When Comparing Strings in JavaScript, Why is \'11\' Lesser Than \'3\'?

Understanding the Logic Behind String Comparison in JavaScript

In JavaScript, when comparing strings, the language employs lexicographical order rather than their length or value. This means that strings are compared character by character, starting from the first character until an inequality is found or there are no characters left to compare.

To address the specific case mentioned in the title, despite "11" appearing numerically larger than "3," JavaScript perceives "11" as less than "3" due to the difference in their first character. The Unicode code point for the character '1' is less than that for the character '3.'

Consider the following examples:

• '31' is less than '3' because the first character of '31' is '3', which is less than '3.'
• '31' is less than '32' because the second character of '31' is '1', which is less than '2.'
• '31' is not less than '30' because the second character of '31' is greater than '0.'
• 'abc' is not less than 'aaa' because 'a' is equal to 'a,' and 'b' is greater than 'a.'
• 'abc' is less than 'abd' because 'b' is less than 'd.'

To explicitly convert strings to numbers, you can use the ' ' operator:

• '11' is not less than '3' because the converted numbers are 11 and 3, respectively.

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