javascript - 检查函数中是否缺少参数

Question

这是检查函数中缺少参数的正确方法吗？这适用于所有浏览器吗？IE怎么样？

function getName(name){
    name = name != null ? name : "default";
    return name;
}

Answer 1

检查参数的方式取决于您传递给函数的信息类型，以及您希望函数如何处理极端情况。

在大多数情况下，您可以使用：

...
bar = bar || ...default value here...
...

但是，当您想要传递虚假值（false, 0, NaN, '', undefined, null）时可能会出现问题：

function foo(bar) {
  bar = bar || 5
  console.log(bar)
}

foo()          // 5
foo(undefined) // 5
foo(null)      // 5
foo(1)         // 1
foo(0)         // 5, probably not what you wanted

相反，您可以检查undefined：

...
if (bar == undefined) {
    bar = 5
}
...

...但是使用松散检查允许null和undefined被覆盖（null == undefined）：

function foo(bar) {
  if (bar == undefined) {
      bar = 5
  }
  console.log(bar)
}

foo()          // 5
foo(undefined) // 5
foo(null)      // 5
foo(1)         // 1

因此，===通常首选严格的相等比较 ( ) ( null !== undefined)：

function foo(bar) {
  if (bar === undefined) { 
      bar = 5
  }
  console.log(bar)
}

foo()          // 5
foo(undefined) // 5
foo(null)      // null
foo(1)         // 1

ES2015 引入了默认参数，本质上等同于严格检查undefined：

function foo(bar = 5) {
  console.log(bar)
}

foo()          // 5
foo(undefined) // 5
foo(null)      // null
foo(1)         // 1

This could lead to trouble if you need to know whether undefined was passed as a parameter.

If you want to be absolutely certain that you're not passing up an argument that was provided, you can check the number of arguments passed to the function:

...
if (arguments.length < 1) {
  bar = 5
}
...

Which means that you can successfully pass undefined as an argument while also choosing to use a different default:

function foo(bar) {
  if (arguments.length < 1) {
    bar = 5
  }
  console.log(bar)
}

foo()          // 5
foo(undefined) // undefined
foo(null)      // null
foo(1)         // 1

---

If you have multiple parameters, you may want to use multiple defaults. I've recently found a use case for fallthrough on a switch statement, although the utility is questionable:

function foo(bar, baz, fizz, buzz) {
  switch (arguments.length) {
    case 0:
      bar = 1;
      //continue; might as well point out that implicit fall-through is desired
    case 1:
      baz = 2;
      //continue;
    case 2:
      fizz = 3;
      //continue;
    case 3:
      buzz = 4;
      //continue;
  }
  console.log(bar, baz, fizz, buzz)
}

foo()               // 1 2 3 4
foo(10)             // 10 2 3 4
foo(10, 20)         // 10 20 3 4
foo(10, 20, 30)     // 10 20 30 4
foo(10, 20, 30, 40) // 10 20 30 40

Answer 2

你可以做：

name = name || 'default';

这表示如果name未定义或虚假（null, 0, "", false, {}, []），请将其设置为"default".

js(h|l)int 会抱怨它，但它至少可以追溯到 IE7。它根本不是无效代码，也不是依赖于一些未记录的行为。

Answer 3

正确的检查方法是

if (typeof name === "undefined") {
    // ...
}

当然，调用者仍然可以通过调用来“愚弄”您getName(undefined)，当提供了参数但检查仍会将其标记为未提供时。但这确实是一种病态的情况。

Answer 4

是的，这适用于所有浏览器，但如果您想查看它是否已定义，您可以使用：

function getName(name){
   name = typeof(name) !== "undefined" ? name : "default";
   return name;
}