3 ms·

Yes, I'm very excited about the spread operator! I've been spoiled by Firefox.

• rplnt 11y ago
  Looking at documentation.. does this only work on lists (or list like object)? I.e. you can't "expand" a k:w object? Or is there another operator for this? Example: function foo(a=1, b=2, c=3) { return a * b * c;} dict = {'b': 4, 'c': 6}; foo(...dict);
  • arcatek 11y ago
    Yes and no. You can still do something like: foo(... Object.keys(dict).map(name => dict[name])) (Maybe someday you might be able to use Object.values() to achieve the exact same thing without needing to use .map) However, the spread operator doesn't match the argument names to the object names. So the above call would mean the following: foo(4, 6) And not what you'd like: foo(undefined, 4, 6) However, note that you can rest an object too: let x = { a : 1, b : 2, c : 3 }; let { a, ... rest } = x; console.log( ... rest ); // { b : 2, c : 3 }
    • msoad 11y ago
      There is a proposal for this but I'm not sure what stage that's in
      • Keithamus 11y ago
        Stage 2 https://github.com/tc39/ecma262 https://github.com/tc39/ecma262 https://github.com/sebmarkbage/ecmascript-rest-spread https://github.com/sebmarkbage/ecmascript-rest-spread
  • cmpb 11y ago
    Technically, the spread operator works with iterables [1]. If you wanted to spread an object `obj`you'd need to define `obj[Symbol.iterator]` [2]. [1] https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_operator https://developer.mozilla.org/en-US/docs/Web/JavaScript/Refe... [2] https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Iteration_protocols https://developer.mozilla.org/en-US/docs/Web/JavaScript/Refe...
  • moretti 11y ago
    You still have to simulate named arguments with an object: const foo = ({ a = 2, b = 2, c = 3 } = {}) => a * b * c; const dict = {b: 4, c: 6}; foo(); // 12 foo(dict); // 48