TLDR:简单定义的变量(没有var, let, const)可以用delete删除。如果你使用var, let, const -它们不能被delete或Reflect.deleteProperty删除。

Chrome 55:

simpleVar = "1";
"1"
delete simpleVar;
true
simpleVar;
VM439:1 Uncaught ReferenceError: simpleVar is not defined
    at <anonymous>:1:1
(anonymous) @ VM439:1
var varVar = "1";
undefined
delete varVar;
false
varVar;
"1"
let letVar = "1";
undefined
delete letVar;
true
letVar;
"1"
const constVar="1";
undefined
delete constVar;
true
constVar;
"1"
Reflect.deleteProperty (window, "constVar");
true
constVar;
"1"
Reflect.deleteProperty (window, "varVar");
false
varVar;
"1"
Reflect.deleteProperty (window, "letVar");
true
letVar;
"1"

Firefox Nightly 53.0a1显示相同的行为。

与简单的属性相比，变量具有属性[[可配置的]]，这意味着不可能通过删除操作符删除变量。

但是，有一个执行上下文不受此规则的影响。它是eval上下文:其中没有为变量设置[[Configurable]]属性。

delete操作符从对象中删除属性。它不能删除变量。所以这个问题的答案取决于全局变量或属性是如何定义的。

(1)如果是用var创建的，则不能删除。

例如:

var g_a = 1; //create with var, g_a is a variable
delete g_a; //return false
console.log(g_a); //g_a is still 1

(2)如果创建时没有使用var，则可以删除。

g_b = 1; //create without var, g_b is a property
delete g_b; //return true
console.log(g_b); //error, g_b is not defined

技术的解释

1. 使用var

在这种情况下，引用g_a是在ECMAScript规范中称为“VariableEnvironment”的地方创建的，它附加到当前作用域-这可能是在函数内部使用var的情况下的函数执行上下文(尽管当你考虑let时可能会变得有点复杂)，或者在“全局”代码的情况下，VariableEnvironment附加到全局对象(通常是窗口)。

VariableEnvironment中的引用通常是不可删除的——ECMAScript 10.5中详细解释了这一点，但只需说明，除非您的代码在eval上下文中执行(大多数基于浏览器的开发控制台使用)，否则使用var声明的变量是不能删除的。

2. 不使用var

When trying to assign a value to a name without using the var keyword, JavaScript tries to locate the named reference in what the ECMAScript spec calls "LexicalEnvironment", and the main difference is that LexicalEnvironments are nested - that is a LexicalEnvironment has a parent (what the ECMAScript spec calls "outer environment reference") and when JavaScript fails to locate the reference in a LexicalEnvironment, it looks in the parent LexicalEnvironment (as detailed in 10.3.1 and 10.2.2.1). The top level LexicalEnvironment is the "global environment", and that is bound to the global object in that its references are the global object's properties. So if you try to access a name that was not declared using a var keyword in the current scope or any outer scopes, JavaScript will eventually fetch a property of the window object to serve as that reference. As we've learned before, properties on objects can be deleted.

笔记

It is important to remember that var declarations are "hoisted" - i.e. they are always considered to have happened in the beginning of the scope that they are in - though not the value initialization that may be done in a var statement - that is left where it is. So in the following code, a is a reference from the VariableEnvironment and not the window property and its value will be 10 at the end of the code: function test() { a = 5; var a = 10; } The above discussion is when "strict mode" is not enabled. Lookup rules are a bit different when using "strict mode" and lexical references that would have resolved to window properties without "strict mode" will raise "undeclared variable" errors under "strict mode". I didn't really understand where this is specified, but its how browsers behave.

详见诺亚的回答

//Option A.) set to null
some_var = null;

//Option B.) set to undefined
some_var = undefined;

//Option C.) remove/delete the variable reference
delete obj.some_var
//if your variable was defined as a global, you'll need to
//qualify the reference with 'window'
delete window.some_var;

引用:

MDN删除API 在严格模式下删除非限定变量名时出现MDN SyntaxError

注意，delete操作成功时返回true。

2021年更新:在Chrome 88和Firefox 84上测试:

implicit_global = 1;
delete implicit_global; // true

window.explicit_global = 1;
delete explicit_global; // true

const _object = {property: 1};
delete _object.property; // true

function_set = function() {};
delete function_set; // true

function function_declaration() {};
delete function_declaration; // false

(function () {
    var _var = 1;
    console.log(delete _var); // false
    console.log(_var); // 1
})()

(function () {
    let _let = 1;
    console.log(delete _let); // false
    console.log(_let); // 1
})()

(function () {
    const _const = 1;
    console.log(delete _const); // false
    console.log(_const); // 1
})()

由于浏览器更新，此答案的先前编辑已不再相关。

如果隐式声明变量时不使用var，正确的方法是使用delete foo。

然而，在你删除它之后，如果你试图在一个操作中使用它，比如添加，一个ReferenceError将被抛出，因为你不能将一个字符串添加到一个未声明的，未定义的标识符。例子:

x = 5;
delete x
alert('foo' + x )
// ReferenceError: x is not defined

在某些情况下，将它赋值为false、null或undefined可能更安全，这样它就被声明了，不会抛出这种类型的错误。

foo = false

请注意，在ECMAScript中，null, false, undefined, 0, NaN或“都将计算为false。只要确保你不使用!==操作符，而是在类型检查布尔值时，你不想进行身份检查(因此null将== false和false == undefined)。

还要注意，delete并没有“删除”引用，而是直接删除对象上的属性，例如:

bah = {}, foo = {}; bah.ref = foo;

delete bah.ref;
alert( [bah.ref, foo ] )
// ,[object Object] (it deleted the property but not the reference to the other object)

如果你用var声明了一个变量，你不能删除它:

(function() {
    var x = 5;
    alert(delete x)
    // false
})();

在Rhino中:

js> var x
js> delete x
false

也不能删除一些预定义的属性，比如Math。PI:

js> delete Math.PI
false

就像任何语言一样，有一些奇怪的例外需要删除，如果你足够在乎，你应该读一下:

https://developer.mozilla.org/en/Core_JavaScript_1.5_Reference/Operators/Special_Operators/delete_Operator http://www.ecma-international.org/publications/files/ECMA-ST/ECMA-262.pdf