ES6解構嵌套的可選參數？

Question

考慮以下ES6代碼：

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

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

第一個電話按預期工作。不過，我想讓address可選（street和postcode應該是undefined）而不是拋出錯誤。這可能嗎？

Answer 1

我找到了解決辦法：

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