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JavaScript JavaScript Foundations Strings Comparing Strings

Alexander Kruglov
Alexander Kruglov
3,148 Points
Posted by Alexander Kruglov
Alexander Kruglov
3,148 Points

Compare function working not as expected

function compare (a,b) { console.log(a+"<"+b+" "+a<b); }

compare('a','b');

What do I expect: a<b true What do I get: true

Why doesn't it output the first half of the argument? I don't get it.

1 Answer

bothxp
bothxp
16,510 Points

So you are doing:

 console.log('a' + "<" + 'b' + " " + 'a' < 'b');

Take a look at: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Operator_Precedence

The additions happen first. So you are actually getting 

('a' + "<" + 'b' + " " + 'a') < 'b'

To get the result that you are expecting you'll need to put some ( ) around less than check

 console.log('a' + "<" + 'b' + " " + ('a' < 'b'));

So that would be:

function compare (a,b) { 
  console.log(a + "<" + b + " " + (a < b)); 
}

compare('a','b');