6 ms·

> I know that after reading a const line I don't have to check if someone rebinds this name before each use site -- it's impossible. This turns out to be almos

by _getify 11y ago

> I know that after reading a const line I don't have to check if someone rebinds this name before each use site -- it's impossible. This turns out to be almost useless information for most of us, since the real problem is not re-assignment, but mutation of the value (in the case of non-primitives). `const x = 2` feels great emotionally, but `const x = [2]` has the feeling of safety without any of the guarantee. This is a classic case (quoted from Crockford regularly) of a utility that it sometimes useful and sometimes harmful (aka not useful), and there's a better option (`let`), so the better option should generally be preferred. I only use `const` (refactor from `let`) when a piece of code is reasonably complete and I'm pretty sure re-assignment will not ever be appropriate. Guessing at that before writing the code is a premature optimization. And guessing wrongly and having to refactor back to `let` later is more risky.

AgentME 11y ago
> `const x = 2` feels great emotionally, but `const x = [2]` has the feeling of safety without any of the guarantee. Const is perfectly sensible with references to mutable objects too. The latter guarantees that x will always point to the same mutable array. If you pass a reference to the x array to a function which needs to mutate that specific array or observe it for changes, then it may be important that x is never rebound to a different array. Consider the following code: const x = [5,6,7]; Object.observe(x, function(changes) { console.log('changes', changes); }); // ... // time to clear the array // WRONG! The code observing changes to the old x array will not see // this change or future changes to this new array. Because we used const, this // will trigger an error and immediately show us our mistake. x = []; // CORRECT. This mutates the array instead of creating a new empty array. x.length = 0;
- _getify 11y ago
  > If you pass a reference to the x array to a function This notion (concern) is meaningless in JS, since when you pass any reference, it's always a reference-copy, so there's no value nor assistance that `const` provides.
  - AgentME 11y ago
    Const usage prevents a bug in my example directly above!

by _getify 11y ago

AgentME 11y ago

> `const x = 2` feels great emotionally, but `const x = [2]` has the feeling of safety without any of the guarantee. Const is perfectly sensible with references to mutable objects too. The latter guarantees that x will always point to the same mutable array. If you pass a reference to the x array to a function which needs to mutate that specific array or observe it for changes, then it may be important that x is never rebound to a different array. Consider the following code: const x = [5,6,7]; Object.observe(x, function(changes) { console.log('changes', changes); }); // ... // time to clear the array // WRONG! The code observing changes to the old x array will not see // this change or future changes to this new array. Because we used const, this // will trigger an error and immediately show us our mistake. x = []; // CORRECT. This mutates the array instead of creating a new empty array. x.length = 0;

_getify 11y ago
> If you pass a reference to the x array to a function This notion (concern) is meaningless in JS, since when you pass any reference, it's always a reference-copy, so there's no value nor assistance that `const` provides.
- AgentME 11y ago
  Const usage prevents a bug in my example directly above!