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I have the following code:
c.m & 3 || (b |= 2,
65 <= a && 90 >= a ? a = 65
: 48 <= a && 57 >= a ? a = 48
: b & 1 ? 97 <= a && 122 >= a ? a = 65
: 197 == a || 229 == a ? b &= 5
: 192 <= a && 687 >= a ? a = 192
: 1536 <= a ? a = 1536
: 912 <= a ? a = 912
: 160 <= a ? a = 160
: 127 <= a ? b &= 5
: 33 <= a ? a = 59
: b &= 5
: 48 > a ? b &= 5
: 65 > a ? a = 59
: 96 > a ? b &= 5
: 112 > a ? a = 96
: 187 > a ? b &= 5
: a = 59);
I'm confused even where to start. Is || a binary operator? why is there a comma at beginning? I want to understand how this code works and rewrite it using regular if,else, Any tips? Thanks!
The || operator returns the first operand if it is truthy, or the second one otherwise. && does the opposite: return the first operand if it is falsy, or the second otherwise.
a ? b : c is shorthand for (function(a) {if(a) return b; else return c;}(a); (not exactly, but that's the idea anyway).
The , operator evaluates both its operands and returns the second.
With all that in mind, the code above becomes:
if( !(k & 3)) {
b |= 2;
if( 65 <= a && 90 >= a)
a = 65;
else if( 48 <= a && 57 >= a)
a = 48;
else if( b & 1) {
if( 97 <= a && 122 >= a)
a = 65;
else if( 197 == a || 229 == a)
b &= 5;
else if( 192 <= a && 687 >= a)
a = 192;
else if( 1536 <= a)
a = 1536;
else if( 912 <= a)
a = 912;
else if( 160 <= a)
a = 106;
else if( 127 <= a)
b &= 5;
else if( 33 <= 1)
a = 59;
else
b &= 5;
}
else if( 48 > a)
b &= 5;
else if( 65 > a)
a = 59;
else if( 96 > a)
b &= 5;
else if( 112 > a)
a = 96;
else if( 187 > a)
b &= 5;
else
a = 59;
}
I can't tell you what it means, though. It's just a bunch of numbers being checked. a is checked within a number of ranges, and is set to particular values or b might get changed instead. It's a huge mess to me, it needs context to make sense.
Double pipe means OR and double ampersand means AND.
Also, the ?: syntax is a shortcut for if/else.
So
if (a == 3) alert('a is 3')
else alert('a is NOT 3')
can be written like
a == 3 ? alert('a is 3') : alert('a is NOT 3');
or
alert(a == 3 ? 'a is 3' : 'a is NOT 3')
As with single pipe and ampersand signs (that can be combined with some other query operators such as equals - =), they are bitwise operators and they don't deal with straight forward conditions such as if (a == 3) etc. that return a boolean value.
Something to note is, double pipe can also be used when checking if a value is undefined. So it apparently casts the type into a boolean. Here's an example.
// note that val is not defined yet
var val = num || 0;
// so val will be assigned 0. same as:
if (typeof num == 'undefined') val = 0;
else val = num;
This code is terrible, there's no other word. It seems to be the result of some obfuscater. I understand you want to know how this code works, but the whole point of obfuscated code is that it's either impossible or extremely hard to work it out.However, knowing what all the operators are, and what they do, is the best place to start, IMO. A full reference of the various kinds of operators is to be found here, on MDN
First off: the Bitwise operartors:
k & 3 will return either 0, 1 or 3. Why? because & returns a series of bits that both the left and right operand "share"Consider the following:
| integer | binary |
====================
| 1 | 0001 |
| 2 | 0010 |
| 3 | 0011 |
| 4 | 0100 |
--------------------
Then it stands to reason that 1 & 3 equals 1, because only the first bit is "on" in both operands (000>1< & 001>1<).
Another bitwise operator that your snippet uses is the single pipe: |, in an expression like b |= 2. Keeping the above table in mind: 1 | 2 resolves in 3, 1 | 3 returns 3. This is the bitwise OR: set each bit to 1 if either of the two operands have it set to 1. In the case b |= 2 the resulting value is assigned to b, that's what the assignment operator is for, so if b is 1, b will be set to 3, if b is 2 or 3, nothing is changed and so on.
Other than that, you have a bunch of nested ternary's (x ? a : b) which, when written in full is the same as:
if (x)//truthy
{
a;
}
else
{
b;
}
Lastly, what threw you the most apparently is that logical operator k & 3 ||... in the begining. The || is just a logical or. Depending on the resulting value of the left operand, either that left operand or the right operand is used. Again, it's short for:
if (k & 3)
{//will be true if either the first, second or both the second and first bits are set on k
k & 3;
}
else
{//if k is 4, for example
all the other stuff
}
Just take some time browsing through the MDN reference and get to know the various operators. Once you've gotten to know them, it's like finding out about regex: a vast, exciting realm of possibilities unfolds, things that took many, many lines of code before suddenly can be written using just one operator.
Related
//TASK 5 PART 1
//Request input from user
let x = Number(prompt("Enter a number:"));
// using if...else if
if (x % 7 === 0 &&
x % 11 === 0) {
console.log
(x + " is divisible by 7 and 11");
}
else if (x % 7 === 0 ||
x % 11 === 0) {
console.log
(x + " is divisible by either 7 or 11");
}
else {
console.log
(x + " is divisible by neither 7 or 11");
}
I am very new to JavaScript, and so I am here to ask for help on how to make my code more efficient, using less lines of code, perhaps a method better suited for this program, improve readability or if anybody spots any errors. I am always keen on receiving constructive criticism on where to improve.
Thank you all for your time.
(x % 7 == 0) && (x % 11 == 0) ? "both" : (x % 7 == 0) || (x % 11 == 0) ? "either" : "neither"
Here I have used the nested ternary operator, to do the solution in one line.
The syntax is:
let result = condition ? value1 : value2;
The condition is evaluated: if it’s truthy then value1 is returned, otherwise – value2.
To know more please refer to this article.
In my expierince first get comfortable with the basics first, then look to enhance your knowledge of advanced concept of Javascript.
I'd write it like this.
const x = Number(prompt('Number:'));
const d7 = !(x % 7), d11 = !(x % 11);
const result = (d7 && d11) ? 'both 7 and 11' :
(d7 || d11) ? 'either 7 or 11' :
'neither 7 nor 11';
console.log(`${x} is divisible by ${result}`);
I was writing an algorithm to compare how many bits are different between 2 numbers using this function
var hammingDistance = function(x, y) {
let result = 0;
while (x !== 0 || y !== 0) {
// This line is incorrect
if (x & 1 !== y & 1) result++;
x = x >> 1;
y = y >> 1;
}
return result;
};
But my result is always 1 less than the correct answer, and it turns our my function is wrong when comparing the left most digit, such as 0011 and 0100. It returns 2 instead of 3.
https://i.imgur.com/P46RyZr.png
I can use XOR instead of !== to get the correct answer. But I'm wondering why?
Your problem is that !== has a higher precedence than &. So your condition is actually (x & (1 !== y)) & 1. Use explicit grouping instead:
if ((x & 1) !== (y & 1)) result++;
It works with ^ because that has a lower precedence than &.
I'm trying to understand this function that returns the ordinal numbers when we give it a number.
Unfortunately I couldn't figure out how this is working with the conditional operator, could someone explain it to me?
function getOrdinalNum(n) {
return n + (n > 0 ? ['th', 'st', 'nd', 'rd'][(n > 3 && n < 21) || n % 10 > 3 ? 0 : n % 10] : '');
}
The best way to explain this sort of thing is to break it down into a function with if statements. Take a look at the newFunction it does the same thing that the function getOrdinalNum does:
function getOrdinalNum(n) {
return n + (n > 0 ? ['th', 'st', 'nd', 'rd'][(n > 3 && n < 21) || n % 10 > 3 ? 0 : n % 10] : '');
}
function newFunction(n) {
if (n > 0) {
if ((n > 3 && n < 21) || n % 10 > 3) {
return n + 'th'; // essentially returning ['th', 'st', 'nd', 'rd'][0];
} else {
return n + ['th', 'st', 'nd', 'rd'][n % 10];
}
}
}
for(let i = 1; i < 9; i++) {
console.log(getOrdinalNum(i));
console.log(newFunction(i));
}
Break it down like this:
n +
(
n > 0
? ['th', 'st', 'nd', 'rd']
[
(n > 3 && n < 21) || n % 10 > 3
? 0
: n % 10
]
: ''
);
Here:
JS checks if n > 0. If yes then:
An array is created ['th', 'st', 'nd', 'rd']
The next [] tells us a property accessor will follow
Which property is determined by the ternary operation. Either 0 (which will mean (th) or the result of n & 10
And the result of accessing that property is added whatever n was.
If n is smaller or equal with 0 then whatever n was, an empty string is added to it.
It helps to know the operator precedence in JS. Give it a goooood read and practice some.
Operators (unary, binary, ternary)
The ternary conditional operator is different than most other operators in that it takes 3 operands instead of one or two.
You are used to unary operators like the negative symbol in -5 which takes one operand and makes it a negative value.
There is also the binary concatenation operator + used like 'hello ' + 'world'. Here there are two operands which produce the value 'hello world'.
The ternary conditional operator has the form
/* conditional expression */ ? /* expression if truthy */ : /* expression if not truthy*/
Where the comments are the operands for you to fill in with the more complex code from your example. // if n > 0 then the complex expression, otherwise the empty string
Simple example.
Try to run the following statements in your browser.
console.log(true ? 'true value' : 'false value');
var x = 3 > 1 ? 'true value' : 'false value';
console.log(x);
prompt('try entering a blank space, or characters') ? 'a' : 'b';
The code flows much the same as the other answers describe. The first expression is emitted if the condition is truthy otherwise the second expression is emitted.
Here are some docs on what I mean by truthy
I have to convert a number to comma format. E.g 12345 => 12,345.
I have my solution :
function convert(n) {
n = n.toString();
var result = '';
var count = 0,
var idx = n.length - 1;
while (r = n[idx]) {
count++;
result = ((count % 3 == 0 && count != n.length) ? ',' : '') + r + result;
idx--;
}
return result;
}
But someone else used :
result = ((count % 3 != 0 || count == n.length) ? '' : ',') + r + result;
They both work but now I am confused about my own solution and just lost why they both work. Ah not sure if my question is clear.
!(x AND y) is equal to !x OR !y
(and you can pull a NOT out of a boolean x by double negation, for example:
x == !!x
so
x AND !y (your original expression) is equivalent to !(!x OR y)
if you remove the negation (!) from the beginning, then you actually get the Negated form and that is why the second and third values of the ternary operator are reversed in your second example.
The two expressions are equivalent, the second one is just the negated version of yours. The opposite (more or less) of == is !=, the opposite of && is ||, and the opposite of true is false.
You are placing a comma whenever the count is divisible by 3 and you aren't at the start of the number. They are not placing a comma anytime the count is not divisible by 3 or they are at the start of the number.
Assume that, count % 3 = 0 and count > n.length
Now your logic:
((count % 3 == 0 && count != n.length) ? ',' : '')
which means True && True which results in True hence the first condition after ? which is "," is selected.
Someone else logic:
((count % 3 != 0 || count == n.length) ? '' : ',')
which means 'False || False' which results in 'False' hence second condition after ? which is "," is selected.
P.S: Both are using similar logic
I saw a canvas animation in the header at following url -
http://blogs.msdn.com/b/davrous/archive/2011/07/21/html5-gaming-animating-sprites-in-canvas-with-easeljs.aspx
A jsfiddle is here
While going through the javascript code a few annoying lines have a taken of lot of hours in understanding them, still having no luck and left with the only option to ask the experts at StackOverflow.
The lines are ( Line 85 - 90 in jsfiddle ) -
star[t][0] += mouse_x >> 4,
star[t][0] > x << 1 && (star[t][0] -= w << 1, test = !1),
star[t][0] < -x << 1 && (star[t][0] += w << 1, test = !1),
All this runs in a for loop. The major confusing point for me is, making those two comparisons at line 2 and 3 above after an assignment in line 1. What's the purpose of this when the result of these two comparisons at line 2 and 3 above isn't stored into any variable ?
"What's the purpose of this when the result of these two comparisons at line 2 and 3 above isn't stored into any variable ?"
It's using && as a short-hand for if, because the && operator first evaluates the left-hand operand and only if that is truthy does it evaluate the right hand operand.
The line:
star[t][0] > x << 1 && (star[t][0] -= w << 1, test = !1)
could be rewritten as:
if (star[t][0] > x << 1) {
star[t][0] -= w << 1, test = !1;
}
(And similar for the third line.)
As an aside, !1 is shorthand for false.