Reference Type - JavaScript Programming Language - Miscellaneous
Reference Type
In-depth language feature
This article covers an advanced topic, to understand certain edge-cases better.
It’s not important. Many experienced developers live fine without knowing it. Read on if you want to know how things work under the hood.
A dynamically evaluated method call can lose this
.
For instance:
let user = { name: "John",hi() { alert(this.name); },bye() { alert("Bye"); } }; user.hi(); // works // now let's call user.hi or user.bye depending on the name (user.name == "John" ? user.hi : user.bye)(); // Error!
On the last line there is a conditional operator that chooses either user.hi
or user.bye
. In this case the result is user.hi
.
Then the method is immediately called with parentheses ()
. But it doesn’t work correctly!
As you can see, the call results in an error, because the value of "this"
inside the call becomes undefined
.
This works (object dot method):
user.hi();
This doesn’t (evaluated method):
(user.name == "John" ? user.hi : user.bye)(); // Error!
Why? If we want to understand why it happens, let’s get under the hood of how obj.method()
call works.
Reference type explained
Looking closely, we may notice two operations in obj.method()
statement:
- First, the dot
'.'
retrieves the propertyobj.method
. - Then parentheses
()
execute it.
So, how does the information about this
get passed from the first part to the second one?
If we put these operations on separate lines, then this
will be lost for sure:
let user = { name: "John",hi() { alert(this.name); } }; // split getting and calling the method in two lines let hi = user.hi; hi(); // Error, because this is undefined
Here hi = user.hi
puts the function into the variable, and then on the last line it is completely standalone, and so there’s no this
.
To make user.hi()
calls work, JavaScript uses a trick – the dot '.'
returns not a function, but a value of the special Reference Type.
The Reference Type is a “specification type”. We can’t explicitly use it, but it is used internally by the language.
The value of Reference Type is a three-value combination (base, name, strict)
, where:
base
is the object.name
is the property name.strict
is true ifuse strict
is in effect.
The result of a property access user.hi
is not a function, but a value of Reference Type. For user.hi
in strict mode it is:
// Reference Type value (user, "hi", true)
When parentheses ()
are called on the Reference Type, they receive the full information about the object and its method, and can set the right this
(user
in this case).
Reference type is a special “intermediary” internal type, with the purpose to pass information from dot .
to calling parentheses ()
.
Any other operation like assignment hi = user.hi
discards the reference type as a whole, takes the value of user.hi
(a function) and passes it on. So any further operation “loses” this
.
So, as the result, the value of this
is only passed the right way if the function is called directly using a dot obj.method()
or square brackets obj['method']()
syntax (they do the same here). There are various ways to solve this problem such as func.bind().
Summary
Reference Type is an internal type of the language.
Reading a property, such as with dot .
in obj.method()
returns not exactly the property value, but a special “reference type” value that stores both the property value and the object it was taken from.
That’s for the subsequent method call ()
to get the object and set this
to it.
For all other operations, the reference type automatically becomes the property value (a function in our case).
The whole mechanics is hidden from our eyes. It only matters in subtle cases, such as when a method is obtained dynamically from the object, using an expression.
Tasks
Syntax check
importance: 2
What is the result of this code?
let user = { name: "John",go: function() { alert(this.name) } } (user.go)()
P.S. There’s a pitfall :)
Explain the value of "this"
importance: 3
In the code below we intend to call obj.go()
method 4 times in a row.
But calls (1)
and (2)
works differently from (3)
and (4)
. Why?
let obj, method; obj = {go: function() { alert(this); } }; obj.go(); // (1) [object Object] (obj.go)(); // (2) [object Object] (method = obj.go)(); // (3) undefined (obj.go || obj.stop)(); // (4) undefined
Original Content at: https://javascript.info/reference-type
© 2007—2024 Ilya Kantor, https://javascript.info