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The question is how is adding an unnecessary backslash a valid key while accessing the property? And by the way it works if I remove the backslash at the key name while accessing as well.

'use strict';
let x  = {
  "2nu^mb$er": "number as key",
}
console.log(x["2nu\^mb$er"]); //prints: number as key
console.log(x["2nu^mb$er"]);  //prints: number as key

What could be the reason behind the 2 console logs able to access the property?

Manoj Bharadwaj
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  • Because `"2nu\^mb$er" === "2nu^mb$er"`. It's just an unnecessary escape character being ignore by string literal syntax. – Bergi Dec 03 '20 at 21:51

2 Answers2

1

The backslash \ is used as an escape character in JavaScript. When it encounters a \ it will try to infer the special meaning of the character after it.

For example \n means carriage return. So when the string "2nu\^mb$er" is evaluated as the key of the object it becomes "2nu^mb$er", which is a valid key as \^ has no special meaning in JavaScript.

So to actually have a \ character in your string you need to escape it using another \ before it:

'use strict';
let x  = {
  "2nu^mb$er": "number as key",
}
//prints: undefined as now the key becomes 2nu\^mb$er
console.log(x["2nu\\^mb$er"]);
 //prints: number as key as the key is 2nu^mb$er
console.log(x["2nu\^mb$er"]);

That implies that \ before a ^ is the same as ^, as \^ in a string has no special meaning:

console.log("\^" == "^")
Fullstack Guy
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0

This is why, from this docs https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Working_with_Objects

An object property name can be any valid JavaScript string, or anything that can be converted to a string...

Please note that all keys in the square bracket notation are converted to string unless they're Symbols, since JavaScript object property names (keys) can only be strings or Symbols

This means that "2nu\^mb$er" is converted to a string. So that's why accessing to an object property name works even with that back slash

For example: 

let a = "abc\d";
console.log(a);
Andres Gardiol
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  • "*This means that `"2nu\^mb$er"` is converted to a string.*" - no, `"2nu\^mb$er"` already *is* a string. There's no conversion going on here. – Bergi Dec 03 '20 at 21:53
  • Well yes, there is a conversion indeed. "Please note that all keys in the square bracket notation are converted to string unless they're Symbols". You can read this here https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Working_with_Objects – Andres Gardiol Dec 04 '20 at 13:27
  • A key that already is a string is not getting converted into a string. I'll edit MDN if you find that confusing. – Bergi Dec 04 '20 at 13:32
  • "...keys in the square bracket notation are converted to string unless they're Symbols...". Maybe you are right. It's only that this part make me think that "2nu\^mb$er" it's converted to a valid javascript string. Even though it is already a string – Andres Gardiol Dec 04 '20 at 13:36