问题

I'm coming from C#/PHP and trying to get my head around Javascript's idea that functions are variables/objects and have quasi constructors, etc.

Can anyone explain why the following code functions as it does, namely:

Why isn't "2" displayed when instantiating the variable/function test?
Why isn't "1" displayed when executing the variable/function test?

code:

var setup = function () {
    console.log(1);
    return function() {
        console.log(2);
    };
};

var test = setup(); // 1
test(); // 2
test(); // 2
test(); // 2

added:

Thanks @thejh @Justin so the function is returning a completely different function that has nothing to do with the first (I was thinking of the second function as a kind of constructor of the first), if I comment it out, it's clearer:

$(document).ready(function() {

    var setup = function () {
        console.log(1);
//        return function() {
//            console.log(2);
//        };
    };

    var test = setup(); // 1
    test(); // "test is not a function"
    test(); // "test is not a function"
    test(); // "test is not a function"

});

回答1:

You're only calling setup() the first time. Once it is called, the new function it returns is assigned to test. From there on, you're calling that new function:

// calls setup which logs 1 and returns a new function.
// setup also returns a new function and assigns that new function to test.
var test = setup(); 

// test now is the equivalent of var test = function(){ console.log(2); };

// call the new function that setup returned which logs 2
test();

// and again
test();

// and again
test();

回答2:

Because you're returning a different function from what gets called when creating it (the one in the return, one you're not calling on the first line)...that's what gets executed on the other calls. For example this would give you 1 then 2:

var test = setup()(); // 1, 2

You can test it here.

回答3:

In your first assignment of test to setup(), you're executing the whole function, so console.log(1) runs, then returns a new function.

Now, you're assigning the return value of that first function to be the next function that runs console.log(2), so now test references that returned function.

Your subsequent calls just run that function which runs console.log(2)

回答4:

Calling setup() prints 1 and returns a reference to that anonymous function. Calling that prints 2.

回答5:

Hello when you do setup() you execute this two lines :

console.log(1);
function() { console.log(2); }

the first line log "1"
the second line doesn't log any thing, because the function is not called, it only create an object (and this object is a function).

when you do test() you actually call the function you've created and so you log "2"

JS is a functional language, you have to see functions like first citizen object.

回答6:

Imagine instead of

var setup = function () {
    console.log(1);
    return function() {
        console.log(2);
    }
}

this equivalent OOP syntax

var setup = new Function("\
    console.log(1);\
    return new Function(\"\
        console.log(2);\
    \");\
");
var test = setup(); // 1
test(); // 2
test(); // 2
test(); // 2

I think things should be more familiar to you now.

来源：https://stackoverflow.com/questions/4462497/why-does-a-javascript-function-act-differently-upon-instantiation-than-execution

标签

javascript

functional-programming

Why does a Javascript function act differently upon instantiation than execution?

问题