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'}); 

第一个电话按预期工作。不过，我想让address可选（street和postcode应该是undefined）而不是抛出错误。这可能吗？

Answer 1

我找到了解决办法：

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