Why does the tostring () method gobble up the leading zero?

Question

There is a variable n which contains the number n = 084746473827;

When I use the tostring() function, tostring() already 84746473827 (that is, without 0 ).

Why it happens?

And how can a number lead to a string without changing it (for example, the number is 084746473827 , but it is necessary to return the string 084746473827 )?

because the number is actually 84746473827 What makes you think that it costs 0 ?
That is, you can not do to return with a zero as at the entrance?
In this case, you should know what should be the final length.
In general, I receive a variable in which 084746473827 is a number.
the variable cannot be 084746473827 because, as you said this number, and it automatically discards leading zeros

Accepted Answer · 2015-12-04T10:42:49

Numbers do not count leading zeros. If you need to fill with zeros to a certain width, you can do this:

 var num, str; num = 84746473827; str = (new Array(12).join('0') + num).slice(-12); console.log(str);

A leading zero means that the number is written in the octal system.

Answer 2 · 2017-08-17T16:29:01

All wrong :). The zero at the beginning of the "literal" of the integer means that the number is written in the octal system. You are lucky that in the number there are numbers more than seven.

 // 017 is treated as octal var a = 017; console.log("a - decimal toString", a.toString()); console.log("a - octal toString", a.toString(8)); // 018 is treated as decimal var b = 018; console.log("b - decimal toString", b.toString());

ishidex2 ishidex2 6.257 2 gold marks 12 silver marks 40 bronze marks · Answer 3 · 2017-08-17T17:18:57

If it's not hard to guess, then 084746473827 == 84746473827 because the first high byte starting from zero has no meaning

 let num = 084746473827; console.log(num)

Why does the tostring () method gobble up the leading zero?

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