CODEWITHSUNDEEP
javascriptarrays
ccnokes
ccnokes

Reputation: 7115

In Javascript, why are \b characters dropped in a .join()?

In Javascript, why does this statement not equal '\b,\b'?

['\b', '\b'].join()
//=> ","

According to MDN docs on join:

If an element is undefined or null, it is converted to the empty string.

So why is \b being evaluated as undefined/null?

Additionally, the\b is dropped from any string prepended with it, e.g:

['\btest', '\btest2'].join()
//=> "test,test2"

Something crazy is going on.

Upvotes: 1

Views: 240

Answers (2)

gurvinder372
gurvinder372

Reputation: 68413

So why is \b being evaluated as undefined/null?

As per the spec

In determining the sequence any occurrences of \ UnicodeEscapeSequence are first replaced with the code point represented by the UnicodeEscapeSequence and then the code points of the entire IdentifierName are converted to code units by UTF16Encoding (10.1.1) each code point.

Also read this part to understand which are escape character and which are not

CharacterEscapeSequence ::

SingleEscapeCharacter NonEscapeCharacter

SingleEscapeCharacter ::

one of ' " \ b f n r t v

NonEscapeCharacter

:: SourceCharacter but not one of EscapeCharacter or LineTerminator

EscapeCharacter ::

SingleEscapeCharacter

DecimalDigit x u

HexEscapeSequence ::

x HexDigit HexDigit UnicodeEscapeSequence :: u Hex4Digits u{ HexDigits }

Which is why \b being a special character is removed while \a is still "a".

Upvotes: 2

kantuni
kantuni

Reputation: 971

\b is a special character, which means backspace.

That is why it is being converted to the 'empty' string.

Reference:

Upvotes: 1

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