Just a quick and easy question which has been confusing me a lot, are function names like:
function drive() {
which are function names considered as variables, is drive considered as a variable?
Function declarations create a variable in the current scope which has the same name as the function and the value of which is the function.
Named function expressions create such a variable in their own scope (not the current scope).
Anonymous function expressions and arrow functions do not create variables.
yes.
drive will be a variable whose value is a named function with the same name. The compiler declares the variable for you, gives it a named function as the value of it -and with the same name as the variable-, and also hoists the variable.
This is done by the compiler as you declare a function.
So, as with any other variable, you can rewrite it:
function bar(){
return 'bar';
}
bar = function(){
return 'foo';
}
console.log(bar() );//foo
However, due to hoisting, changing the order doesn't changes the result:
bar = function(){
return 'foo';
}
function bar(){
return 'bar';
}
console.log(bar() );//also foo
Which doesn't mean that the function bar()... block isn't creating a bar variable, but just that it is created before the bar = function... statement.
You can prevent the compiler to create the variable just by wrapping the function declaration with ():
( function bar(){
return 'foo';
});
bar(); //error
Which is really common to launch IIFEs:
( function bar(){
alert('foo');
})();
//this will launch the alert in a browser.
Related
So I am confused as to when I would use an anonymous function such as:
let foo = function () {
//code
}
versus a named anonymous function such as:
let foo = function foo () {
//code
}
Besides browser support, namely IE, are there any differences between the two? When should I use one over the other?
In this case, where the function declaration name is the same as the variable it is assigned to, it doesn't make much difference.
If you used a different name for the definition and assignment, the name on the right takes precedence in naming the function:
foo = function bar() {}
foo.name // "bar"
In both cases you assign your function to a variable (function expression), but in the first case you assign an unnamed/anonymous function, whereas in the second case you assign a named function. When assigning an anonymous function to a variable in such a simple expression, the JS engine is able to name the function properly.
Consider the following case where this assignment is non-obvious for the engine:
function p(fun) { return fun; }
foo = p(function() {})
foo.name // empty string
TL;DR; with named functions you often get better stack traces.
I'm not really a javascript noob at all, although in my whole life I've never came across this, but am I right in assuming that javascript must assign functions before running anything or something?
In all my experience, I expected this to return 'undefined', but obviously it returns 'function'.
function bar() {
return foo;
foo = 10;
function foo() {}
var foo = '11';
}
alert(typeof bar());
Is someone able to explain this for me?
This behaviour of JavaScript is called hoisting. There is a good explanation on the MDN (https://developer.mozilla.org/en-US/docs/Glossary/Hoisting)
In JavaScript, functions and variables are hoisted. Hoisting is JavaScript's behavior of moving declarations to the top of a scope (the global scope or the current function scope).
That means that you are able to use a function or a variable before it has been declared, or in other words: a function or variable can be declared after it has been used already.
Basically, if you declare a variable like this:
console.log(s); // s === undefined
var s = 'some string';
s declaration will be "hoisted" to the beginning of the scope (i.e. there will be no ReferenceError on the line with console.log). The variable's value will not be defined at that moment though.
The same goes with assigning an anonymous function to a variable, so:
console.log(f); // f === undefined
f(); // TypeError: f is not a function
var f = function () {}; // assigning an anonymous function as a value
f(); // ok, now it is a function ;)
The variable will be hoisted and thus visible in the entire scope, but it's value, even if it's a function, will still be undefined - hence the error if you try to execute it.
On the other hand if you declare a named function:
console.log(f); // f === function f()
f(); // we can already run it
function f() {}; // named function declaration
It's definition will also be hoisted so you can run it even in the first line of the scope you've declared it.
This is quite easily tested;
foo(1);
function foo(i) {
if (bar()) {
alert("foo called, bar true, i = " + i);
};
}
foo(2);
function bar() {
return true;
}
foo(3);
DEMO
This shows that Javascript loads all functions before executing anything. Therefor it does not matter what order functions are defined.
Well in JavaScript function is nothing but Object.
When you say typeof bar(),in the bar function you are returning foo which is another function.You just returning name of the function,So,it return the constructor of the foo function.So,your typeof get the value of foo constructor which is type of function.So,it alert function.It still refer the foo because of closure
Again in the bar defination,you returning foo,but its definition is still not encountered.In JavaScript when while parsing the instruction,declaration of variable and function is placed on top in current function scope.
So,
your statement
function bar() {
return foo;
foo = 10;
function foo() {}
var foo = '11';
}
is equivalent to
function bar() {
function foo() {}
return foo;
foo = 10;
var foo = '11';
This is called JavaScript top hoist
}
I came across a interesting quiz
function bar() {
return foo;
foo = 10;
function foo() {}
var foo = '11';
}
alert(typeof bar());
My interpretation is like this (Which is wrong according to console :) ):
var foo; // global variable
function bar(){
function foo(){}
var foo; // Here variable foo should override foo function
return foo; // (according to me foo should be variable with undefined value) What is going on here, How JavaScript resolve naming order ?
foo = 10;
foo = "11";
}
Here is a reference which I am reading this
In JavaScript, a name enters a scope in one of four basic ways:
1.Language-defined: All scopes are, by default, given the names this and arguments.
2. Formal parameters: Functions can have named formal parameters, which are scoped to the body of that function.
3. Function declarations: These are of the form function foo() {}.
4. Variable declarations: These take the form var foo;
He later quoted :
The most important special case to keep in mind is name resolution order. Remember that there are four ways for names to enter a given scope. The order I listed them above is the order they are resolved in. In general, if a name has already been defined, it is never overridden by another property of the same name. This means that a function declaration takes priority over a variable declaration. This does not mean that an assignment to that name will not work, just that the declaration portion will be ignored.
Which is confusing for me, Can anyone simplify this with reference to above example ?
Main point I want to know :
How variables without var inside a function hoisted ?
Does variable overriding occurs during hoisting ?
Lets leave the function foo for the moment. Within a function, if a variable is declared anywhere inside that function, the declaration will be moved to the top of the function. So, it is actually evaluated like this
function bar() {
var foo;
return foo;
foo = 10;
foo = '11';
}
But when you have a function declared by the same name, it will take precedence and it will be evaluated similar to this
function bar() {
var foo = function() {};
return foo;
foo = 10;
foo = '11';
}
That is why you are getting function in the alert box.
Any variable without var inside a function becomes a global variable by default.
when you have a function declaration inside another function(like in your example), it gets hoisted first followed by the variable declarations.
examples to demonstrate variable overriding.
function bar() {
var foo = 10;
function foo() {}
return foo;
}
bar(); //--> returns 10;
function bar() {
var foo;
function foo() {}
return foo;
}
bar(); //--> returns the function object foo.
The var keyword in javascript causes a variable to be stored in the local scope. Without var variables belong to the global scope. What about functions? It's clear what happens when functions are declared like variables
var foo = function() {...}
but what scope does
function foo() {...}
belong to?
EDIT:
I realized I didn't ask quite the right question so as a follow up. In the outer most nesting is there a difference between the above two declarations and the following declaration?
foo = function() {...}
It belongs to the current scope, always. For example:
// global scope
// foo is a global function
function foo() {
// bar is local to foo
function bar() {
}
}
Regarding your second question, this:
foo = function() {...}
is an anonymous function expression assigned to a global variable (unless you're running is strict mode, then foo would be undefined). The difference between that and function foo() {} is that the latter is a function declaration (versus a variable declaration, which is assigned an anonymous function expression).
You might be interested in this excellent article about function declarations and function expressions: Named function expressions demystified.
Function declarations are always local to the current scope, like a variable declared with the var keyword.
However, the difference is that if they are declared (instead of assigned to a variable) their definition is hoisted, so they will be usable everywhere in the scope even if the declaration comes in the end of the code. See also var functionName = function() {} vs function functionName() {}.
Noteworthy distinction taking implicit globals into account:
var foo = function() {
// Variables
var myVar1 = 42; // Local variable
myVar2 = 69; // Implicit global (no 'var')
// Functional Expressions
var myFn1 = function() { ... } // Local
myFn2 = function() { ... } // Implicit global
function sayHi() {
// I am a function declaration. Always local.
}
}
Hopefully that clarifies a little. Implicit globals are defined if you forget a var before your assignment. Its a dangerous hazard that applies to variable declarations and functional expressions.
Your first example (var foo = function() {...}) is called an anonymous function. It is dynamically declared at runtime, and doesn't follow the same rules as a normal function, but follows the rules of variables.
I'm trying to learn JS on codeacademy and I can't understand/get past this thing. Can someone please provide an answer and also an explanation of why is it so? Would deeply appreciate.
// This function tries to set foo to be the
// value specified.
function setFoo(val) {
// foo is declared in a function. It is not
// accessible outside of the function.
var foo = val;
}
setFoo(10);
// Now that we are outside the function, foo is
// not defined and the program will crash! Fix this
// by moving the declaration of foo outside of the
// function. Make sure that setFoo will still update
// the value of foo.
alert(foo);
You can see scope as a term meaning what variables you can reach at a specific "level" in the code. In JavaScript, these "levels" are defined by functions. Each function introduces a new level.
For example, take this sample code:
var a;
// you can access a at this level
function function1() {
var b;
// you can access a, b at this level
function function2() {
var c;
// you can access a, b, c at this level
}
}
So in your case, you should declare var foo; outside the function, preferably above it. Then you can set it inside setFoo with foo = val;. foo then refers to the one you declared in the level above setFoo.
foo is accessible both in setFoo and in the alert call that way; compare it with the above sample code (function1 is setFoo, a is foo and the alert call is in the top-most level. function2, b and c are not used in your case.).
// Create globale variable
// (You should not use globale variables!)
var foo;
// set value
function setFoo(val) {
foo = val;
}
setFoo(10);
// show value
alert(foo);
Just declare foo outside any function then it will be global:
var foo = null;
function setFoo(val) {
foo = val;
}
setFoo(10);
alert(foo);
Try it !
When you declare a variable in Javascript it is only visible to code that is in the same function as it is declared, or a function inernal to that function. Because foo is originally declared in the SetFoo function nothing outside of SetFoo is able to see it, so the call to alert fails as foo does not exist in the gloabl scope.
As the comments suggest, moving the declaration of foo out of the function and into the global scope (which you can think of as a catch-all function that contains everything) would allow you to use foo when calling alert.
var foo;
function setFoo(val) {
foo = val;
}
setFoo(10);
alert(foo); // No longer crashes
Every function in Javascript has it's own scope. That means that every variable you define there with the var keyword, will only be available within that function. That means that when you call setFoo(10), you create the variable foo, give it a value of five, after which it is immediately destroyed because it went out of scope.
There are multiple ways to solve this problem. The first would be to remove the var keyword. This would put foo in the global scope, which means that it's available everywhere. However, this is discouraged, you want to keep the global scope as uncluttered as possible, so that if you have javascript code provided by multiple people on the same page, they can't overwrite other people's variables. Another way to do it would be this:
function setFoo(val){
var foo = val;
alertfoo = function(){
alert(foo)
}
}
In this example, the only thing you're putting in the global scope is the alertfoo function, because you want that to be available everywhere. The alertfoo function is defined inside the setFoo function, this means that although foo should have gone out of scope after setfoo has been executed, it is kept in memory, because alertfoo has access to it.
This makes for some nice tricks. For example, let's say you're making a javascript library that will be included on other people's pages, you'll want to create a scope inside of which you can define variables, without polluting the global scope. The most common way to do this, is by declairing a self-executing function. This is a function which is executed immediately after being defined, it looks like this:
(function(){
//set variables you want to be global in your own code
var mainpage = document.getElementById('main');
//define functions you want to make available to other people in a way that puts them in the global scope
setMainElement = function(newmain){mainpage = newmain;}
})();
You can make this even better by making only one object global, and provide your interfae through the methods of that object, this way, you create a namespace with all the functions that your library contains. The next example uses an object literal to do this. In javascript, you can create an object by putting key/value pairs petween curly braces. the key/value pairs are properties of the object. for example:
(function(){
var privatevar = 10,otherprivate=20;
publicInterface = {
'addToPrivate': function(x){privatevar+=x;},
'getPrivate': function(){return private}
};
})();
Original code:
function setFoo(val) {
var foo = val;
}
setFoo(10);
alert(foo); // Crash!
Their advice to fix the crash:
Fix this by moving the declaration of foo outside of the function
I'm guessing you're confused as to what they mean by "outside of the function".
Try this edited code:
var foo = 5; // "var" declares the variable to be in this outer scope
function setFoo(val) {
foo = val; // but we can still access it in this inner scope...
}
setFoo(10);
alert(foo); // Displays a dialog box that says "10"
Variables defined in the function is valid only in the function
function setFoo(val) {
foo = val;
}
In JavaScript, new scopes are only created by functions