Why Does (0 < 5 < 3) Return True

Why does (0 5 3) return true?

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).

Why 542 is returning true?

This is because of the Operator precedence.

Less than (<) operator is evaluated from left-to-right.

First 5<4 is evaluated to false then false is converted to 0 in the next evaluation. Finally 0<2 is evaluated to true

console.log(5<4);// falseconsole.log(0<2);// true

Why does this .every() function return true and then false?

You need to use return inside .every, currently you are just logging num < 0 which is true for num[0] which is (-1 < 0), but false for num[1] which is (50 >= 0). And since you do not return is instantly exits the loop as this is considered false.

let numsA = [-1, 50, 75, 200, 350, 525, 1000];console.log("should return false, because 50 is >= 0");console.log(numsA.every(function(num) {  return (num < 0);}));
let numsB = [1, 50, -75, 200, 350, 525, 1000];
console.log(numsB.every(function(num) {  return (num < 0);}));
let numsC = [-1, 50, 75, 200, 350, 525, 1000];
console.log(numsC.every(num => num < 0));
let numsD = [-5, 0, 5, 10, 15];console.log("should return true because for each num in numsD, num %5 === 0");console.log(numsD.every(num => {  return num % 5 === 0}));//you can omit return with arrow function syntax, as this automatically returns the value within ()console.log(numsD.every(num => (num % 5 === 0)));

Why does my function keep returning True?

You can try any of the following functions:

Function 01:

def in_order(nums):
    flag = 0
    if all(nums[i] <= nums[i + 1] for i in range(len(nums)-1)):
        flag = 1

    return flag

Function 02:

def in_order(nums):
    i=1
    while i< len(nums):
         if nums[i] < nums[i-1]:
            return False
         i=i+1
    return True

Driver Code:

if __name__ == '__main__':
    # Test out-of-order example
    nums1 = [5, 6, 7, 8, 3]
    if in_order(nums1):
        print('In order')
    else:
        print('Not in order')
        
    # Test in-order example
    nums2 = [5, 6, 7, 8, 10]
    if in_order(nums2):
        print('In order')
    else:
        print('Not in order')

Output:

Not in order
In order

Javascript - (10310) return true

10<3<10 is equivalent to (10<3)<10. 10<3 is false, false<10 is true.

Why does not my function math.abs return true?

You need to Math.abs the limiet too

Also you had a logical error in your (2) - you did not have more than one minute's difference and you have < not <= so we need a difference

const ms2MMSS = ms => {   let minutes = Math.floor(ms / 60000);   let seconds = ((ms % 60000) / 1000).toFixed(0);  return minutes + ":" + (seconds < 10 ? '0' : '') + seconds; }

function timeout(date1, date2, timeLimit) {
  const limiet = Math.abs(timeLimit * 60000);
  console.log(ms2MMSS(limiet),ms2MMSS(Math.abs(date1.getTime()-date2.getTime())));
  return Math.abs(date1 - date2) < Math.abs(limiet);
}

console.log(timeout(new Date("2019-09-01T08:30:00"), new Date("2019-09-01T11:30:00"), 60), "should return false")
console.log(timeout(new Date("2019-09-01T11:30:00"), new Date("2019-09-01T08:30:00"), 60), "should return false")
console.log(timeout(new Date("2019-09-01T11:30:00"), new Date("2019-09-01T08:30:00"), 190), "should return true")
console.log(1,timeout(new Date("2019-08-31T23:59:00"), new Date("2019-09-01T00:01:00"), -5), "should return true")
console.log(2,timeout(new Date("2019-08-31T23:59:00"), new Date("2019-09-01T00:00:00"), -2), "should return true")
console.log(timeout(new Date("2019-08-31T23:59:00"), new Date("2019-09-01T00:01:00"), 0), "should return false")
console.log(timeout(new Date("2019-08-31T23:59:00"), new Date("2019-09-01T00:01:00"), 1), "should return false")
console.log(timeout(new Date("2019-08-31T23:59:00"), new Date("2019-09-01T00:01:00"), 2), "should return false")
console.log(timeout(new Date("2019-08-31T23:59:00"), new Date("2019-09-01T00:01:00"), 3), "should return true")

Why is 1x3 always true?

1 < x < 3 actually is doing this:

(1 < x) < 3

Or even more long form:

const tempVarA = 1 < x
const tempVarB = tempVarA < 3

So 1 < x is either true or false. Then the next step is true < 3 or false < 3. Those don't make much sense as comparisons, but let's see what javascript does with that:

console.log(true < 3) // true
console.log(false < 3) // true

Weird, but let's dig deeper:

console.log(true >= 0) // true
console.log(true >= 1) // true
console.log(true >= 2) // false

console.log(false >= 0) // true
console.log(false >= 1) // false
console.log(false >= 2) // false

It seems that true is being treated as 1 and false is being treated as 0. To verify that let's compare with == (instead of ===) so that it coerces the type of the data for us.

console.log(true == 1) // true
console.log(true == 0) // false
console.log(false == 1) // false
console.log(false == 0) // true

So 1 < x < 3 is always true because false becomes 0 or true becomes 1, and both 0 and 1 always less than 3.

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Explanation:

in javascript, comparison operators <, <=, >, >=, ==, and != coerce their operands to make them comparable when they are different types. So when comparing a boolean to a number it coverts the boolean to a number, 0 or 1.

This is why you should almost always use === instead of ==, and why this is a type error in typescript:

const a = true < 3
// Operator '<' cannot be applied to types 'boolean' and 'number'.(2365)

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Short version

Javascript and typescript lack a chainable comparison operator.

Did you mean to do this?

1 < x && x < 3

In JavaScript why 2 1 0 true while 8 1 1 false?

I'd like to clear this problem using following example.

1 < 2 < 3 // evaluates to true

But

3 > 2 > 1 // evaluates to false

As we know js execute from left to right. Therefor, in first example when statement

1 < 2 

is executed it evalutes to true. In programming true refers to 1 and false refers to zero. Then after executing the above statement we have true means 1. Now when this result is combined with next statement

true < 3 

Then in short this statement means

1 < 3 

because true refers to 1 in programming. Since 1 < 3 is true that's why we have the final result as true.

Coming to the next example 3 > 2 > 1 which evaluates to false because

3 > 2 // evaluates to true 

Since above statement is true then in programming sense we got 1(true) and then by combining this result with next statement

true > 1 // evaluates  to false

The above statement is in short

1 > 1 

Since 1 is not greater than 1 therefor it returns the final result to false.

3 > 2 = true
true > 1 = false

So finally we conclude that by programming sense both statement evaluates to different results but in sense of mathematics both are same. Its a interview question. Hopefully you understand this.

Please let me know if needs any other information. In other case, It might be accepted answer.

Why does 3 2 1 return false while 1 2 3 returns true?

Since 1 < 2 evaluates to true and 3 > 2 evaluates to true, you're basically doing :

console.log(true < 3);console.log(true > 1);

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