Bitwise operations but what is the meaning (Espruino) - javascript

What is the meaning of this line of code
n = (n<<1) | ((d>=0.0004)?1:0);
Trying to understand code from here in function sigOff()
http://www.espruino.com/Remote+Control+Sockets

This snippet seems to use the bitwise OR (|) and left shift (<<) operators:
Bitwise OR: a | b;
Returns a one in each bit position for which the corresponding bits of either or both operands are ones.
Left shift: a << b;
Shifts a in binary representation b (< 32) bits to the left, shifting in zeros from the right.
The left shift by 1 (<< 1) basically doubles the value of n.
Then, the or (|) basically "adds" 1 to the result to make it uneven, if d >= 0.0004.
If d < 0.0004, the result from the left shift isn't changed.
So, for n == 3 and d == 0.0004, this happens:
n << 1 // 6
(d>=0.0004)?1:0 // 1
6 | 1 // 7
For n == 5 and d == 0.0002, this happens:
n << 1 // 10
(d>=0.0004)?1:0 // 0
10 | 0 // 10

Related

What does OR 0x80 do?

In Bitwise operation, what does | 0x80 do? I know (& 0xFF) is convert value to 8 bit integer but how about (| 0x80) ?
I have the following code:
const buf = createHash('sha256').update('test').digest()
for (let i = 0; i < n; i++) {
const ubyte = buf.readUInt8(i)
const shifted = (ubyte >> 1) | mask
destBuf.writeUInt8(shifted, i)
mask = (ubyte & 1) * 0x80 // mask is 0 or 128
}
Can anyone explain that for me?
0x... means that what comes next is an hexadecimal number.
0x80 is the hexadecimal representation of the number 128. In binary, this equals 10000000.
The | character is the bitwise or operator. Suppose you have a 8-bit number:
a = xxxxxxxx
with x being either a 0 or a 1. Now, masking this number with 0x80 means:
xxxxxxxx | 10000000 = 1xxxxxxx
So it basically means you will have a 1 for your leftmost significant bit, while keeping all the other bits the same.
Now, in your code you use this mask in the line:
const shifted = (ubyte >> 1) | mask
What this does is takes the number ubyte:
ubyte = xxxxxxxy // x and y can be either 1 or 0
It shifts it right by onw digit:
ubyte >> 1 = zxxxxxxx // y gets lost, and z is a 0 if ubyte was unsigned.
Now it masks this number with your mask. When the mask is 128, the result is:
(ubyte >> 1) | 10000000 = 1xxxxxxx
So you will have a 1 as your most significant bit, and the other bits are unchanged.
It combines the bits of both participating numbers with the logical "or":
const b= 0x7;
function tst(a,b){
console.log(a.toString(2).padStart(8)+" first number: "+a)
console.log(b.toString(2).padStart(8)+" second number: "+b)
console.log((a | b).toString(2).padStart(8)+" bitwise overlay: "+(a|b))
console.log("----")
}
[0x80,0x6A,0x70,0x8f].forEach(a=>tst(a,b))

How to add bits in JavaScript

Say you have two integers 10 and 20. That is 00001010 and 00010100. I would then like to just basically concat these as strings, but have the result be a new integer.
00001010 + 00010100 == 0000101000010100
That final number is 2580.
However, I am looking for a way to do this without actually converting them to string. Looking for something more efficient that just does some bit twiddling on the integers themselves. I'm not too familiar with that, but I imagine it would be along the lines of:
var a = 00001010 // == 10
var b = 00010100 // == 20
var c = a << b // == 2580
Note, I would like for this to work with any sequences of bits. So even:
var a = 010101
var b = 01110
var c = a + b == 01010101110
You basic equation is:
c = b + (a << 8).
The trick here is that you need to always shift by 8. But since a and b do not always use all 8 bits in the byte, JavaScript will automatically omit any leading zeros. We need to recover the number of leading zeros (of b), or trailing zeros of a, and prepend them back before adding. This way, all the bits stay in their proper position. This requires an equation like this:
c = b + (a << s + r)
Where s is the highest set bit (going from right to left) in b, and r is the remaining number of bits such that s + r = 8.
Essentially, all you are doing is shifting the first operand a over by 8 bits, to effectively add trailing zeros to a or equally speaking, padding leading zeros to the second operand b. Then you add normally. This can be accomplishing using logarithms, and shifting, and bitwise OR operation to provide an O(1) solution for some arbitrary positive integers a and b where the number of bits in a and b do not exceed some positive integer n. In the case of a byte, n = 8.
// Bitwise log base 2 in O(1) time
function log2(n) {
// Check if n > 0
let bits = 0;
if (n > 0xffff) {
n >>= 16;
bits = 0x10;
}
if (n > 0xff) {
n >>= 8;
bits |= 0x8;
}
if (n > 0xf) {
n >>= 4;
bits |= 0x4;
}
if (n > 0x3) {
n >>= 2;
bits |= 0x2;
}
if (n > 0x1) {
bits |= 0x1;
}
return bits;
}
// Computes the max set bit
// counting from the right to left starting
// at 0. For 20 (10100) we get bit # 4.
function msb(n) {
n |= n >> 1;
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
n = n + 1;
// We take the log here because
// n would otherwise be the largest
// magnitude of base 2. So, for 20,
// n+1 would be 16. Which, to
// find the number of bits to shift, we must
// take the log base 2
return log2(n >> 1);
}
// Operands
let a = 0b00001010 // 10
let b = 0b00010100 // 20
// Max number of bits in
// in binary number
let n = 8
// Max set bit is the 16 bit, which is in position
// 4. We will need to pad 4 more zeros
let s = msb(b)
// How many zeros to pad on the left
// 8 - 4 = 4
let r = Math.abs(n - s)
// Shift a over by the computed
// number of bits including padded zeros
let c = b + (a << s + r)
console.log(c)
Output:
2580
Notes:
This is NOT commutative.
Add error checking to log2() for negative numbers, and other edge cases.
References:
https://www.geeksforgeeks.org/find-significant-set-bit-number/
https://github.com/N02870941/java_data_structures/blob/master/src/main/java/util/misc/Mathematics.java
so the problem:
a is 10 (in binary 0000 1010)
b is 20 (in binary 0100 0100)
you want to get 2580 using bit shift somehow.
if you right shift a by 8 using a<<=8 (this is the same as multiplying a by 2^8) you get 1010 0000 0000 which is the same as 10*2^8 = 2560. since the lower bits of a are all 0's (when you use << it fills the new bits with 0) you can just add b on top of it 1010 0000 0000 + 0100 0100 gives you 1010 0001 0100.
so in 1 line of code, it's var result = a<<8 + b. Remember in programming languages, most of them have no explicit built-in types for "binary". But everything is binary in its nature. so int is a "binary", an object is "binary" ....etc. When you want to do some binary operations on some data you can just use the datatype you have as operands for binary operations.
this is a more general version of how to concatenate two numbers' binary representations using no string operations and data
/*
This function concate b to the end of a and put 0's in between them.
b will be treated starting with it's first 1 as its most significant bit
b needs to be bigger than 0, otherwise, Math.log2 will give -Infinity for 0 and NaN for negative b
padding is the number of 0's to add at the end of a
*/
function concate_bits(a, b, padding) {
//add the padding 0's to a
a <<= padding;
//this gets the largest power of 2
var power_of_2 = Math.floor(Math.log2(b));
var power_of_2_value;
while (power_of_2 >= 0) {
power_of_2_value = 2 ** power_of_2;
a <<= 1;
if (b >= power_of_2_value) {
a += 1;
b -= power_of_2_value;
}
power_of_2--;
}
return a;
}
//this will print 2580 as the result
let result = concate_bits(10, 20, 3);
console.log(result);
Note, I would like for this to work with any sequences of bits. So even:
var a = 010101
var b = 01110
var c = a + b == 01010101110
This isn't going to be possible unless you convert to a string or otherwise store the number of bits in each number. 10101 010101 0010101 etc are all the same number (21), and once this is converted to a number, there is no way to tell how many leading zeroes the number originally had.

Why does var bug = 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 equal 15 [duplicate]

console.log(0.5 | 0); // 0
console.log(-1 | 0); // -1
console.log(1 | 0); // 1
Why does 0.5 | 0 return zero, but any integer (including negative) returns the input integer? What does the single pipe ("|") do?
This is a bitwise or.
Since bitwise operations only make sense on integers, 0.5 is truncated.
x | 0 is x, if x is an integer.
Bit comparison is so simple it's almost incomprehensible ;) Check out this "nybble"
8 4 2 1
-------
0 1 1 0 = 6 (4 + 2)
1 0 1 0 = 10 (8 + 2)
=======
1 1 1 0 = 14 (8 + 4 + 2)
Bitwise ORing 6 and 10 will give you 14:
alert(6 | 10); // should show 14
Terribly confusing!
A single pipe is a bit-wise OR.
Performs the OR operation on each pair
of bits. a OR b yields 1 if either a
or b is 1.
JavaScript truncates any non-integer numbers in bitwise operations, so its computed as 0|0, which is 0.
This example will help you.
var testPipe = function(input) {
console.log('input => ' + input);
console.log('single pipe | => ' + (input | 'fallback'));
console.log('double pipe || => ' + (input || 'fallback'));
console.log('-------------------------');
};
testPipe();
testPipe('something');
testPipe(50);
testPipe(0);
testPipe(-1);
testPipe(true);
testPipe(false);
This is a Bitwsie OR (|).
The operands are converted to 32-bit integers and expressed by a series of bits (zeroes and ones). Numbers with more than 32 bits get their most significant bits discarded.
So, in our case decimal number is converted to interger 0.5 to 0.
= 0.5 | 0
= 0 | 0
= 0

What kind of checksum is this function calculating?

This is a function where bytes comes from a string, and each letter's char code was grabbed to create an array of bytyes
function checksum(bytes) {
var a = 0;
var b = 0;
for (var i = 0; i < bytes.length; i++) {
a = (a + bytes[i]) % 0xff;
b = (b + a) % 0xff;
}
return (b << 8) | a;
}
I'm mostly confused at the role of % 0xff and b << 8. Could someone help me break this down?
a is equal to the sum of all the elements modulo 255
b is equal to the sum of all the values that a assumes modulo 255 (so 1 time the last element + 2 times the one before that ....)
The final value is a 16 bit number where the higher 8 bits are b and the lower 8 bits are a.
That is actually a Fletcher's checksum
https://en.wikipedia.org/wiki/Fletcher%27s_checksum
This function calculates a and b, and generates the checksum from it.
a is calculated as follows:
0+the value of the current char
a must be between 0 and 254, hence modulo 255 is applied
b is calculated as the cumulative value of a modulo 255
increase the value of b by a
b must be between 0 and 254, hence modulo 255 is applied
at the end the checksum is generated by concatenating a to b
take b's bits and move them 8 places to the left
set a to the right side of b
The result will have a length of 2 bytes (16 bits), where the first byte is b and the second a.
Example:
(c is the current char value)
c | a |b
----------------
8 |8 |8
13 |21 |29
5 |26 |55
0 |26 |81
180 |206 |287
100 |51 |83
cheksum = 51,83 = 0x3353 = 0011 0011 0101 0011

What does the "|" (single pipe) do in JavaScript?

console.log(0.5 | 0); // 0
console.log(-1 | 0); // -1
console.log(1 | 0); // 1
Why does 0.5 | 0 return zero, but any integer (including negative) returns the input integer? What does the single pipe ("|") do?
This is a bitwise or.
Since bitwise operations only make sense on integers, 0.5 is truncated.
x | 0 is x, if x is an integer.
Bit comparison is so simple it's almost incomprehensible ;) Check out this "nybble"
8 4 2 1
-------
0 1 1 0 = 6 (4 + 2)
1 0 1 0 = 10 (8 + 2)
=======
1 1 1 0 = 14 (8 + 4 + 2)
Bitwise ORing 6 and 10 will give you 14:
alert(6 | 10); // should show 14
Terribly confusing!
A single pipe is a bit-wise OR.
Performs the OR operation on each pair
of bits. a OR b yields 1 if either a
or b is 1.
JavaScript truncates any non-integer numbers in bitwise operations, so its computed as 0|0, which is 0.
This example will help you.
var testPipe = function(input) {
console.log('input => ' + input);
console.log('single pipe | => ' + (input | 'fallback'));
console.log('double pipe || => ' + (input || 'fallback'));
console.log('-------------------------');
};
testPipe();
testPipe('something');
testPipe(50);
testPipe(0);
testPipe(-1);
testPipe(true);
testPipe(false);
This is a Bitwsie OR (|).
The operands are converted to 32-bit integers and expressed by a series of bits (zeroes and ones). Numbers with more than 32 bits get their most significant bits discarded.
So, in our case decimal number is converted to interger 0.5 to 0.
= 0.5 | 0
= 0 | 0
= 0

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