Why JavaScript treats a number as octal if it has a leading zero

Question

var x = 010;
console.log(x); //8

JS engine convert the number x to octal number. Why it happens? How can I prevent it?

Answer 1

I think my answer here answers the question, but the question is not exactly a duplicate, so I include a copy of my answer.

History

The problem is that decimal integer literals can't have leading zeros:

DecimalIntegerLiteral ::
    0
    NonZeroDigit DecimalDigits(opt)

However, ECMAScript 3 allowed (as an optional extension) to parse literals with leading zeros in base 8:

OctalIntegerLiteral ::
    0 OctalDigit
    OctalIntegerLiteral OctalDigit

But ECMAScript 5 forbade doing that in strict-mode:

A conforming implementation, when processing strict mode code (see 10.1.1), must not extend the syntax of NumericLiteral to include OctalIntegerLiteral as described in B.1.1.

ECMAScript 6 introduces BinaryIntegerLiteral and OctalIntegerLiteral, so now we have more coherent literals:

• BinaryIntegerLiteral, prefixed with 0b or 0B.
• OctalIntegerLiteral, prefixed with 0o or 0O.
• HexIntegerLiteral, prefixed with 0x or 0X.

The old OctalIntegerLiteral extension has been renamed to LegacyOctalIntegerLiteral, which is still allowed in non-strict mode.

Conclusion

Therefore, if you want to parse a number in base 8, use the 0o or 0O prefixes (not supported by old browsers), or use parseInt.

And if you want to be sure your numbers will be parsed in base 10, remove leading zeros, or use parseInt.

Examples

• 010
  • In strict mode (requires ECMAScript 5), it throws.
  • In non strict mode, it may throw or return 8 (implementation dependent).
• 0o10, 0O10
  • Before ECMAScript 6, they throw.
  • In ECMAScript 6, they return 8.
• parseInt('010', 8)
  • It returns 8.
• parseInt('010', 10)
  • It returns 10.