So I wondered to myself if there is a way to do a double greater than, like this:
if(x > y > z) { ... }
Then I saw this
Expression for "more than x and less than y"?
But then I tried the following expression in the console and got a weird result:
(5 < 2 < 1) // returned true
(5 > 2 > 1) // returned false
How?
Update: I know that you can't do this "(x > y > z)", just wanted explanation on the weird result.
You need two separate conditions, such as 5<2 && 2<1 for this to do what you're after. Without two conditions, you are comparing the result of the first condition against the right side of the second condition.
Why?
For the unexplained behaviour, I believe that the explanation for why it's returning what it's returning is the way javascript handles booleans (among other things) when used in numerical operations, in other words, javascript will cast your booleans to 0 or 1, there are a lot of examples of this in a few questions here at so, for example this one, you could do +false and get a 0 for instance.
(5 < 2 < 1) is true because:
5<2 is resolved first, returning false
false is cast to a number, returning 0
0<1 will return true
(5 > 2 > 1) is false because:
5>2 is resolved first, will return true
true is cast to a number, will return 1
1>1 will return false
No. You can't do this in JavaScript. Much as it would be useful, you're stuck with x > y && y > z.
That said, to explain your result:
5 < 2 gives false
false casts to 0
0 < 1 is true
And:
5 > 2 gives true
true casts to 1
1 > 1 is false
As far as a "way to do a double greater than", you could define a function:
function gt () {
var prev = arguments[0];
return !([].slice.call(arguments, 1).some(function (arg) {
var ret = prev <= arg;
prev = arg;
return ret;
}));
}
...then call like this:
if(gt(x, y, z)) { ... }
I would just try if((5>2) && (2>1)) {
} not sure why the other one didnt work but thus implementation should be a proper substitute for code that could be written in the other way you have it.
Related
I was going through some Javascript code used to validate form entry and I noticed an if statement that read if (!(x < y && y > x)) {...}
My initial thought was that this sort of tautological construction is completely redundant, and one of the two comparisons should be dropped. Just in case there's a chance I'm wrong and there's actually more to it, though, I thought I'd ask.
My other thought was that it might be a case of some idiom that is necessary in another language, which the programmer has here just brought along with them to Javascript out of habit (although I'd be surprised to find a language where something like this is required that is also used in any sort of environment).
EDIT
The specific code is in a function to test that the start and end dates for a submitted event are possible (i.e. the end date is after the start date). The actual example reads if(!(start_time < end_time && end_time > start_time)) {...} where both start_time and end_time are DateTime values.
EDIT 2
Not a duplicate of this question as in this case the issue is the need to test two conditions that appear mutually inclusive in an if statement, whilst in that case the issue is how to make an if statement resolve that appears to require two mutually exclusive conditions to be simultaneously true.
It looks like a pattern to allow falsy values, which convert to a number (without NaN, like '' or null).
function f(x, y) {
return [!(x > y && y < x), x <= y, x < y || x === y].join(' ');
}
console.log(f(1, 2)); // true
console.log(f(2, 1)); // false
console.log(f(1, 1)); // true
console.log(f('', 1)); // true
console.log(f(1, '')); // false
console.log(f('', '')); // true
console.log(f(undefined, undefined)); // true \
console.log(f(1, undefined)); // true different values by using other comparisons
console.log(f(undefined, 1)); // true /
console.log(f(null, null)); // true
console.log(f(1, null)); // false
console.log(f(null, 1)); // true
.as-console-wrapper { max-height: 100% !important; top: 0; }
I had been going through the ES6 assuming that it would be easy to switch to EcmaScript 2017.
While going through, I got confused about this code
function f (x, y = 7, z = 42) {
return x + y + z
}
f(1) === 50
Which has ES5 equivalent
function f (x, y, z) {
if (y === undefined)
y = 7;
if (z === undefined)
z = 42;
return x + y + z;
};
f(1) === 50;
I did understand the default parameter thing from it.
But what does f(1)===50 mean in both the codes? Whats the use of it?
Here is another example
function f (x, y, ...a) {
return (x + y) * a.length
}
f(1, 2, "hello", true, 7) === 9
What does f(1, 2, "hello", true, 7) === 9 mean?
I understand that === for comparison between the LHS and RHS of the operator including the type of both and not just value.
But why have it been used like that??
Kindly explain its usage.
This is the link from where I got this. http://es6-features.org/#RestParameter
This is a strict comparison test whether function f(x,y,z), when called with an x parameter value of 1 returns the value 50. This would be true when the default parameter values added to the value of x are 7 and 42.
These function calls and comparisons are purely to provide usage examples and possibly test examples for the functions they call.
The code
function f (x, y, ...a) {
return (x + y) * a.length
}
f(1, 2, "hello", true, 7) === 9
is an example of extended parameter handling. the ...a variables length property equates to 3 thus the test is confirming the number of parameters passed to the function after x and y.
According to me, you almost took it in a correct way.
Just put that function call along with the triple equals sign in if condition.
if ( f(1) === 50 ){
console.log(true);
}
else {
console.log(false);
}
Thats it.
The triple equal to is simply a comparison operator. And the function call on one side of the triple equal to operator means the value returned from that function.
Hence just treat it as any other comparison operator in javascript.
And please correct me if I have misinterpreted your question.!
All the best!
The f(1)===50 checks if f(1) is equal to 50. If this expression is evaluated to true, then the result of this expression is true. Otherwise it is false. Since you don't assign this value to a variable, as it is, you can't use it anywhere.
Formally, the === is called strict equality operator. For further info please have a look here.
The point here is that its example code. They are showing you that the function's results when called with those arguments is equal to something. The expression itself won't do anything, unless you paste it into a console.
They could have just as easily used a comment.
f(1, 2, "hello", true, 7) // 9
The identity (===) operator behaves identically to the equality (==) operator except no type conversion is done, and the types must be the same to be considered equal.
In your example if you will put all three arguments of numeric type you will get number as the result and then to check if the result has correct type you have to use === operator.
Maybe this example will be more clear in your case:
f(1,1,1) // returns 3 - numeric type
f(1,1,"1") // returns "111" - string type
//so now if you will write
f(1,1,1) == "3" // true
f(1,1,1) == 3 // true
f(1,1,1) === "3" // false, because the result is 3 not "3" as string.
f(1,1,1) === 3 // true
f(1,1,"1") == "111" // true
f(1,1,"1") == 111 // true
f(1,1,"1") === "111" // true
f(1,1,"1") === 111 // false, because the result is string "111" not 111 number.
So in your case this === operator is used to double check if the result is really what you expect it to be.
Much apologies for the vague title, but I need to elaborate. Here is the code in question, which I read on http://ariya.ofilabs.com/2013/07/prime-numbers-factorial-and-fibonacci-series-with-javascript-array.html:
function isPrime(i) {
return (i > 1) && Array.apply(0, Array(1 + ~~Math.sqrt(i))).
every(function (x, y) {
console.log(x + ' ' + i % y);
return (y < 2) || (i % y !== 0)
});
}
isPrime(23);
isPrime(19);
isPrime(188);
Just for fun, I added those logs so we can see some output:
undefined NaN
undefined 0
undefined 1
undefined 2
undefined 3
undefined NaN
undefined 0
undefined 1
undefined 1
undefined 3
undefined NaN
undefined 0
undefined 0
This is the first time I have every seen apply and every, so bear with me, but my understanding is that apply basically calls the Array function, where the first argument is the substitution for its this and the second is the output...Never would think that would be useful, but this function seems to work, so...
Here, they seem to be creating an array of length equal to the square root of the number in question. I suppose that makes sense because the square root would be the largest possible factor of the number in question.
OK, so from here, if we were to log that array for, say, the first number, it would look like this:
> var i = 23;
undefined
> Array.apply(0, Array(1 + ~~Math.sqrt(i)));
[ undefined, undefined, undefined, undefined, undefined ]
Great, so it is an array of five undefined. Ok, fine, so from here, the every method is supposed to check whether every element in that array passes the callback function test (or whatever).
The Microsoft documentation specifies three possible arguments for the every method:
value
index
array
Therefore, in this example x is the value, i.e. undefined, and y is the index.
Our output agrees with that conclusion. However, I'm still fuzzy about nested return statements (if the lowest one returns, does its parent also return?), the || operator here (if the first test passes, does the every loop stop?), and just generally how this works.
EDIT
the log should be with an x, not a y. my mistake:
console.log(y + ' ' + i % y); -> console.log(x + ' ' + i % y);
EXPLANATION
So, how did I come across this code, you ask? Well, of course, the simplest way to check for a prime in Java would be like this:
public static boolean isPrime(double num) {
for (double i = 2.0; i < sqrt(num); i++) {
if (num % i == 0.0) {
return true;
}
}
return false;
}
or Python
def isPrime(num):
x = 2
isPrime = True
while x < math.sqrt(num):
if num % x == 0:
isPrime = False
break
x = x + 1
return isPrime
or js
function isPrime(n) {
for (var i = 2.0; i < Math.sqrt(n); i++) {
if (n % i === 0.0) {
return false;
}
}
return true;
}
But say I wanted to check for the largest prime factor of a number like 600851475143 These looping methods would take too long, right? I think this "hack", as we are describing it, may be even less efficient, because it is using arrays instead of integers or floats, but even still, I was just looking for a more efficient way to solve that problem.
The code in that post is basically crap. Teaching people to write code while simultaneously using hacks is garbage. Yes, hacks have their place (optimization), but educators should demonstrate solutions that don't depend on them.
Hack 1
// the 0 isn't even relevant here. it should be null
Array.apply(0, Array(1 + ...))
Hack 2
// This is just Math.floor(x), but trying to be clever
~~x
Hack 3
// this is an outright sin; totally unreadable code
// I bet most people don't know the binding precedence of % over +
y + ' ' + i % y
// this is evaluated as
y + ' ' + (i % y)
// example
2 + ' ' + (5 % 2) //=> "2 1"
I'm still fuzzy about nested return statements (if the lowest one returns, does its parent also return?),
No. A return only return the function the statement exists in
the || operator here (if the first test passes, does the every loop stop?)
No. Array.prototype.every will return false as soon as the callback returns a false. If a false is never returned from the callback, .every will return `true.
function isEven(x) { return x % 2 === 0; }
[2,4,5,6].every(isEven); //=> false, stops at the 5
[2,4,6].every(isEven); //=> true
Here's an example of .every short circuiting
[1,2,3,4,5,6].every(x=> {console.log(x, x<4); return x<4;});
// 1 true
// 2 true
// 3 true
// 4 false
//=> false
See how it stops once the callback returns false? Elements 5 and 6 aren't even evaluated.
... and just generally how this works.
&& kind of works like Array.prototype.every and || kind of works like Array.prototype.some.
&& will return false as soon as the first false is encountered; in other words, it expects every arguments to be true.
|| will return true as soon as the first true is encountered; in other words, it expects only some argument to be true.
Relevant: short circuit evaluation
I am trying to understand this code from Eloquent JavaScript:
function unless(test, then) {
if (!test) then();
}
function repeat(times, body) {
for (var i = 0; i < times; i++) body(i);
}
repeat(3, function(n) {
unless(n % 2, function() {
console.log(n, "is even");
});
});
// → 0 is even
// → 2 is even
I get that it says, run the following code 3 times testing with 0,1,2:
if (!n%2) function(n) {
console.log(n, "is even");
});
What I don't get is how we get true/false from (!n%2)?
Is (!n%2) the same as (!n%2 == 0)?
Logical NOT ! has higher precedence than modulo operator %.
Thus, (!n%2) is equivalent to (!n) % 2 which will always return 0 a falsy value except when n = 0.
Whereas (!n%2 == 0) will return true(again except 0).
They both are not equal, in fact they are opposite of each other(falsy vs truthy value).
You need !(n % 2).
Or simply to check if number is even
n % 2 === 0
Your test code you wrote is not equivalent to the sample code from the article.
In the sample code, n % 2 is evaluated first, and the result is passed into the unless function. There, you are performing a Logical Not operation against the result.
If n is even, n % 2 will pass 0 to unless. A Boolean comparison of 0 returns false, and the ! negates the result (logical not), so !0 == true. this, in turn, causes the then function to fire.
If n is odd, the opposite occurs. Some value other than 0 is passed, which evaluates to false, causing then to not fire.
In contrast, your attempt to reproduce the sample code without using Higher-Order functions won't work the same way. !n % 2 will perform the Logical Not on n first, then try to modulo the result. !(n % 2) is a better representation of the sample code.
!Boolean(n%2) should work as a way to determine whether one is even or odd.
Remember that Boolean does not follow BIDMAS.
It does !n first where n is treated as a Boolean so that the numerical value of 0 is treated as false and any other numerical value is treated as true. The exclamation mark inverts the logic state of n.
Now the %2 converts the Boolean back to an integer where true is 1 and 0 is false. !n = 0 returns 0 and !n = 1 returns 1.
Using !n%2 in an if statement converts it back to a Boolean (1 = true, 0 = false).
Thus if n = 0 then the if statements proceeds because it returned true. If n != 0 (!= meaning not equal) then if statement is skipped because it returned false.
No, it's not. As stated here the "!" operator has highest precedence than "%" so the expression will return true if n is 0 and false if n is different from 0.
More in detail, suppose n is 2. the execution of the expression is:
(!n)%2
(!2)%2
0%2
0
so it is false, now for n=0
(!0)%2
1%2
1
so it is true. It is the opposite behavior of (!n%2 == 0) that returns true if n is different from 0 and false otherwise since == has less precendence and the comparison with 0 is executed at the end of the calculation above. You can easily convince yourself by using this simple javascript with different values of n:
n = 1;
if(!n%2)
{
document.getElementById("par1").innerHTML="true";
}
if(!n%2 == 0)
{
document.getElementById("par2").innerHTML="true";
}
I was playing around in jsfiddle.net and I'm curious as to why this returns true?
if(0 < 5 < 3) {
alert("True");
}
So does this:
if(0 < 5 < 2) {
alert("True");
}
But this doesn't:
if(0 < 5 < 1) {
alert("True");
}
Is this quirk ever useful?
Order of operations causes (0 < 5 < 3) to be interpreted in javascript as ((0 < 5) < 3) which produces (true < 3) and true is counted as 1, causing it to return true.
This is also why (0 < 5 < 1) returns false, (0 < 5) returns true, which is interpreted as 1, resulting in (1 < 1).
My guess is because 0 < 5 is true, and true < 3 gets cast to 1 < 3 which is true.
probably because true is assumed as 1 so
0 < 5 < 3 --> true < 3 --> 1 < 3 --> true
Because true < 3, because true == 1
As to your question whether this quirk is ever useful: I suppose there could be some case where it would useful (if condensed code is what you are after), but relying on it will (most likely) severely reduce the understandability of your code.
It's kind of like using post/pre increment/decrement as a part of bigger expressions. Can you determine what this code's result is at a glance?
int x = 5;
int result = ++x + x++ + --x;
Note: with this code, you can sometimes even get different results depending on the language and compiler.
It's a good idea to make life easy for yourself and the next guy who will read your code. Clearly write out what you actually want to have happen rather then relying on side effects like the implicit conversion of booleans.
The answer to the second part of the question, "is this quirk ever useful?" is perhaps no, as noted by a previous answer, if it is indeed a quirk of the language (Javascript) that true is cast to 1, but that the programmer does not in general view 1 and true (and 0 and false) as the same thing.
If however you have a mental model of 1 being true and 0 being false, then it leads to all sorts of nice boolean techniques that are extremely useful, powerful, and direct. For example, you could increment a counter directly with the result of A > 100, which would increment the counter if A is greater than 100. This technique might be viewed as a quirk or a trick in Java, but in an array or functional language may be idiomatic.
A classic example in the array language APL would be to count the number of items in an array that are (say) greater than 100:
+/A>100
Where if A is the 5 item array 107 22 256 110 3 then:
A>100
yields the 5 item boolean array:
1 0 1 1 0
and summing this boolean result:
+/1 0 1 1 0
yields the final answer:
3
This question is a perfect example of where this technique would be very useful, especially if the problem is generalized to determine if n out of m boolean values are true.
Check if at least two out of three booleans are true
That's easy.
(0 < 5 < 3)
Start with left to right so it evaluates the first 0 < 5. Is it true? Yes. Since TRUE=1, it evaluates 1 < 3. Since 1 is less than 3 so it's true.
Now with this
(0 < 5 < 1)
Is 0 less than 5? Yes. So make it TRUE which also means 1. Now with that fact in mind, it evaluates to (1 < 1). Is 1 less than 1? No, therefore it's false. It has to be equal.
is it evaluating 0<5 which would return 1 for true when 1<3 which is true?
C# want let you do this "Operator '<' cannot be applied to operands of type 'bool' and 'int'"
I ran into this a little while ago in Obj-C and was very puzzled by it. I got the results I wanted by doing something like this:
if(0 < 5 && 5 < 3) {
alert("True");}
Which of course is false so you wouldn't get that "true" alert.
Glad I read this, I now know why.
In addition to python, CoffeeScript is another language that supports chained comparisons, thus 3 < x < 10 would be converted to (3 < x && x < 10) in vanilla JS
0 < 5 < 3
==> ( ( 0 < 5 ) < 3 )
==> true < 3
==> 1 < 3
==> true
A boolean operand when operated over a math operator returns a number.
to check this we do
true + 1 which gives you 2.
So 0 < 5, the returned boolean(true) operated with math operator(<) will return a number. So it boils to 1<3 which returns true
because 0 is less then 5 then that returns true, and by default true is anything including and can be evaluated to 1 which is still less than 3 which again returns true
try phrasing your results as Number()
if(Number(0) < Number(5) < Number(3)) {
alert("True");
}
or try this:
if(Number(0) < Number(5) && Number(5) < Number(3)) {
alert("True");
}
I googled this because I was getting (3 >= 20) //returning true and I guess javascript was trying to check 3 as a boolean because I was getting this value from the elm.getAttribute(); function which console.log(); was printing in String form.