我理解闭包定义为:

[A] stack-frame which is not deallocated when the function returns. (as if a 'stack-frame' were malloc'ed instead of being on the stack!)

但我不明白这个答案如何适合 JavaScript 的存储机制。解释器如何跟踪这些值？浏览器的存储机制是不是像Heap和Stack一样分段的？

这个问题的答案:How do JavaScript closures work?解释说:

[A] function reference also has a secret reference to the closure

这个神秘的“ secret 引用”背后的底层机制是什么？

编辑 许多人说这取决于实现，因此为了简单起见，请在特定实现的上下文中提供解释。

最佳答案

这是slebetman's answer的一部分问题javascript can't access private properties这很好地回答了你的问题。

The Stack:

A scope is related to the stack frame (in Computer Science it's called the "activation record" but most developers familiar with C or assembly know it better as stack frame). A scope is to a stack frame what a class is to an object. By that I mean that where an object is an instance of a class, a stack frame is an instance of scope.

Let's use a made-up language as an example. In this language, like in javascript, functions define scope. Lets take a look at an example code:

var global_var

function b {
    var bb
}

function a {
    var aa
    b();
}

When we read the code above, we say that the variable aa is in scope in function a and the variable bb is in scope in function b. Note that we don't call this thing private variables. Because the opposite of private variables are public variables and both refer to properties bound to objects. Instead we call aa and bb local variables. The opposite of local variables are global variables (not public variables).

Now, let's see what happens when we call a:

a() gets called, create a new stack frame. Allocate space for local variables on the stack:

The stack:
 ┌────────┐
 │ var aa │ <── a's stack frame
 ╞════════╡
 ┆        ┆ <── caller's stack frame

a() calls b(), create a new stack frame. Allocate space for local variables on the stack:

The stack:
 ┌────────┐
 │ var bb │ <── b's stack frame
 ╞════════╡
 │ var aa │
 ╞════════╡
 ┆        ┆

In most programming languages, and this includes javascript, a function only has access to its own stack frame. Thus a() cannot access local variables in b() and neither can any other function or code in global scope access variables in a(). The only exception are variables in global scope. From an implementation point of view this is achieved by allocating global variables in an area of memory that does not belong to the stack. This is generally called the heap. So to complete the picture the memory at this point looks like this:

The stack:     The heap:
 ┌────────┐   ┌────────────┐
 │ var bb │   │ global_var │
 ╞════════╡   │            │
 │ var aa │   └────────────┘
 ╞════════╡
 ┆        ┆

(as a side note, you can also allocate variables on the heap inside functions using malloc() or new)

Now b() completes and returns, it's stack frame is removed from the stack:

The stack:     The heap:
 ┌────────┐   ┌────────────┐
 │ var aa │   │ global_var │
 ╞════════╡   │            │
 ┆        ┆   └────────────┘

and when a() completes the same happens to its stack frame. This is how local variables gets allocated and freed automatically - via pushing and popping objects off the stack.

Closures:

A closure is a more advanced stack frame. But whereas normal stack frames gets deleted once a function returns, a language with closures will merely unlink the stack frame (or just the objects it contains) from the stack while keeping a reference to the stack frame for as long as it's required.

Now let's look at an example code of a language with closures:

function b {
    var bb
    return function {
        var cc
    }
}

function a {
    var aa
    return b()
}

Now let's see what happens if we do this:

var c = a()

First function a() is called which in turn calls b(). Stack frames are created and pushed onto the stack:

The stack:
 ┌────────┐
 │ var bb │
 ╞════════╡
 │ var aa │
 ╞════════╡
 │ var c  │
 ┆        ┆

Function b() returns, so it's stack frame is popped off the stack. But, function b() returns an anonymous function which captures bb in a closure. So we pop off the stack frame but don't delete it from memory (until all references to it has been completely garbage collected):

The stack:             somewhere in RAM:
 ┌────────┐           ┌╶╶╶╶╶╶╶╶╶┐
 │ var aa │           ┆ var bb  ┆
 ╞════════╡           └╶╶╶╶╶╶╶╶╶┘
 │ var c  │
 ┆        ┆

a() now returns the function to c. So the stack frame of the call to b() gets linked to the variable c. Note that it's the stack frame that gets linked, not the scope. It's kind of like if you create objects from a class it's the objects that gets assigned to variables, not the class:

The stack:             somewhere in RAM:
 ┌────────┐           ┌╶╶╶╶╶╶╶╶╶┐
 │ var c╶╶├╶╶╶╶╶╶╶╶╶╶╶┆ var bb  ┆
 ╞════════╡           └╶╶╶╶╶╶╶╶╶┘
 ┆        ┆

Also note that since we haven't actually called the function c(), the variable cc is not yet allocated anywhere in memory. It's currently only a scope, not yet a stack frame until we call c().

Now what happens when we call c()? A stack frame for c() is created as normal. But this time there is a difference:

The stack:
 ┌────────┬──────────┐
 │ var cc    var bb  │  <──── attached closure
 ╞════════╤──────────┘
 │ var c  │
 ┆        ┆

The stack frame of b() is attached to the stack frame of c(). So from the point of view of function c() it's stack also contains all the variables that were created when function b() was called (Note again, not the variables in function b() but the variables created when function b() was called - in other words, not the scope of b() but the stack frame created when calling b(). The implication is that there is only one possible function b() but many calls to b() creating many stack frames).

But the rules of local and global variables still applies. All variables in b() become local variables to c() and nothing else. The function that called c() has no access to them.

What this means is that when you redefine c in the caller's scope like this:

var c = function {/* new function */}

this happens:

                     somewhere in RAM:
                           ┌╶╶╶╶╶╶╶╶╶┐
                           ┆ var bb  ┆
                           └╶╶╶╶╶╶╶╶╶┘
The stack:
 ┌────────┐           ┌╶╶╶╶╶╶╶╶╶╶╶╶╶╶╶╶╶╶╶╶┐
 │ var c╶╶├╶╶╶╶╶╶╶╶╶╶╶┆ /* new function */ ┆
 ╞════════╡           └╶╶╶╶╶╶╶╶╶╶╶╶╶╶╶╶╶╶╶╶┘
 ┆        ┆

As you can see, it's impossible to regain access to the stack frame from the call to b() since the scope that c belongs to doesn't have access to it.

关于javascript - JavaScript 闭包如何在底层工作？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/31735129/