javascript - The issue about closures in js. I struggled with it last night and am almost dead now.

Question

Because I just started learning about closures, I didn’t understand many things. How do you get undefined in the console as shown in the picture? I only executed the return function, why are there two execution results? Guys please explain in detail~

Answer 1

You can repeat the following two pieces of code.
var result = f1(); The variable points to the function
console.log(result()) In fact, it can be converted to f1()()
That is The function returned by f1() is f2(), so the f2() function under f1() will be executed first, and then f1()
, so first console.log(n ) That is, 1
return f2 when executing function f1(), but since the function does not return a value, it prints out undefined

Answer 2

Essentially

var n = 1;
function f2() {
    console.log(n);
}

console.log(f2())

Because your f2 does not return a value, so it is undefined

Answer 3

console.log(result())
First output 1, because result() calls f2()
Then output is undefined, because result() has no return value

Answer 4

http://www.liaoxuefeng.com/wi...

I suggest you read this

Answer 5

First result=f1(); At this time, result=f2;
Then console.log(result()); First execute result, which is f2, and print the value of n. Because you did not execute test, n is 1, so what is printed is 1 Then execute console.log(result()); because result() has no return value, it is undefined.

Answer 6

console.log(console.log()) must be undefined, big brother