why does '\\\[' equals '\\[' ? How does backslash work in string?

As the title

console.log('\\\[' === '\\[');

A backslash before most characters will only be parsed as an unnecessary escape character - the backslash will be ignored. This is what's happening in the second part of the first string. Before a certain few characters though, such as another backslash in \\, or \n, it will be parsed as a escape sequence. \\ is the escape sequence for a single literal backslash:


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In strings, the backslash (\) is a special character used to encode other special characters, including the backslash.

'\\[' is a JavaScript string literal that contains a backslash (\\) and an open square bracket ([). In the compiled program the string is \[.

'\\\[' is a JavaScript string literal that contains a correctly encoded backslash (\\) followed by the combination of characters \[ that looks like an escape sequence but doesn't mean anything. Because this combination is not defined and \ by itself does not mean anything, the JavaScript interpreter ignores the backslash and corrects the string; it becomes identical to the first one (\[).

The behaviour is documented:

For characters not listed in the table, a preceding backslash is ignored, but this usage is deprecated and should be avoided.

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Backslash is a special character. Literally, JS talk to browser to interpret the symbol after \ as is. Sometimes it calls screening or shielding. That is why we can write smth like that: console.log("Double \"quotes\" inside another one."); with the result of Double "quotes" inside another one. without any error. Although that is not the way we need to use anywhere.

  1. "\\\[" separates into 2 parts: \\ and \[. First returns \ and the second returns [. Finally it is \[.
  2. "\\[" separates into 2 parts: \\ and [. First returns \ and the second returns [. Finally it is \[.

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  • because [ doesn't need to be escaped, i.e. '\[' === '['