How would you explain JavaScript closures to someone with a knowledge of the concepts they consist of (for example functions, variables and the like), but does not understand closures themselves?
I have seen the Scheme example given on Wikipedia, but unfortunately it did not help.
转载于:https://stackoverflow.com/questions/111102/how-do-javascript-closures-work
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Submitted by Morris on Tue, 2006-02-21 10:19. Community-edited since.
This page explains closures so that a programmer can understand them — using working JavaScript code. It is not for gurus or functional programmers.
Closures are not hard to understand once the core concept is grokked. However, they are impossible to understand by reading any theoretical or academically oriented explanations!
This article is intended for programmers with some programming experience in a mainstream language, and who can read the following JavaScript function:
function sayHello(name) {
var text = 'Hello ' + name;
var say = function() { console.log(text); }
say();
}
sayHello('Joe');When a function (foo) declares other functions (bar and baz), the family of local variables created in foo is not destroyed when the function exits. The variables merely become invisible to the outside world. Foo can therefore cunningly return the functions bar and baz, and they can continue to read, write and communicate with each other through this closed-off family of variables ("the closure") that nobody else can meddle with, not even someone who calls foo again in future.
A closure is one way of supporting first-class functions; it is an expression that can reference variables within its scope (when it was first declared), be assigned to a variable, be passed as an argument to a function, or be returned as a function result.
The following code returns a reference to a function:
function sayHello2(name) {
var text = 'Hello ' + name; // Local variable
var say = function() { console.log(text); }
return say;
}
var say2 = sayHello2('Bob');
say2(); // logs "Hello Bob"Most JavaScript programmers will understand how a reference to a function is returned to a variable (say2) in the above code. If you don't, then you need to look at that before you can learn closures. A programmer using C would think of the function as returning a pointer to a function, and that the variables say and say2 were each a pointer to a function.
There is a critical difference between a C pointer to a function and a JavaScript reference to a function. In JavaScript, you can think of a function reference variable as having both a pointer to a function as well as a hidden pointer to a closure.
The above code has a closure because the anonymous function function() { console.log(text); } is declared inside another function, sayHello2() in this example. In JavaScript, if you use the function keyword inside another function, you are creating a closure.
In C and most other common languages, after a function returns, all the local variables are no longer accessible because the stack-frame is destroyed.
In JavaScript, if you declare a function within another function, then the local variables of the outer function can remain accessible after returning from it. This is demonstrated above, because we call the function say2() after we have returned from sayHello2(). Notice that the code that we call references the variable text, which was a local variable of the function sayHello2().
function() { console.log(text); } // Output of say2.toString();
Looking at the output of say2.toString(), We can see that the code refers to the variable text. The anonymous function can reference text which holds the value 'Hello Bob' because the local variables of sayHello2() have been secretly kept alive in a closure.
The genius is that in JavaScript a function reference also has a secret reference to the closure it was created in — similar to how delegates are a method pointer plus a secret reference to an object.
For some reason, closures seem really hard to understand when you read about them, but when you see some examples, it becomes clear how they work (it took me a while). I recommend working through the examples carefully until you understand how they work. If you start using closures without fully understanding how they work, you would soon create some very weird bugs!
This example shows that the local variables are not copied — they are kept by reference. It is as though the stack-frame stays alive in memory even after the outer function exists!
function say667() {
// Local variable that ends up within closure
var num = 42;
var say = function() { console.log(num); }
num++;
return say;
}
var sayNumber = say667();
sayNumber(); // logs 43All three global functions have a common reference to the same closure because they are all declared within a single call to setupSomeGlobals().
var gLogNumber, gIncreaseNumber, gSetNumber;
function setupSomeGlobals() {
// Local variable that ends up within closure
var num = 42;
// Store some references to functions as global variables
gLogNumber = function() { console.log(num); }
gIncreaseNumber = function() { num++; }
gSetNumber = function(x) { num = x; }
}
setupSomeGlobals();
gIncreaseNumber();
gLogNumber(); // 43
gSetNumber(5);
gLogNumber(); // 5
var oldLog = gLogNumber;
setupSomeGlobals();
gLogNumber(); // 42
oldLog() // 5The three functions have shared access to the same closure — the local variables of setupSomeGlobals() when the three functions were defined.
Note that in the above example, if you call setupSomeGlobals() again, then a new closure (stack-frame!) is created. The old gLogNumber, gIncreaseNumber, gSetNumber variables are overwritten with new functions that have the new closure. (In JavaScript, whenever you declare a function inside another function, the inside function(s) is/are recreated again each time the outside function is called.)
This example shows that the closure contains any local variables that were declared inside the outer function before it exited. Note that the variable alice is actually declared after the anonymous function. The anonymous function is declared first, and when that function is called it can access the alice variable because alice is in the same scope (JavaScript does variable hoisting).
Also sayAlice()() just directly calls the function reference returned from sayAlice() — it is exactly the same as what was done previously but without the temporary variable.
function sayAlice() {
var say = function() { console.log(alice); }
// Local variable that ends up within closure
var alice = 'Hello Alice';
return say;
}
sayAlice()();// logs "Hello Alice"Tricky: also note that the say variable is also inside the closure, and could be accessed by any other function that might be declared within sayAlice(), or it could be accessed recursively within the inside function.
This one is a real gotcha for many people, so you need to understand it. Be very careful if you are defining a function within a loop: the local variables from the closure may not act as you might first think.
You need to understand the "variable hoisting" feature in Javascript in order to understand this example.
function buildList(list) {
var result = [];
for (var i = 0; i < list.length; i++) {
var item = 'item' + i;
result.push( function() {console.log(item + ' ' + list[i])} );
}
return result;
}
function testList() {
var fnlist = buildList([1,2,3]);
// Using j only to help prevent confusion -- could use i.
for (var j = 0; j < fnlist.length; j++) {
fnlist[j]();
}
}
testList() //logs "item2 undefined" 3 timesThe line result.push( function() {console.log(item + ' ' + list[i])} adds a reference to an anonymous function three times to the result array. If you are not so familiar with anonymous functions think of it like:
pointer = function() {console.log(item + ' ' + list[i])};
result.push(pointer);
Note that when you run the example, "item2 undefined" is logged three times! This is because just like previous examples, there is only one closure for the local variables for buildList (which are result, i and item). When the anonymous functions are called on the line fnlist[j](); they all use the same single closure, and they use the current value for i and item within that one closure (where i has a value of 3 because the loop had completed, and item has a value of 'item2'). Note we are indexing from 0 hence item has a value of item2. And the i++ will increment i to the value 3.
It may be helpful to see what happens when a block-level declaration of the variable item is used (via the let keyword) instead of a function-scoped variable declaration via the var keyword. If that change is made, then each anonymous function in the array result has its own closure; when the example is run the output is as follows:
item0 undefined
item1 undefined
item2 undefined
If the variable i is also defined using let instead of var, then the output is:
item0 1
item1 2
item2 3
In this final example, each call to the main function creates a separate closure.
function newClosure(someNum, someRef) {
// Local variables that end up within closure
var num = someNum;
var anArray = [1,2,3];
var ref = someRef;
return function(x) {
num += x;
anArray.push(num);
console.log('num: ' + num +
'; anArray: ' + anArray.toString() +
'; ref.someVar: ' + ref.someVar + ';');
}
}
obj = {someVar: 4};
fn1 = newClosure(4, obj);
fn2 = newClosure(5, obj);
fn1(1); // num: 5; anArray: 1,2,3,5; ref.someVar: 4;
fn2(1); // num: 6; anArray: 1,2,3,6; ref.someVar: 4;
obj.someVar++;
fn1(2); // num: 7; anArray: 1,2,3,5,7; ref.someVar: 5;
fn2(2); // num: 8; anArray: 1,2,3,6,8; ref.someVar: 5;If everything seems completely unclear, then the best thing to do is to play with the examples. Reading an explanation is much harder than understanding examples. My explanations of closures and stack-frames, etc. are not technically correct — they are gross simplifications intended to help to understand. Once the basic idea is grokked, you can pick up the details later.
function inside another function, a closure is used.eval() inside a function, a closure is used. The text you eval can reference local variables of the function, and within eval you can even create new local variables by using eval('var foo = …')
new Function(…) (the Function constructor) inside a function, it does not create a closure. (The new function cannot reference the local variables of the outer function.)myFunction = Function(myFunction.toString().replace(/Hello/,'Hola'));), it won't work if myFunction is a closure (of course, you would never even think of doing source code string substitution at runtime, but...).If you have just learned closures (here or elsewhere!), then I am interested in any feedback from you about any changes you might suggest that could make this article clearer. Send an email to morrisjohns.com (morris_closure @). Please note that I am not a guru on JavaScript — nor on closures.
Original post by Morris can be found in the Internet Archive.
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