Why is 1 == '1\n' true in Javascript? - javascript

The same goes for '1\t' (and probably others).
if (1 == '1\n') {
console.log('Equal');
}
else {
console.log('Not Equal');
}

As said before if you compare number == string, it will automatically try to convert the string to a number. the \n and \t are simply whitespace characters and therefore ignored.
This and similar behaviour can be rather confusing leading to situations like this:
(Picture taken from: https://www.reddit.com/r/ProgrammerHumor/comments/3imr8q/javascript/)

The equality operator(==) converts the operands if they are not of the same type, then applies strict comparison. If both operands are objects, then JavaScript compares internal references which are equal when operands refer to the same object in memory.
1 is of type Number hence '1\n' is converted to Number first then comparison takes place!
And Number() constructor will convert the string('1\n') to 1 :-
Number('1\n') === 1
In Strict equality using ===, Neither value is implicitly converted to some other value before being compared. If the values have different types, the values are considered unequal.
Hence 1 === '1\n' will be expressed as false

Related

What does the preceding `+` do when defining SVG varaibles in D3? [duplicate]

I was perusing the underscore.js library and I found something I haven't come across before:
if (obj.length === +obj.length) { ... }
What is that + operator doing there? For context, here is a direct link to that part of the file.
The unary + operator can be used to convert a value to a number in JavaScript. Underscore appears to be testing that the .length property is a number, otherwise it won't be equal to itself-converted-to-a-number.
According to MDN:
The unary plus operator precedes its operand and evaluates to its
operand but attempts to converts it into a number, if it isn't
already. For example, y = +x takes the value of x and assigns that to
y; that is, if x were 3, y would get the value 3 and x would retain
the value 3; but if x were the string "3", y would also get the value
3. Although unary negation (-) also can convert non-numbers, unary plus is the fastest and preferred way of converting something into a
number, because it does not perform any other operations on the
number. It can convert string representations of integers and floats,
as well as the non-string values true, false, and null. Integers in
both decimal and hexadecimal ("0x"-prefixed) formats are supported.
Negative numbers are supported (though not for hex). If it cannot
parse a particular value, it will evaluate to NaN.
It's a way of ensuring that obj.length is a number rather than a potential string. The reason for this is that the === will fail if the length (for whatever reason) is a string variable, e.g. "3".
It's a nice hack to check whether obj.length is of the type number or not. You see, the + operator can be used for string coercion. For example:
alert(+ "3" + 7); // alerts 10
This is possible because the + operator coerces the string "3" to the number 3. Hence the result is 10 and not "37".
In addition, JavaScript has two types of equality and inequality operators:
Strict equality and inequality (e.g. 3 === "3" expresses false).
Normal equality and inequality (e.g. 3 == "3" expresses true).
Strict equality and inequality doesn't coerce the value. Hence the number 3 is not equal to the string "3". Normal equality and inequality does coerce the value. Hence the number 3 is equal to the string "3".
Now, the above code simply coerces obj.length to a number using the + operator, and strictly checks whether the value before and after the coercion are the same (i.e. obj.length of the type number). It's logically equivalent to the following code (only more succinct):
if (typeof obj.length === "number") {
// code
}

Javascript Simple Boolean Arithemetic

I’ve heard of boolean arithmetic and thought of giving it a try.
alert (true+true===2) //true
alert (true-true===0) //true
So algebra tells me true=1
alert (true===1) //false :O
Could someone explain why this happens?
=== is the strict equality operator. Try == operator instead.
true==1 will evaluate to true.
The strict equality operator === only considers values equal if they
have the same type. The lenient equality operator == tries to
convert values of different types, before comparing like strict
equality.
Case 1:
In case of true===1, The data type of true is boolean whereas the type of 1 is number. Thus the expression true===1 will evaluate to false.
Case 2:
In case of true+true===2 and true-true===0 the arithmetic operation is performed first(Since + operator takes precedence over ===. See Operator Precedence) and then the result is compared with the other operand.
While evaluating expression (true+true===2), the arithmetic operation true+true performed first producing result 2. Then the result is compered with the other operand. i.e. (2==2) will evaluate to true.
Because comparing data TYPE and value (that's what operator '===' does ), TRUE is not exactly the same as 1. If you changed this to TRUE == 1, it should work fine.
At the beginning, you're doing bool + bool. The + operator takes precedence over the === operator so it's evaluated first. In this evaluation, it's converting the booleans to their number forms. Run console.log(true + true); and this will return 2. Since you're comparing the number 2 to the number 2, you get a return value true with the strict equality.
When you're just comparing true === 1, like everyone else said you're comparing the boolean true to the number 1 which is not strictly equal.
first 2 expression are true because you are using expression (true+true) (true-true) it convert type of a value first due to expression and check equality with "===", toNumber and toPrimitive are internal methods which convert their arguments (during expression ) this is how conversion take place during expression
That's why true+true equal to the 2
In your third expression you are using === this not convert arguments just check equality with type, to make it true both values and there type must be same.
Thats all

Javascript 0 == '0'. Explain this example

I found these examples in another thread but I don't get it.
0 == '0' // true
0 to the left I converted to false(the only number that does that). The right is a non empty string which converts to true.
So how can
false == true --> true
What have I missed?
Here is an official answer to your question (the quoted parts, link at the bottom) and analysis:
Equality (==)
The equality operator converts the operands if they are not of the same type, then applies strict comparison. If either operand is a number or a boolean, the operands are converted to numbers if possible; else if either operand is a string, the string operand is converted to a number if possible. If both operands are objects, then JavaScript compares internal references which are equal when operands refer to the same object in memory.
Syntax
x == y
Examples
3 == 3 // true
"3" == 3 // true
3 == '3' // true
This means, as I read it, that the first 3 (integer) is converted to string to satisfy the comparison, so it becomes '3' == '3', which is true, same as in your case with zeroes.
NOTE: I assume that the converion may vary in different JS engines, although they have to be unified under ECMAScript specifiction - http://www.ecma-international.org/ecma-262/5.1/#sec-11.9.3 (quoted #Derek朕會功夫). This assumption is made on a subjective and imperative opinion that not all browsers and JavaScript engines out there are ECMAScript compliant.
Identity / strict equality (===)
The identity operator returns true if the operands are strictly equal (see above) with no type conversion.
The Identity / strict equality (===) found on the same resource at the end of the answer will skip the automatic type conversion (as written above) and will perform type checking as well, to ensure that we have exact match, i.e. the expression above will fail on something like:
typeof(int) == typeof(string)
This is common operator in most languages with weak typing:
http://en.wikipedia.org/wiki/Strong_and_weak_typing
I would say that one should be certain what a function/method will return, if such function/method is about to return numbers (integers/floating point numbers) it should stick to that to the very end, otherwise opposite practices may cut your head off by many reasons and one lays in this question as well.
The above is valid for other languages with weak typing, too, like PHP for example.
Read more of this, refer second head (Equality operators):
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Comparison_Operators
When you use == JavaScript will try very hard to convert the two things you are trying to compare to the same type. In this case 0 is converted to '0' to do the comparison, which then results in true.
You can use ===, which will not do any type coercion and is best practice, to get the desired result.
Equality operator
The equality operator converts the operands if they are not of the same type, then applies strict comparison. If either operand is a number or a boolean, the operands are converted to numbers if possible; else if either operand is a string, the string operand is converted to a number if possible. If both operands are objects, then JavaScript compares internal references which are equal when operands refer to the same object in memory.
Source: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Comparison_Operators
JavaScirpt Table Equality: http://dorey.github.io/JavaScript-Equality-Table/

A javascript statement that I haven't seen before

I am inspecting some Javascript code written in UnderscoreJS. In this, there are quite a few instances of JS code being written like this:
if(length === +length){ //do something}
or
if(length !== +length){ //do something }
What exactly does this compute to? I have never seen this before.
if (length === +length)
It makes sure that the length is actually a numeric value.
Two things to understand here,
The Strict equality operator, will evaluate to true only when the objects are the same. In JavaScript, the numbers and strings are immutables. So, when you compare two numbers, the Number values will be compared.
Unary + operator will try to get the numeric value of any object. If the object is already of type Number, it will not make any changes and return the object as it is.
In this case, if length is actually a number in string format, lets say "1", the expression
"1" == +"1"
will evaluate to be true, because "1" is coerced to numeric 1 internally and compared. But
"1" === +"1"
which will be transformed to
"1" === 1
and that will not undergo coercion and since the types are different, === will evaluate to false. And if length is of any other type, === will straight away return false, as the right hand side is a number.
You can think of this check as a shorthand version of this
if (Object.prototype.toString.call(length) === "[object Number]")
It is a way of testing whether a value is numeric. See this for example:
> length="string"
> length === +length
false
> length=2
> length === +length
true
The unary + converts the variable length to the Number type, so if it was already numeric then the identity will be satisfied.
The use of the identity operator === rather than the equality operator == is important here, as it does a strict comparison of both the value and the type of the operands.
Perhaps a more explicit (but slightly longer to write) way of performing the same test would be to use:
typeof length === "number"

jsHint error saying I should use "===" [duplicate]

This question already has answers here:
Which equals operator (== vs ===) should be used in JavaScript comparisons?
(48 answers)
Closed 8 years ago.
My code is giving me an error with jsHint. I am trying to do this:
if (data.result == 't' || task == 'show') {
But it tells me I should replace "==" with "===" can someone tell me why it gives this message?
=== is the strict equality operator.
== is the normal equality operator, == converts its operands to the same type if they are not already of the same type.
So there is a danger that something like "" == 0 will give you true although they are of different types.
Since there is implicit conversion involved which you might not no about because it is happening automatically, there is some danger and potential for errors or bugs that are hard to trace.
=== won't convert its operands, it will just compare them.
Silent type conversion can be a source of bugs. If you avoid converting between data types at comparison time, then you avoid many of those bugs.
=== means equality without type coercion. === operator will not do the conversion, it will only compare the operands.
The example given by sdfx here is very helpful in understanding this:
0==false // true
0===false // false, because they are of a different type
1=="1" // true, auto type coercion
1==="1" // false, because they are of a different type
From MDN:
JavaScript has both strict and Type–converting (abstract) comparisons.
A strict comparison (e.g., ===) is only true if the operands are the
same Type. The more commonly used abstract comparison (e.g., ==)
converts the operands to the same Type before making the comparison.
For relational abstract comparisons (e.g., <=), the operands are first
converted to primitives, then the same Type, before comparison.
Features of comparisons:
Two strings are strictly equal when they have the same sequence of characters, same length, and same characters in corresponding
positions.
Two numbers are strictly equal when they are numerically equal (have the same number value). NaN is not equal to anything,
including NaN. Positive and negative zeros are equal to one
another.
Two Boolean operands are strictly equal if both are true or both are false.
Two distinct objects are never equal for either strictly or abstract comparisons.
An expression comparing Objects is only true if the operands reference the same Object.
Null and Undefined Types are == (but not ===).

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