Question

a="12345"
a[2]=3
a[2]='9'
console.log(a) //=> "12345"

What is going on?? This quirk caused me 1 hour painful debugging. How to avoid this in a sensible way?

Answer 1

You cannot use brackets to rewrite individual characters of the string; only ‘getter’ (i.e. read) access is available. Quoting the doc (MDN):

For character access using bracket notation, attempting to delete or
assign a value to these properties will not succeed. The properties
involved are neither writable nor configurable.

That’s for “what’s going on” part of the question. And for “how to replace” part there’s a useful snippet (taken from an answer written long, long ago):

String.prototype.replaceAt = function(index, char) {
    return this.slice(0, index) + char + this.slice(index+char.length);
}

You may use as it is (biting the bullet of extending the JS native object) – or inject this code as a method in some utility object (obviously it should be rewritten a bit, taking the source string as its first param and working with it instead of this).