Deliberately "locking" a JavaScript variable with a immediately-invoked function expression

Question

Deliberately "locking" a JavaScript variable with a immediately-invoked function expression

While browsing through a blog post here that discusses creating a web scraper using node.js, I stumbled upon an intriguing piece of javascript that has left me somewhat perplexed. This particular snippet of code seems like something I could implement in my own scripts, but as a newcomer, I want to understand its functionality before blindly incorporating it.

Take a look at this function:

function main()
{
    var a = 1;
    var f = function() { console.log(a); }
    a = 2;
    f();
}
main();

In the above code, the output is 2 because var a is updated before calling f().

On the other hand, consider this function:

function main()
{
    var a = 1;
    var f = ( function(a) { return function() { console.log(a); } } )(a);
    a = 2;
    f();
}
main();

In this case, the output is 1. The blog post linked earlier provides a detailed explanation, but I'm still struggling to grasp the concept completely.

The blog mentions something about the scope of var a being passed into the function – can someone shed more light on this? Furthermore, why is it necessary to include the final (a) at the end of the var f function?

javascript

Answer 1

Answer №1

Understanding the code might be clearer when presented in the following way:

function main()
{
    var a = 1;
    var f = ( function(b) { return function() { console.log(b); } } )(a);
    a = 2;
    f();
}
main();

What's happening here is known as variable shadowing - where the parameter of the function is named the same as the variable, essentially hiding it from the function scope. In this example, without shadowing the variable, the code is straightforward. We define a function (1) that returns another function (2), which prints the value passed to function (1). We pass the current value of 'a' to the function, resulting in:

var f = ( function(b) { return function() { console.log(b); } } )(1);

or

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

Answer 2

Understanding the code might be clearer when presented in the following way:

function main()
{
    var a = 1;
    var f = ( function(b) { return function() { console.log(b); } } )(a);
    a = 2;
    f();
}
main();

What's happening here is known as variable shadowing - where the parameter of the function is named the same as the variable, essentially hiding it from the function scope. In this example, without shadowing the variable, the code is straightforward. We define a function (1) that returns another function (2), which prints the value passed to function (1). We pass the current value of 'a' to the function, resulting in:

var f = ( function(b) { return function() { console.log(b); } } )(1);

or

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

Answer 3

Answer №2

JavaScript provides two ways to access variables within a function:

Variables can be accessed when they are within the same scope where the function is declared.
Variables can also be passed as arguments when the function is called.

When a variable is passed as an argument, it takes precedence over any variable with the same name in the outer scope. This means that changes to the outer variable do not affect the inner variable.

In the provided code snippet, the inner function f is immediately invoked within the outer function, ensuring that the variable a maintains a consistent value.

It's important to note that objects are more difficult to shadow than scalar values. Consider the following example:

var a = { x: 1 },
f = (function(a) { 
    return function() {
        console.log(a);
    };
}(a));

a.x = 123;
f(); // Object { x: 123 }

Despite preserving the value of a, any changes made to its properties are reflected inside the f() function. However, if the variable a is completely reassigned, a different outcome will occur:

a = { x: 456 };
f(); // Object { x: 456 }

In this case, the previous value of a remains intact, while the variable outside of the function is assigned a new value.

Answer 4

JavaScript provides two ways to access variables within a function:

Variables can be accessed when they are within the same scope where the function is declared.
Variables can also be passed as arguments when the function is called.

When a variable is passed as an argument, it takes precedence over any variable with the same name in the outer scope. This means that changes to the outer variable do not affect the inner variable.

In the provided code snippet, the inner function f is immediately invoked within the outer function, ensuring that the variable a maintains a consistent value.

It's important to note that objects are more difficult to shadow than scalar values. Consider the following example:

var a = { x: 1 },
f = (function(a) { 
    return function() {
        console.log(a);
    };
}(a));

a.x = 123;
f(); // Object { x: 123 }

Despite preserving the value of a, any changes made to its properties are reflected inside the f() function. However, if the variable a is completely reassigned, a different outcome will occur:

a = { x: 456 };
f(); // Object { x: 456 }

In this case, the previous value of a remains intact, while the variable outside of the function is assigned a new value.

Answer 5

Answer №3

When we talk about the scope of a variable, we're essentially referring to the boundary within which that variable can operate. Returning a closure is like taking a snapshot of the environment in which a function exists. It's like putting everything inside a bubble, where nothing from the outside can interfere with what's on the inside. In this scenario, the scope of variable 'a' is limited to just that one function. Closures are a valuable concept and one of the reasons why I find JavaScript so appealing.

Answer 6

When we talk about the scope of a variable, we're essentially referring to the boundary within which that variable can operate. Returning a closure is like taking a snapshot of the environment in which a function exists. It's like putting everything inside a bubble, where nothing from the outside can interfere with what's on the inside. In this scenario, the scope of variable 'a' is limited to just that one function. Closures are a valuable concept and one of the reasons why I find JavaScript so appealing.

Answer 7

Answer №4

Consider the following function implementation in the clearest way possible:

   var f = (function(a) { 
       return function() { 
         console.log(a); } 
       } 
    )(a);

When you wrap the function inside parentheses, it causes the function to be directly executed. This means that the function will execute after assigning the value of 1 to a, but before assigning the value of 2.

The (a) following the wrap represents the parameter passed to the function.

Therefore, the value passed to the function is a=1.

Within this function, you are returning another function that will log the value of a (the parameter), which is 1.

By assigning this to the variable f, you are preserving the state of the variables within the function at the moment it was executed, before assigning a=2.

Answer 8

Consider the following function implementation in the clearest way possible:

   var f = (function(a) { 
       return function() { 
         console.log(a); } 
       } 
    )(a);

When you wrap the function inside parentheses, it causes the function to be directly executed. This means that the function will execute after assigning the value of 1 to a, but before assigning the value of 2.

The (a) following the wrap represents the parameter passed to the function.

Therefore, the value passed to the function is a=1.

Within this function, you are returning another function that will log the value of a (the parameter), which is 1.

By assigning this to the variable f, you are preserving the state of the variables within the function at the moment it was executed, before assigning a=2.

Answer 9

Answer №5

function start()
{

   var x = ( function(b) { return function() { console.log(b); } } )(b);
   b = 3;
   x();
}
start();

var b = 1; // Introducing a basic variable. var x =

( function(b) { return function() { console.log(b); } } )(b);

Here, x is a self-executing function with b as input. When executed, a function is returned, taking b as an input variable. It's important to note that a copy of b is sent, not b itself. This essentially extends the scope of b for use beyond the function's execution. However, as per the function x, it was passed a parameter without knowing that b has been changed externally. This is because a copy of b was passed, not b. Thus, it extends the scope of the copied b.

This is the underlying mechanism at play. For further insight, delve into the concept of "Currying" in JavaScript.

Answer 10

function start()
{

   var x = ( function(b) { return function() { console.log(b); } } )(b);
   b = 3;
   x();
}
start();

var b = 1; // Introducing a basic variable. var x =

( function(b) { return function() { console.log(b); } } )(b);

Here, x is a self-executing function with b as input. When executed, a function is returned, taking b as an input variable. It's important to note that a copy of b is sent, not b itself. This essentially extends the scope of b for use beyond the function's execution. However, as per the function x, it was passed a parameter without knowing that b has been changed externally. This is because a copy of b was passed, not b. Thus, it extends the scope of the copied b.

This is the underlying mechanism at play. For further insight, delve into the concept of "Currying" in JavaScript.

Deliberately "locking" a JavaScript variable with a immediately-invoked function expression

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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