When a global variable is defined after a return statement within a function declaration, it does

Question

When a global variable is defined after a return statement within a function declaration, it does

Check out this Javascript code snippet: http://jsfiddle.net/ramchiranjeevi/63uML/

var foo = 1;
function bar() {
    foo = 10;
    return;
    function foo() {}   
}

bar();
console.log(foo);  // should return 10, but returns 1

It's interesting to note that even though the variable is overwritten within the bar() function, the log still prints the original value of 1 instead of 10.

javascript closures

Answer 1

Answer №1

Function declarations are "hoisted", which means they are moved to the top of the scope.

As a result, a new context for foo is created within the local scope. Any later assignment of 10 will affect this localized scope and not the variable from the parent scope.

A similar behavior can be observed when declaring a block-local variable named foo using the var keyword:

var foo = 1;
function bar() {
    var foo = 10; // effectively creates a function named foo in this scope
    return;
}

bar();
console.log(foo); // output: 1

http://jsfiddle.net/GRMule/8F5K3/

Another example elucidating this concept:

var foo = 1;
function bar() {
    console.log(foo); // due to hoisting, "function" will be the output
    foo = 10;         // overwrites the block-local foo
    return;
    function foo () {} // hoisted to the top of this scope, creating a new "foo" context
}

To prevent hoisting, you can use the var method while declaring functions, though it's recommended to maintain code readability by avoiding reusing names like this:

var foo = 1;
function bar() {
    console.log(foo); // undefined
    foo = 10;
    return;
    var foo = function () {};
}
bar();
console.log(foo); // 1

http://jsfiddle.net/znrG2/

Using the var word in a scope block causes the existence of that local context to be hoisted, not the value. Variables with the same name from the parent scope won't be accessible or affected in the current scope.

Functions and variables declared using this, such as this.foo = function () {};, do not hoist. Details here: http://jsfiddle.net/8F5K3/3/

Declaring functions or variables with this doesn't overwrite variable contexts from the parent scope. This can be utilized to circumvent this behavior when needing to reuse the name "foo":

var foo = 1;
function bar() {
    console.log(foo); // 1
    foo = 10;
    return;
    this.foo = function () {};
}
bar();
console.log(foo); // 10

http://jsfiddle.net/znrG2/1/

Additional Resources

Insight into function and variable hoisting in JavaScript -
Scope Cheatsheet on MDN - https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Scope_Cheatsheet
Functions and function scope on MDN - https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions_and_function_scope

Answer 2

Function declarations are "hoisted", which means they are moved to the top of the scope.

As a result, a new context for foo is created within the local scope. Any later assignment of 10 will affect this localized scope and not the variable from the parent scope.

A similar behavior can be observed when declaring a block-local variable named foo using the var keyword:

var foo = 1;
function bar() {
    var foo = 10; // effectively creates a function named foo in this scope
    return;
}

bar();
console.log(foo); // output: 1

http://jsfiddle.net/GRMule/8F5K3/

Another example elucidating this concept:

var foo = 1;
function bar() {
    console.log(foo); // due to hoisting, "function" will be the output
    foo = 10;         // overwrites the block-local foo
    return;
    function foo () {} // hoisted to the top of this scope, creating a new "foo" context
}

To prevent hoisting, you can use the var method while declaring functions, though it's recommended to maintain code readability by avoiding reusing names like this:

var foo = 1;
function bar() {
    console.log(foo); // undefined
    foo = 10;
    return;
    var foo = function () {};
}
bar();
console.log(foo); // 1

http://jsfiddle.net/znrG2/

Using the var word in a scope block causes the existence of that local context to be hoisted, not the value. Variables with the same name from the parent scope won't be accessible or affected in the current scope.

Functions and variables declared using this, such as this.foo = function () {};, do not hoist. Details here: http://jsfiddle.net/8F5K3/3/

Declaring functions or variables with this doesn't overwrite variable contexts from the parent scope. This can be utilized to circumvent this behavior when needing to reuse the name "foo":

var foo = 1;
function bar() {
    console.log(foo); // 1
    foo = 10;
    return;
    this.foo = function () {};
}
bar();
console.log(foo); // 10

http://jsfiddle.net/znrG2/1/

Additional Resources

Insight into function and variable hoisting in JavaScript -
Scope Cheatsheet on MDN - https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Scope_Cheatsheet
Functions and function scope on MDN - https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions_and_function_scope

Answer 3

Answer №2

Understanding JavaScript hoisting behavior is crucial. You can find more information about it in this resource. In the scenario you presented, the script is executed as follows:

var foo = 1;
function bar() {
    function foo() {}
    foo = 10;
    return; 
}

bar();
console.log(foo);

This happens because:

Function declarations and variable declarations are always effortlessly moved (“hoisted”) to the top of their containing scope by the JavaScript interpreter

In your case, when you set foo = 10;, you are actually assigning to the local variable. This explains why the global variable remains unchanged.

Answer 4

Understanding JavaScript hoisting behavior is crucial. You can find more information about it in this resource. In the scenario you presented, the script is executed as follows:

var foo = 1;
function bar() {
    function foo() {}
    foo = 10;
    return; 
}

bar();
console.log(foo);

This happens because:

Function declarations and variable declarations are always effortlessly moved (“hoisted”) to the top of their containing scope by the JavaScript interpreter

In your case, when you set foo = 10;, you are actually assigning to the local variable. This explains why the global variable remains unchanged.

Answer 5

Answer №3

When the function foo() is defined within the local scope of bar(), any redefinition of foo() within bar() will only affect the local scope variable and not the global scope.

Answer 6

When the function foo() is defined within the local scope of bar(), any redefinition of foo() within bar() will only affect the local scope variable and not the global scope.

Answer 7

Answer №4

function bar() {
  const baz = 10;  // let's create a local variable named "baz"
  return;    // the return value of this function is undefined
  function buzz() {}   // we define a function called "buzz" inside the bar function, which means that the global variable "baz" does not change.
}

Therefore, the global variable baz remains unaffected.

Answer 8

function bar() {
  const baz = 10;  // let's create a local variable named "baz"
  return;    // the return value of this function is undefined
  function buzz() {}   // we define a function called "buzz" inside the bar function, which means that the global variable "baz" does not change.
}

Therefore, the global variable baz remains unaffected.

When a global variable is defined after a return statement within a function declaration, it does

Answer №1

Answer №2

Answer №3

Answer №4

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