Category: JavaScript

Global To-String
JavaScript Corners - Part 9

Global To-String
JavaScript Corners - Part 9

(This is Part 9 in my series on JavaScript corner cases).

Here’s another one.

In JavaScript, global variables are properties of the global object. By default, the global object is like any other, and inherits from the Object.prototype  object.  Object.prototype comes with a number of its own properties, such as the  toString method. So, that means that toString is also a global variable1.

Everything seems expected, except the last line, which might seem a little confusing. The toString() call is clearly invoking a function using a reference to that function, where the base of the reference is the global object, right? (Take a look at my posts on references). So surely  toString() and  global.toString() mean the same thing?

Wrong.

There’s a subtlety here. The unqualified  toString reference actually has a base value2 that is the global environment, which “knows about” the global object, but is not exactly the global object. The base object for the global environment is actually always the value  undefined. See here in the spec. This is why it prints "[object Undefined]" .

 


  1. To qualify as a global variable, there is actually an additional criterion. The property of the global object must not be listed in the set of unscopables on the global object. In this case, toString is not listed as an unscopable, since it was introduced into JavaScript before the existence of the unscopables feature, and for backwards compatibility it remains that way. 

  2. Recall that a reference has two components: the thing being referred on, and the name of the thing being referred to. For example, referring to the property named x on the object obj, in the case of obj.x 

JavaScript Corners – Part 9
Node.js With-Statement Bug

JavaScript Corners – Part 9
Node.js With-Statement Bug

What does the following evil code print?

If you’re not sure, don’t worry — neither are current JavaScript engines. Firefox prints “after”, while Edge, IE, and Node.js print “before” (node v7.9.0). I believe that Firefox is correct in this case.

The tricky statement is obviously the following one, which sets a property on an object in the same statement that deletes the property:

(Side note: if you’re not very familiar with JavaScript, the relevant language features that are being used here are the delete operator, comma operator, and the good ol’ evil with statement).

What we expect to happen

The statement  var x introduces a new variable at the script scope1.

The { x } expression creates a new object with a single property2  x, where the value of x  is copied from the variable  x in the outer scope, so it has the initial value of 'before'.

The with  statement brings the properties of the object obj into scope in a new lexical environment.

The statement x = (delete x, 'after') should perform the following steps:

  1. Evaluate the left hand side
  2. Evaluate the right hand side
  3. Assign the value from the right hand result, to the reference created when evaluating the left hand side

When the left hand side is evaluated, the property x will be found in object obj. The base value of the reference is the object, not the script variable scope.

The right hand side evaluates to 'after', but in the process it deletes the property x  from obj. However, the reference on the left hand side should still refer to “the property named 'x' on the object obj”, even though the property with that name is now deleted.

When the assignment happens, it should create a new property named 'x' on object obj, with value 'after'. The variable x in the outer scope should be left unaffected.

In this case, I think Node.js gets the wrong answer.


  1. Theoretically, the script scope is the global scope. But in Node.js, scripts are wrapped in a module wrapper that changes the behavior of global vars. This doesn’t affect the outcome of this experiment though 

  2. Bonus fact. Object literals inherit from the global intrinsic object Object.prototype, which has other properties on it, such as toString. So when I say that it has a single property, it would be more accurate to instead say that it has a single own property 

JavaScript Corners – Part 8
References (Continued)

JavaScript Corners – Part 8
References (Continued)

Given an object o  with a member function f  that prints out what the this value is:

We know what the following prints:

And we know what the following prints1:

I always thought that the difference came down to the fact that o.f()  is actually invoking a different operator — something like a “member call operator”.

However, what do you think the following prints?

My guess, up until today, would have been that this prints “global”, since with the parentheses, this is no longer invoking the member call operator, but is instead invoking the call operator.

But I was wrong. There is no such thing as a “member call operator”. Rather, the “call” operator just behaves differently depending on whether the target of the call is a value or a reference2.

So this actually prints “o”.

But hang on. Why didn’t the parentheses coerce o.f to a value?

One might have expected the parentheses to automatically dereference  o.f, something like the following examples that use the logical OR and comma operators to coerce the target to a value instead of a reference:

Indeed, this could have been the case for bare parentheses as well, but the language designers chose not to do it that way, so that the  delete and  typeof operators still work when extraneous parentheses are provided:

 


  1. assuming the use strict directive isn’t provided in this case 

  2. To be more accurate, the target also behaves differently depending on whether the target reference refers to a property of an object vs a variable in an environment record 

JavaScript Corners – Part 7
Calls and With Statements

JavaScript Corners – Part 7
Calls and With Statements

Here’s a quick one. What does the following print? (Assuming not in strict mode)

In non-strict mode, the naked function call foo() gets a  this value that is the global object. So the first case prints “Global”.

In the second case, we’re invoking foo as a member of  bar, and so the this value is bar (it prints “Bar”).

The last case is the most interesting, and the most useless (since with statements are strongly discouraged, and cannot be used outside of non-strict mode). The this object in this case is actually bar. JavaScript recognizes that the function foo here is being invoked within the context of a with statement, and implicitly uses the bar object. This prints “Bar”.

JavaScript Corners – Part 6

JavaScript Corners – Part 6

In what order does the following evaluate?

TL;DR Answer


Step 1: Variable Access

First off, what does this code even mean? If you’re not intimate with JavaScript, this might seem like a very confusing line of code. In fact, even if you’re familiar with JavaScript, this can be confusing.

So let’s break it down, starting with:

The expression a loads the value a from the surrounding scope1. This is done by searching up the scope chain until a is found.

There are a number of different types of scopes in JavaScript, including those that refer to blocks (like the inside of a for-loop), functions (the contents of a function), objects (scopes that are created using a with statement, or the global scope).

For our purposes, let’s define a at the global scope. You’ll see why in a moment. Assuming we’re working in Node.js, the global object is called global, and properties of the global object are part of the global scope2.

But, since we’re interested in the order of evaluation, it would be useful to know when the value a is accessed. Luckily, in JavaScript, you can define properties that have a getter and/or setter, which we can use to log when the global variable is accessed:

Great! We can now see when the global variable “a” is accessed. There aren’t many languages where you can do that. Hooray for JavaScript.

We may want to define more globals this way, so lets refactor this to use a helper:

Step 2: Calling the function

Now let’s look at the following statement:

This is, unsurprisingly, a function call. It first evaluates a, as indicated above, by fetching a from the current scope. Then it calls a as a function. Nothing special going on here.

But to make this work with our a, we’re going to need to make sure that a is defined as a function, and not the value 42. So let’s change our getter to return a function:

To answer our original question, we’re going to need to create a whole bunch of functions. So let’s again refactor this into a helper:

Step 3: Member access

The expression x[y], in JavaScript, is a property lookup. It evaluates the expressions x and y, and then finds the property on the object x that has the name resulting from the expression y. Here’s a snippet that illustrates this:

If you’re not very familiar with JavaScript, it’s important to note here that the property name used here is "myProp", and not "y". The property name is the result of evaluating y.

Again, it will be useful to know exactly when the property is accessed, so let’s use a getter instead:

Here I’ve just used the ES6 getter syntax, rather than using defineProperty.

As before, we’re going to need to do this a few times, so let’s create a helper function:

Step 4: Assignment

The last piece of the puzzle is the assignment operator. Consider the following code:

The assignment operator, like the other operators so far, will evaluate the each operand, and then perform some operation on the results. In the above case, x is evaluated, and then y is evaluated, and then the result of y is assigned to the result of x.

But wait. What do you mean “the result” of x?

The model here that JavaScript uses internally, is that x actually evaluates to a reference. This is a type in JavaScript which you’ve probably never heard of. A reference value consists of two components:

  • A base value, that tells you what container the value is stored in
  • A name, that tells you which value in the container is being referred to

In this case, the expression x evaluates to a reference that has the following attributes:

  • A base value that is the global object
  • A name that is the string "x"

In other words, the reference value is something like the English description “the property x on the global object”. When you assign to x, you are assigning to “the property x on the global object”. When you delete x, you are deleting “the property x on the global object”.

The expression y also evaluates to a reference, but the assignment operator coerces that reference to the actual referenced value. The same thing is done in expressions such as x + y or x(y).

Here’s another example of an assignment:

In this case, the base value of the reference is the object x, and the name is y.  The assignment sets the value referred to as “the property ‘y’ of the object x”. Similarly, you can do delete x.y to delete “the property ‘y’ of the object x”.

In a more detailed consideration of the above example, x and z evaluate to references. Both x and z are then coerced to values (dereferenced, by fetching the property or variable), and then a third reference is created refers to the property y on the base object x.

But, what order does this occur in? To find out, let’s use our trusty helper functions:

This might come as a little bit of a surprise. The expression x is evaluated before the expression y, and then the assignment takes place. In some ways, one expects the opposite — one expects that the left hand side of an assignment is not considered until the right hand side.

This seems to be a general rule in JavaScript. Operands are evaluated from left to right, and then the operator is executed. Perhaps an exception to this rule-of-thumb, is that the short-circuiting operators such as && must necessarily execute part of the operation without all the operands fully evaluated.

Side note: in languages such as C++, the order of the left and right hand side of a most operators is not defined. The compiler can chose to evaluate them in whatever order it thinks is best, or even evaluate them simultaneously (e.g. if the CPU has multiple cores). JavaScript is different, in that the specification lays out a specific, unambiguous ordering.

We can follow this to its logical conclusion, and determine the order of execution of the whole of the original program in question:

Can we abuse it? (Advanced)

The reason I started looking into this at all, is that I was trying to discover a way to “see” references. They are objects that exist in the execution model, but are never shown explicitly to the user of the language, so do they really need to exist at all?

This is import to me, because I’m writing a JavaScript compiler, and need to know whether references are best left as just a description mechanism in the ECMAScript specification, or if they should be considered to be real entities with real allocated memory in the runtime.

So, can we design an example, that unequivocally proves that there must be a reference allocated in memory at some point?

Here’s my attempt:

What I’ve done here is break up the x[y] = z assignment using the await operator. The await operator will suspend the statement (and the rest of the async function), allowing us to swap out various things in the environment to see if we can mess with the operation while it is suspended. What we’re trying to prove here, is that the reference itself must be preserved in memory, from the time that the operation is suspended, to the time that it is resumed (when z is resolved).

To make it even more apparent, I’ve executed the async function multiple times, trying different ways to “mess” with the pending operations.

Conclusions

This experiment has proven to me that references are “almost” tangible objects. We can see that they must exist in memory under some circumstances, and that they are not simple “pointer” values — they must refer to both the object and the property name.

This leads to some interesting results when it comes to the order of evaluation of various expressions. While this knowledge isn’t needed for everyday programming scenarios, it helps to have a deeper understanding of what’s going on so that we know where the limit lies.

 

 

 

 


  1. Known in ECMAScript as a Lexical Environment 

  2. There is an interesting recursion here, since the value global here is also a globally scoped binding, which means the global property on the global object points to itself. You can see this if you have a statement like console.log(global.global.a) 

JavaScript Corners – Part 5

JavaScript Corners – Part 5

Here’s a quick one:

This was quite unexpected to me. It’s the only time I’ve ever seen where the left-hand side of an assignment can affect the right-hand side.

This only happens once. Once the anonymous function has a name, it can’t be re-named:

Interestingly, this doesn’t seem to work with destructuring:

This implies that somewhere between when the anonymous function is instantiated, and when it added to the array literal, the anonymous function acquires a name. It’s not the destructuring itself that suppresses the name, as can be seen in the following example:

It also doesn’t seem to work with anonymous functions passed as parameters:

It also doesn’t seem to work with anonymous functions evaluated from more complicated expressions:

 

JavaScript Corners – Part 4

JavaScript Corners – Part 4

Here’s another quick one. It relates to the scoping rules with parameter expressions.

Function foo has three parameters: x, f, and y. The key thing here is that the default value for f is a function that encloses the local lexical scope surrounding the function, and we’ve designed the function in such a way that when we call it it “tells” us the values of x, y, and z in the surrounding scope.

So the question is, which values of x, y, and z does f close over? There are global variables x, y, and z, there are parameters x and y, and there are local variables x, y and z. The parameter x is intentionally ordered before the function f is initialized, while parameter y is positioned to be after the function f is initialized.

I’ll spare you the time of running the code yourself. On my machine, in node.js 7.0.0, the output is [ 1, 2, 'c']. Clearly no local variables have been used, and f is capable of seeing parameter y even though it’s declared after f. The output is the same whether or not the code is executed in strict mode.

One of the most interesting things here, to me, is that the local variable x is not an alias for the parameter x.  This is strange when you consider the following code:

Given that local variables are apparently not aliases of the parameters with the same name, you would expect the output of the above to be undefined, since variable x is not initialized, even though parameter x is initializedHowever, the output is actually 42.

Where did we go wrong in our logic? Let’s see if we can get an example that explores what’s going on:

The output of this is:

This clearly tells me that variable x is not an alias of parameter x, but rather a separate variable that is initialized to the same value as parameter x.

Also, as a side note, it tells me that there is a subtle semantic distinction between initializing a variable to undefined and not initializing a variable at all, which is very interesting.

I should note that we’ve explored this empirically, not deducing the behavior from the spec. When I run the same experiment in Firefox I get a different result — both instances of y are undefined.

Which is correct?

I believe the V8 (node.js) implementation is more accurate in this case. There’s a note in the ECMAScript specification1 that says:

NOTE A separate Environment Record is needed to ensure that closures created by expressions in the formal parameter list do not have visibility of declarations in the function body.

I think this note is self explanatory, and tell us that the closure function in the parameter list is not meant to be able to see the local variables of its parent function.

P.S.

I’m going add one last investigation to this post. The above quote applies only if there are parameter expressions (if some parameters have default values). If there are no parameter expressions, then apparently this second “environment” is not created, and so the variable x should be an alias for the parameter x.

At first I thought that there was no way that we could ever test this distinction. Can you think of a way to test this difference?

Spoiler alert, I did think of a way. If the code is not executing in strict mode, then the arguments variable holds aliases for the parameters, which makes it possible to mutate parameters without referring to them by name. This is important, because once you have a local variable with the same name, there is no way to know if the local variable is a copy of the parameter, or an alias of the parameter, based purely on code that identifies it by name.

Take a look at the following experiment:

First note that y is not used in this code sample, but you’ll see why I added it in a moment.

The first log output of “1” shows that the variable x is either a copy of parameter x or an alias of parameter x. Then we continue by modifying the parameter x, without touching the variable x. Then we log the value of the variable x and see that it’s also changed — because in fact the variable and parameter x are both symbols for the same binding.

But now look at this slight modification to the code:

Note that again y is not used. In fact, to keep this a controlled experiment, y is even initialized to the same value that is passed as a parameter. The contents of the value initializer is not relevant though, because the initializer is never evaluated. What is important in this case is that there is an initializer expression at all, regardless of whether it executes or not.

The surprising thing, as I noted in the comment on the snippet, is that the second output is now 1 instead of 2, resulting from the fact that the mutation of the argument does not change the variable.

This is not a bug in node.js, but just an interesting quirk of the spec, in another corner of JavaScript.

 


  1. https://tc39.github.io/ecma262/#sec-functiondeclarationinstantiation 

JavaScript Corners – Part 3

JavaScript Corners – Part 3

I was busy reading ES spec today, and the following question popped into my head: at what point are default parameters evaluated?

To answer this question, I wrote the following script:

If default parameters are evaluated when the function is declared, then we would expect 'a' to appear once in the terminal output.

On the other hand, if the default parameter expressions are evaluated when the function is called, then we would expect 'a' to appear two or three times, corresponding to each time the function is called, and whether the default parameter expression is evaluated even when an argument is provided.

What do you think the output is? I won’t show you here… you can run the snippet yourself in a REPL to see if you’re correct or not.

As I did last time, I’ll leave you with this  mind-bending challenge: what does the following do?

 

JavaScript Corners – Part 2

JavaScript Corners – Part 2

I’m continuing my series on JavaScript corner cases.

Last time I looked how function names interact with each other and with variable names. Today I’m looking at how parameters interact with variables.

Take a look at the following code snippet:

It creates a function that takes a parameter named x, and then calls that function with an argument ((You may have heard of parameters being called “formal parameters”, and arguments being called “actual parameters”, but I think this originates largely from people who didn’t understand the distinction and so confused things for everyone else)) with a value of 1.  The curve ball here is that we’ve then also declared a variable with exactly the same name as a parameter. Does the symbol x then refer to the parameter or the variable? Do variables shadow parameters?

The variable hasn’t been given a value, so it must be undefined, right? If x refers to the variable, then the console output will be “undefined“, while if x refers to the parameter, then the console output will be “1“. So which is it…

On my computer, using node, this outputs 1. So it seems x refers to the parameter, and not the variable… right?

But now consider the following code snippet:

This creates a variable, and then gives it the value 2. According to our theory thus far in this post, the console will still output the value 1, since the x refers to the parameter and not the variable, as we saw from the previous snippet.

But if you run this code, it actually outputs 2. So it seems that it’s not that variables or a parameters “win” the battle to get into the namespace. Rather, it seems that parameters actually are themselves variables, and in the same namespace. The parameter x and the variable x are actually one and the same.

I’ll leave you with this bonus challenge: what does the following do?

 

 

JavaScript Corners – Part 1

JavaScript Corners – Part 1

Recently I’ve been trying to write a simple JavaScript compiler, and it’s lead me to think more deeply about some JavaScript behavior that I previously would not have thought about, and I’d like to share that with my JavaScript readers.

Take a look at the following code JavaScript code, and try to figure out what it outputs to the console (I’ll give you a hint: it doesn’t output any errors):

The function code first calls foo – but which foo?

Perhaps it executes the foo 1. After all, there can only be one function called foo, and so the others may not bind correctly after the first one is declared, so the first remains the “real one”.

On the other hand, perhaps it’s not the first definition of foo that “wins”, but the last one. So could it be foo 5?

But the last definition of foo a variable named “foo”, not a function. So which wins when it comes to binding a symbol: variables or functions? Or are they treated equally?

If variable are somehow considered “second prize” to functions when it comes to finding which value matches which name, then it wouldn’t be foo 5 that wins, but rather foo 4, since foo 4 actually  has a function named foo, whereas foo 5 is only a variable named foo that holds an anonymous function.

But foo 4 is not declared at the outer scope of the function. Could it be that declarations at the outer scope win against declarations that are declared in some kind of nested scope? Perhaps then the answer might then be foo 2, since foo 2 is the last function that is actually named foo and declared in the outer block?

The only one we’ve completely ruled out, that it can’t possibly be, is foo 3. The function foo 3 is not the first foo, and not the last. It is not declared in the outer block (and if non-outer-block declarations could win then foo 4 would win). It is also nested inside a block that is only executed on a condition, and the condition is always false so the block is clearly never executed, and thus the function is never declared anyway.

So it can’t be foo 3, but it could be any of the others. Which do you think is it?


You may have guessed it based on my harpings-on. The answer, when I execute the script using node, is indeed foo 3.

In JavaScript, the nested scope is not really a nested scope. All local var and function declarations are at the function scope of the whole parent function. If you’ve been working in JavaScript for a while, you probably already know this. The fact that it’s in a “false” condition block has nothing to do with it, since function declarations are not “statements” to be executed in the sequence of the program (and if they were, then this whole snippet would fail).

What was interesting to me is that it must be the case when function and variable names clash, the functions seem to always win.

The foo 4 function is a little misleading. Just because the function has a name, doesn’t mean that it’s attached to the function scope, because in the case of foo 4 the function is a function expression. These are not even part of the function’s namespace, as you can see if you execute the following snippet, and see that it gives you an error:

Does "use strict" cure this strange behavior, and somehow give a parse failure? On my machine when I change the function to have "use strict", it doesn’t give any errors, but it does change the output to foo 2 instead of foo 3. I found this quite unexpected.

This is all interesting behavior. These things don’t affect everyday JavaScript much, but they are fun and interesting corner cases to consider, and help us understand the language more deeply.

Let me know what you think about this in the comments. Hopefully I’ll continue this series with more of the interesting corner cases that I find along the way.