CODEWITHSUNDEEP
javascriptcomparisonoperators
harsh atal
harsh atal

Reputation: 409

Explain the behaviour of >= and <= operators in the following case

function foo(){
console.log( function(){return 1} <= function(){return 1} );
}
foo();

The above code prints "true". It prints true for both <= and >= , For all other operators it is "false".

What trick going on here?

Upvotes: 2

Views: 76

Answers (3)

Dmitri Pavlutin
Dmitri Pavlutin

Reputation: 19080

function(){return 1} <= function(){return 1} transforms the functions to a primitive. A function transformed to a primitive evaluates to it's code in a string (calling toString()).
Then "function (){return 1}" <= "function (){return 1}" performs an less or equal comparison on strings, which is true.

In function(){return 1} == function(){return 1} JavaScript compares the function objects. But because these are different instances, you'll have false.

You can check more about equality operator in this article.

Upvotes: 2

Sterling Archer
Sterling Archer

Reputation: 22395

Because it's not comparing the returned 1, it's comparing the functions. When you run an operator on a function, it implicitly calls toString.

As for ==, Dmitri explained it well enough that toString is not being called, but it runs an object comparison, which will be false because 2 different objects are never the same.

Here is the ECMA spec On relational operators, which explains how it operates with comparison operations. With objects, instanceOf is used.

Upvotes: 3

Patty
Patty

Reputation: 11

Because it´s always is comparing 1 <= 1, and it´s true. If you compare 1 >= 1 is the same always it´s true. But if you compare 1 > 1 it´s false.

Upvotes: -1

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