").value );
function handleStuff(evt) {
if (evt.target.nodeName.toLowerCase() === "button") {
if (evt.target.id === "combine") {
numbers = combine(
+document.querySelector("#num1").value,
+document.querySelector("#num2").value );
if (numbers.combined > Number.MAX_SAFE_INTEGER) {
log (`${numbers.combined} too large, unraveled will be unreliable`);
} else {
log (`Combined ${numbers.a} and ${numbers.b} to ${numbers.combined}`);
}
} else {
log(`${numbers.combined} unraveled to ${numbers.unraveled}`);
}
}
}
})();
input[type=number] {width: 100px;}
<p>
<input type="number" id="num1"
value="12315" min="1"> first number
</p>
<p>
<input type="number" id="num2"
value="231091" min="1"> second number
</p>
<p>
<button id="combine">combine</button>
<button id="unravel">unravel</button>
</p>
<pre id="result"></pre>
Note: #RallFriedl inspired this answer
JSFiddle

Yes, you can, assuming your two integers don't contain more information than the one integer can handle.
Let's assume your tasks and sub tasks are in the range 1..255. Then you can encode
combined = (task * 256) + subtask
And decode
task = combined / 256
subtask = combined % 256

At first, you don't have to convert an integer to hexadecimal to do this. An integer is a value and decimal, hexadecimal or binary is a representation to visualize that value. So all you need is integer arithmetics to achieve your goal.
According to this answer the maximum allowed integer number in javascript would be 9007199254740991. If you write this down in binary you'll get 53 ones, which means there are 53 bits available to store within an integer. Now you can split up this into two or more smaller ranges as you need.
For example let's say you need to save three numbers, the first is always lower 4.294.967.296 (32-bit), the second always lower 65.536 (16-bit) and the third always lower 32 (5-bit). If you sum up all the bits of these three values, you'll get 53 bits which means it would perfectly match.
To pack all these values into one, all you need is to move them at the right bit position within the integer. In my example I'd like to let the 32 bit number on the lowest position, then the 16 bit number and at the highest position the 5 bit number:
var max32bitValue = 3832905829; // binary: 1110 0100 0111 0101 1000 0000 0110 0101
var max16bitValue = 47313; // binary: 1011 1000 1101 0001
var max5bitValue = 17; // binary: 1000 1
var packedValue = max32bitValue // Position is at bit 0, so no movement needed.
+ max16bitValue << 32 // Move it up next to the first number.
+ max5bitValue << 48; // Move it up next to the second number (32 + 16)
This single integer value can now be stored, cause is a perfectly valid javascript integer value, but for us it holds three values.
To get all three values out of the packed value, we have to pick the correct bits out of it. This involves two steps, first remove all unneeded bits on the lower side (by using shift right), then remove all unneeded bits on the higher side (by masking out):
var max32bitValueRead = packedValue & Math.pow(2, 32); // No bits on the lower side, just mask the higher ones;
var max16bitValueRead = (packedValue >> 32) & Math.pow(2, 16); // Remove first 32 bits and set all bits higher then 16 bits to zero;
var max5bitValueRead = (packedValue >> 48); // Remove first 48 bits (32 + 16). No higher bits there, so no mask needed.
So hope this helps to understand, how to put multiple integer values into one, if the ranges of these values don't exceed the maximum bit range. Depending on your needs you could put two values with 26 bits each into this or move the range like one 32 bit value and one 21 bit value or a 48 bit value and a 5 bit value. Just be sure what your maximum value for each one could be and set the width accordingly (maybe add one to three bits, just to be sure).

I wouldn't suggest using hexadecimal if you can not have 2 sequential numbers. Try converting to an ASCII character and then back. So if you wanted to send:
{ 'task': 21 }
You could set the 21 to a character like:
var a = 55; var b = String.fromCharCode(a); var send2 = { 'task': b };
And to convert it back it would be: var res = { 'task': b }; var original = res.task.charCodeAt();

Node.js with Geddy: Does geddy.string.uuid(x) make sure the string is unique?

I am following a tutorial on making a simple app with Node.js and Geddy. For each new entry it sets the entry id with geddy.string.uuid(10) and doesn't seem to run any other checks. My concern is when creating a new entry this could generate the same ID as a previous entry and the save script in the tutorial will overwrite/update an entry if the id already exists.
I know this would be a rare occurrence if it could happen, but I feel it is not responsible to leave this to chance if it can. I know in PHP I would be running a check of some sort to make sure it is unique, and just wanted to know if maybe Geddy was doing this already for me being the tutorial seems to not worry about it?
Thank you.

There is no real chance of a uuid occurring twice.
Technically, there is a one in 340,282,366,920,938,463,463,374,607,431,768,211,456 (16 to the 32nd power) chance that there will be a duplicate. Geddy's uuid function (like most others) uses 32 hex digits (128 bits) of random bits. See http://en.wikipedia.org/wiki/Universally_unique_identifier#Random_UUID_probability_of_duplicates.
Basically, you're not going to get a duplicate.
As a side note, it's interesting that you're passing 10 uuid here: geddy.string.uuid(10). In the latest version of getty string.js, uuid does not take an argument, so the 10 will be ignored.
148 this.uuid = function () {
149 var chars = _UUID_CHARS, uuid = [], rnd=0, r;
150 for (var i = 0; i < 36; i++) {
151 if (i==8 || i==13 || i==18 || i==23) {
152 uuid[i] = '-';
153 } else {
154 if (rnd <= 0x02) rnd = 0x2000000 + (Math.random()*0x1000000)|0;
155 r = rnd & 0xf;
156 rnd = rnd >> 4;
157 uuid[i] = chars[(i == 19) ? (r & 0x3) | 0x8 : r];
158 }
159 }
160 return uuid.join('');
161 };
If the 10 in your version of geddy is limiting the number of hex digits to 10, then you only have a 1 in 1,099,511,627,776 (1 trillion) chance of a collision. That's still quite unlikely, though it could be brute forced to find a duplicate id.

How do I swap endian-ness (byte order) of a variable in javascript

I am receiving and sending a decimal representation of two little endian numbers. I would like to:
shift one variable 8 bits left
OR them
shift a variable number of bits
create 2 8 bit numbers representing the first and second half of the 16 bit number.
javascript (according to https://developer.mozilla.org/en/JavaScript/Reference/Operators/Bitwise_Operators) uses big endian representation when shifting...
endianness is a bit foreign to me (I am only 90 percent sure that my outlined steps are what i want.) so swapping is a bit dizzying. please help! I only really need to know how to swap the order in an efficient manner. (I can only think of using a for loop on a toString() return value)

function swap16(val) {
return ((val & 0xFF) << 8)
| ((val >> 8) & 0xFF);
}
Explanation:
Let's say that val is, for example, 0xAABB.
Mask val to get the LSB by &ing with 0xFF: result is 0xBB.
Shift that result 8 bits to the left: result is 0xBB00.
Shift val 8 bits to the right: result is 0xAA (the LSB has "dropped off" the right-hand side).
Mask that result to get the LSB by &ing with 0xFF: result is 0xAA.
Combine the results from steps 3 and step 5 by |ing them together:
0xBB00 | 0xAA is 0xBBAA.
function swap32(val) {
return ((val & 0xFF) << 24)
| ((val & 0xFF00) << 8)
| ((val >> 8) & 0xFF00)
| ((val >> 24) & 0xFF);
}
Explanation:
Let's say that val is, for example, 0xAABBCCDD.
Mask val to get the LSB by &ing with 0xFF: result is 0xDD.
Shift that result 24 bits to the left: result is 0xDD000000.
Mask val to get the second byte by &ing with 0xFF00: result is 0xCC00.
Shift that result 8 bits to the left: result is 0xCC0000.
Shift val 8 bits to the right: result is 0xAABBCC (the LSB has "dropped off" the right-hand side).
Mask that result to get the second byte by &ing with 0xFF00: result is 0xBB00.
Shift val 24 bits to the right: result is 0xAA (everything except the MSB has "dropped off" the right-hand side).
Mask that result to get the LSB by &ing with 0xFF: result is 0xAA.
Combine the results from steps 3, 5, 7 and 9 by |ing them together:
0xDD000000 | 0xCC0000 | 0xBB00 | 0xAA is 0xDDCCBBAA.

Such function can be used to change endianness in js:
const changeEndianness = (string) => {
const result = [];
let len = string.length - 2;
while (len >= 0) {
result.push(string.substr(len, 2));
len -= 2;
}
return result.join('');
}
changeEndianness('AA00FF1234'); /// '3412FF00AA'

Use the << (bit shift) operator. Ex: 1 << 2 == 4.
I really think that the underlying implementation of JavaScript will use whatever endianess the platform it is running on is using. Since you cannot directly access memory in JavaScript you won't ever have to worry about how numbers are represented physically in memory. Bit shifting integer values always yield the same result no matter the endianess. You only see a difference when looking at individual bytes in memory using pointers.

Here is a oneliner for arrays to swap between big and little endian (and vise versa). The swapping is done using reverse on byte level. I guess for large arrays, it is more efficient than looping over scalar swap function.
function swapbyte(x) {
return new Float64Array(new Int8Array(x.buffer).reverse().buffer).reverse()
}
// Example
buf = new ArrayBuffer(16); // for 2 float64 numbers
enBig = new Float64Array(buf);
enBig[0] = 3.2073756306779606e-192;
enBig[1] = 2.7604354232023903e+199;
enLittle = swapbyte(enBig)
// two famous numbers are revealed
console.log(enLittle)
// Float64Array [ 6.283185307179586, 2.718281828459045 ]
// swapping again yields the original input
console.log(swapbyte(enLittle))
// Float64Array [ 3.2073756306779606e-192, 2.7604354232023903e+199 ]