The conundrum of JavaScript closures

Question

The conundrum of JavaScript closures

function f1(){
    var n=999;

    nAdd=function(){n+=1;};

    function f2(){
        alert(n);
    }
    return f2;
}

var result = f1();
var result2 = f1();

result();  // Output: 999
nAdd();
result2(); // Output: 1000
result2(); // Output: 1000
result();  // Output: 999

I am currently studying JavaScript closures, and the code snippet above has left me feeling perplexed. The initial call to result() displays 999, which is understandable.

After invoking nAdd(), the subsequent call to result2() yields 1000. This behavior can be attributed to both result2() and result() pointing to the same instance of f1().

However, I am puzzled as to why the final call to result() outputs 999 instead of 1000.

javascript closures

Answer 1

Answer №1

Whenever f1() is invoked, a new closure is generated with its own unique local n variable.

Nevertheless, the global nAdd variable gets overwritten each time f1() is called, resulting in nAdd() only incrementing the n variable within the most recent closure.

UPDATE: To independently increase the values of n in each closure, you can implement something like this:

function f1(){
    var n=999;
    return {
        incrementN : function(){n+=1;},
        getN : function f2(){console.log(n);}
    }
}    
var result = f1();
var result2 = f1();
result.getN(); // 999
result.incrementN();
result2.getN();//999
result2.incrementN();
result2.getN();//1000
result.getN();//1000

In this scenario, f1() returns an object with two methods that are not globally declared and both interact with the local n variable specific to their respective closure.

Answer 2

Whenever f1() is invoked, a new closure is generated with its own unique local n variable.

Nevertheless, the global nAdd variable gets overwritten each time f1() is called, resulting in nAdd() only incrementing the n variable within the most recent closure.

UPDATE: To independently increase the values of n in each closure, you can implement something like this:

function f1(){
    var n=999;
    return {
        incrementN : function(){n+=1;},
        getN : function f2(){console.log(n);}
    }
}    
var result = f1();
var result2 = f1();
result.getN(); // 999
result.incrementN();
result2.getN();//999
result2.incrementN();
result2.getN();//1000
result.getN();//1000

In this scenario, f1() returns an object with two methods that are not globally declared and both interact with the local n variable specific to their respective closure.

Answer 3

Answer №2

While the existing answers are solid, I believe incorporating a visual aid would enhance comprehension.

Answer 4

While the existing answers are solid, I believe incorporating a visual aid would enhance comprehension.

Answer 5

Answer №3

Every time you invoke the function f1(), several actions are taken:

A new local variable named n is created with a value of 999
An unnamed function is generated and assigned to the global variable nAdd, which updates the value of n (replacing any existing function stored in nAdd)
A new function is constructed and returned, which displays an alert containing the current value of n

Since you call f1() twice, these steps are repeated twice. The second invocation replaces the previous function assigned to nAdd with one that alters the value of the second n.

After this process, your setup includes:

result() triggering an alert displaying the value of the first n
result2() prompting an alert showing the value of the second n
nAdd() incrementing the value of the second n

When result() is executed at the end, it alerts 999 since it retrieves the unaltered value of the first n.

Answer 6

Every time you invoke the function f1(), several actions are taken:

A new local variable named n is created with a value of 999
An unnamed function is generated and assigned to the global variable nAdd, which updates the value of n (replacing any existing function stored in nAdd)
A new function is constructed and returned, which displays an alert containing the current value of n

Since you call f1() twice, these steps are repeated twice. The second invocation replaces the previous function assigned to nAdd with one that alters the value of the second n.

After this process, your setup includes:

result() triggering an alert displaying the value of the first n
result2() prompting an alert showing the value of the second n
nAdd() incrementing the value of the second n

When result() is executed at the end, it alerts 999 since it retrieves the unaltered value of the first n.

Answer 7

Answer №4

result and result2 store the outcomes of separate calls to f1, resulting in distinct instances of the local variable n. The values of local variables within a function can vary with each call, even without closures being part of the equation.

Answer 8

result and result2 store the outcomes of separate calls to f1, resulting in distinct instances of the local variable n. The values of local variables within a function can vary with each call, even without closures being part of the equation.

Answer 9

Answer №5

The line of code nAdd=function(){n+=1;}; establishes a global function that acts as a closure within the f1() function. A closure in JavaScript has access to all variables in the scope where it was created. Therefore, each time you invoke f1(), a new instance of nAdd() is generated with the value of n being linked to the value of the variable var n specific to that particular call of f1().

In the provided code snippet:

var result = f1();
var result2 = f1();
result(); // Produces output: 999
nAdd();         // This was created by invoking "var result2 = f1();" and shares the same 'n' value as the function in result2
result2();//Produces output: 1000
result2();//Produces output: 1000
result();//Produces output: 999

Answer 10

The line of code nAdd=function(){n+=1;}; establishes a global function that acts as a closure within the f1() function. A closure in JavaScript has access to all variables in the scope where it was created. Therefore, each time you invoke f1(), a new instance of nAdd() is generated with the value of n being linked to the value of the variable var n specific to that particular call of f1().

In the provided code snippet:

var result = f1();
var result2 = f1();
result(); // Produces output: 999
nAdd();         // This was created by invoking "var result2 = f1();" and shares the same 'n' value as the function in result2
result2();//Produces output: 1000
result2();//Produces output: 1000
result();//Produces output: 999

Answer 11

Answer №6

Two distinct closures are formed by the result and result2, each with its own value of n. By defining n as a global variable outside of the f1() function, you ensure that both results access the same global n:

var n = 999; function f1(){
nAdd = function(){n+=1;};
function f2(){
console.log(n);
}
return f2;
}
var result = f1();
var result2 = f1();
result(); // 999
nAdd();
result2();//1000
result2();//1000
result();//1000

Answer 12

Two distinct closures are formed by the result and result2, each with its own value of n. By defining n as a global variable outside of the f1() function, you ensure that both results access the same global n:

var n = 999; function f1(){
nAdd = function(){n+=1;};
function f2(){
console.log(n);
}
return f2;
}
var result = f1();
var result2 = f1();
result(); // 999
nAdd();
result2();//1000
result2();//1000
result();//1000

Answer 13

Answer №7

Here is how it works：

var nAdd;
function f1(){
    var num=999;

    nAdd=function(){num+=1;};

    function f2(){
        alert(num);
    }
    return f2;
}

var result = f1();//var nAdd=function(){num+=1;} num=result.num=999

var result2 = f1();//var nAdd=function(){num+=1;} num=result2.num=999

var result3 = f1();//var nAdd=function(){num+=1;} num=result3.num=999

nAdd();

result();  // 999

result2(); // 999

result3(); // 1000

var r1 = f1();//var nAdd=function(){num+=1;} num=r1.num=999

var r2 = f1();//var nAdd=function(){num+=1;} num=r2.num=999

nAdd();

var r3 = f1();//var nAdd=function(){num+=1;} num=r3.num=999

r1();  // 999

r2(); // 1000

r3(); // 999

var res1 = f1();//var nAdd=function(){num+=1;} num=res1.num=999

var res2 = f1();//var nAdd=function(){num+=1;} num=res2.num=999

nAdd();

var res3 = f1();//var nAdd=function(){num+=1;} num=res3.num=999
nAdd(); 
nAdd();
nAdd();   

res1();  // 999

res2(); // 1000

res3(); // 1002

Answer 14

Here is how it works：

var nAdd;
function f1(){
    var num=999;

    nAdd=function(){num+=1;};

    function f2(){
        alert(num);
    }
    return f2;
}

var result = f1();//var nAdd=function(){num+=1;} num=result.num=999

var result2 = f1();//var nAdd=function(){num+=1;} num=result2.num=999

var result3 = f1();//var nAdd=function(){num+=1;} num=result3.num=999

nAdd();

result();  // 999

result2(); // 999

result3(); // 1000

var r1 = f1();//var nAdd=function(){num+=1;} num=r1.num=999

var r2 = f1();//var nAdd=function(){num+=1;} num=r2.num=999

nAdd();

var r3 = f1();//var nAdd=function(){num+=1;} num=r3.num=999

r1();  // 999

r2(); // 1000

r3(); // 999

var res1 = f1();//var nAdd=function(){num+=1;} num=res1.num=999

var res2 = f1();//var nAdd=function(){num+=1;} num=res2.num=999

nAdd();

var res3 = f1();//var nAdd=function(){num+=1;} num=res3.num=999
nAdd(); 
nAdd();
nAdd();   

res1();  // 999

res2(); // 1000

res3(); // 1002

The conundrum of JavaScript closures

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

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