ES6解构嵌套的可选参数？

Question

考虑以下ES6代码：

function foo({name, address: {street, postcode}}) {
  console.log(name, street, postcode);
}

foo({name: 'John', address: {street: 'Foo', postcode: 1234}});
foo({name: 'Bob'});

第一个呼叫按预期工作。不过，我希望address可选（street和postcode为undefined），而不是抛出错误。这可能吗？

Answer 1

我找到了解决方案：

function foo({name, address: {street, postcode} = {}}) {
  console.log(name, street, postcode);
}