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Recently, I ran some of my JavaScript code through Crockford's JSLint, and it gave the following error:

Problem at line 1 character 1: Missing "use strict" statement.

Doing some searching, I realized that some people add "use strict"; into their JavaScript code. Once I added the statement, the error stopped appearing. Unfortunately, Google did not reveal much of the history behind this string statement. Certainly it must have something to do with how the JavaScript is interpreted by the browser, but I have no idea what the effect would be.

So what is "use strict"; all about, what does it imply, and is it still relevant?

Do any of the current browsers respond to the "use strict"; string or is it for future use?

Answered By: Pascal MARTIN ( 836)

This article about that might interest you: John Resig - ECMAScript 5 Strict Mode, JSON, and More

To quote some interesting parts:

Strict Mode is a new feature in ECMAScript 5 that allows you to place a program, or a function, in a "strict" operating context. This strict context prevents certain actions from being taken and throws more exceptions.

And:

Strict mode helps out in a couple ways:

  • It catches some common coding bloopers, throwing exceptions.
  • It prevents, or throws errors, when relatively "unsafe" actions are taken (such as gaining access to the global object).
  • It disables features that are confusing or poorly thought out.

Also note you can apply "strict mode" to the whole file... Or you can use it only for a specific function (still quoting from John Resig's article):

// Non-strict code...

(function(){
  "use strict";

  // Define your library strictly...
})();

// Non-strict code... 

Which might be helpful if you have to mix old and new code ;-)

So, I suppose it's a bit like the "use strict" you can use in Perl (hence the name?): it helps you make fewer errors, by detecting more things that could lead to breakages.

694
Richard

I've recently started maintaining someone else's JavaScript code. I'm fixing bugs, adding features and also trying to tidy up the code and make it more consistent.

The previous developer uses two ways of declaring functions and I can't work out if there is a reason behind it or not.

The two ways are:

var functionOne = function() {
    // Some code
};
function functionTwo() {
    // Some code
}

What are the reasons for using these two different methods and what are the pros and cons of each? Is there anything that can be done with one method that can't be done with the other?

Answered By: Greg ( 629)

The difference is that functionTwo is defined at parse-time for a script block, whereas functionOne is defined at run-time. For example:

<script>
  // Error
  functionOne();

  var functionOne = function() {
  }
</script>

<script>
  // No error
  functionTwo();

  function functionTwo() {
  }
</script>
488
JohnJohnGa
++[[]][+[]]+[+[]]

is valid and returns "10" in JavaScript (more examples here).

Can you explain why? I don't understand what's happening here.

Answered By: pimvdb ( 895)

If we split it up, the mess is equal to:

++[[]][+[]]
+
[+[]]

In JavaScript, it is true that +[] === 0. + converts something into a number, and in this case it will come down to +"" or 0 (see specification details below).

Therefore, we can simplify it (++ has precendence over +):

++[[]][0]
+
[0]

Because [[]][0] means: get the first element from [[]], it is true that:

  • [[]][0] returns the inner array ([]). Due to references it's wrong to say [[]][0] === [], but let's call the inner array A to avoid wrong notation.
  • ++[[]][0] == A + 1, since ++ means 'increment by one'.
  • ++[[]][0] === +(A + 1); in other words, it will always be a number (+1 does not necessarily return a number, whereas ++ always does - thanks to Tim Down for pointing this out).

Again, we can simplify the mess into something more legible. Let's substitute [] back for A:

+([] + 1)
+
[0]

In JavaScript, this is true as well: [] + 1 === "1", because [] == "" (joining an empty array), so:

  • +([] + 1) === +("" + 1), and
  • +("" + 1) === +("1"), and
  • +("1") === 1

Let's simplify it even more:

1
+
[0]

Also, this is true in JavaScript: [0] == "0", because it's joining an array with 1 element. Joining will concatenate the elements separated by ,. With one element, you can deduce that this logic will result in the first element itself.

So, in the end we obtain (number + string = string):

1
+
"0"

=== "10" // Yay!

Specification details for +[]:

This is quite a maze, but to do +[], first it is being converted to a string because that's what + says:

11.4.6 Unary + Operator

The unary + operator converts its operand to Number type.

The production UnaryExpression : + UnaryExpression is evaluated as follows:

  1. Let expr be the result of evaluating UnaryExpression.

  2. Return ToNumber(GetValue(expr)).

ToNumber() says:

Object

Apply the following steps:

  1. Let primValue be ToPrimitive(input argument, hint String).

  2. Return ToString(primValue).

ToPrimitive() says:

Object

Return a default value for the Object. The default value of an object is retrieved by calling the [[DefaultValue]] internal method of the object, passing the optional hint PreferredType. The behaviour of the [[DefaultValue]] internal method is defined by this specification for all native ECMAScript objects in 8.12.8.

[[DefaultValue]] says:

8.12.8 [[DefaultValue]] (hint)

When the [[DefaultValue]] internal method of O is called with hint String, the following steps are taken:

  1. Let toString be the result of calling the [[Get]] internal method of object O with argument "toString".

  2. If IsCallable(toString) is true then,

a. Let str be the result of calling the [[Call]] internal method of toString, with O as the this value and an empty argument list.

b. If str is a primitive value, return str.

The .toString of an array says:

15.4.4.2 Array.prototype.toString ( )

When the toString method is called, the following steps are taken:

  1. Let array be the result of calling ToObject on the this value.

  2. Let func be the result of calling the [[Get]] internal method of array with argument "join".

  3. If IsCallable(func) is false, then let func be the standard built-in method Object.prototype.toString (15.2.4.2).

  4. Return the result of calling the [[Call]] internal method of func providing array as the this value and an empty arguments list.

So +[] comes down to +"", because [].join() === "".

Again, the + is defined as:

11.4.6 Unary + Operator

The unary + operator converts its operand to Number type.

The production UnaryExpression : + UnaryExpression is evaluated as follows:

  1. Let expr be the result of evaluating UnaryExpression.

  2. Return ToNumber(GetValue(expr)).

ToNumber is defined for "" as:

The MV of StringNumericLiteral ::: [empty] is 0.

So +"" === 0, and thus +[] === 0.

I have two Python dictionaries, and I want to write a single expression that returns these two dictionaries, merged. The update() method would be what I need, if it returned its result instead of modifying a dict in-place.

>>> x = {'a':1, 'b': 2}
>>> y = {'b':10, 'c': 11}
>>> z = x.update(y)
>>> print z
None
>>> x
{'a': 1, 'b': 10, 'c': 11}

So I want that final merged dict in z, not x. How can I do this?

Answered By: Thomas Vander Stichele ( 422)

In your case, what you can do is:

z = dict(x.items() + y.items())

This will, as you want it, put the final dict in z, and make the value for b be properly overridden by the second (y) dict's value:

>>> x = {'a':1, 'b': 2}
>>> y = {'b':10, 'c': 11}
>>> z = dict(x.items() + y.items())
>>> z
{'a': 1, 'c': 11, 'b': 10}

If you use Python 3, it is only a little more complicated. To create z:

>>> z = dict(list(x.items()) + list(y.items()))
>>> z
{'a': 1, 'c': 11, 'b': 10}

I often hear claims that C++ is a context-sensitive language. Take the following example:

a b(c);

Is this a variable definition or a function declaration? That depends on the meaning of the symbol c. If c is a variable, then a b(c); defines a variable named b of type a. It is directly initialized with c. But if c is a type, then a b(c); declares a function named b that takes a c and returns an a.

If you look up the definition of context-free languages, it will basically tell you that all grammar rules must have left-hand sides that consist of exactly one non-terminal symbol. Context-sensitive grammars, on the other hand, allow arbitrary strings of terminal and non-terminal symbols on the left-hand side.

Browsing through Appendix A of "The C++ Programming Language", I couldn't find a single grammar rule that had anything else besides a single non-terminal symbol on its left-hand side. That would imply that C++ is context-free. (Of course, every context-free language is also context-sensitive in the sense that the context-free languages form a subset of the context-sensitive languages, but that is not the point.)

So, is C++ context-free or context-sensitive?

Answered By: rici ( 171)

Below is my (current) favorite demonstration of why parsing C++ is (probably) Turing-complete, since it shows a program which is syntactically correct if and only if a given integer is prime.

So I assert that C++ is neither context-free nor context-sensitive.

If you allow arbitrary symbol sequences on both sides of any production, you produce an Type-0 grammar ("unrestricted") in the Chomsky hierarchy, which is more powerful than a context-sensitive grammar; unrestricted grammars are Turing-complete. A context-sensitive (Type-1) grammar allows multiple symbols of context on the left hand side of a production, but the same context must appear on the right hand side of the production (hence the name "context-sensitive"). [1] Context-sensitive grammars are equivalent to linear-bounded Turing machines.

In the example program, the prime computation could be performed by a linear-bounded Turing machine, so it does not quite prove Turing equivalence, but the important part is that the parser needs to perform the computation in order to perform syntactic analysis. It could have been any computation expressible as a template instantiation and there is every reason to believe that C++ template instantiation is Turing-complete. See, for example, Todd L. Veldhuizen's 2003 paper.

Regardless, C++ can be parsed by a computer, so it could certainly be parsed by a Turing machine. Consequently, an unrestricted grammar could recognize it. Actually writing such a grammar would be impractical, which is why the standard doesn't try to do so. (See below.)

The issue with "ambiguity" of certain expressions is mostly a red herring. To start with, ambiguity is a feature of a particular grammar, not a language. Even if a language can be proven to have no unambiguous grammars, if it can be recognized by a context-free grammar, it's context-free. Similarly, if it cannot be recognized by a context-free grammar but it can be recognized by a context-sensitive grammar, it's context-sensitive. Ambiguity is not relevant.

But in any event, like line 21 (i.e. auto b = foo<IsPrime<234799>>::typen<1>();) in the program below, the expressions are not ambiguous at all; they are simply parsed differently depending on context. In the simplest expression of the issue, the syntactic category of certain identifiers is dependent on how they have been declared (types and functions, for example), which means that the formal language would have to recognize the fact that two arbitrary-length strings in the same program are identical (declaration and use). This can be modelled by the "copy" grammar, which is the grammar which recognizes two consecutive exact copies of the same word. It's easy to prove with the pumping lemma that this language is not context-free. A context-sensitive grammar for this language is possible, and a Type-0 grammar is provided in the answer to this question: http://math.stackexchange.com/questions/163830/context-sensitive-grammar-for-the-copy-language .

If one were to attempt to write a context-sensitive (or unrestricted) grammar to parse C++, it would quite possibly fill the universe with scribblings. Writing a Turing machine to parse C++ would be an equally impossible undertaking. Even writing a C++ program is difficult, and as far as I know none have been proven correct. This is why the standard does not attempt to provide a complete formal grammar, and why it chooses to write some of the parsing rules in technical English.

What looks like a formal grammar in the C++ standard is not the complete formal definition of the syntax of the C++ language. It's not even the complete formal definition of the language after preprocessing, which might be easier to formalize. (That wouldn't be the language, though: the C++ language as defined by the standard includes the preprocessor, and the operation of the preprocessor is described algorithmically since it would be extremely hard to describe in any grammatical formalism. It is in that section of the standard where lexical decomposition is described, including the rules where it must be applied more than once.)

The various grammars (two overlapping grammars for lexical analysis, one which takes place before preprocessing and the other, if necessary, afterwards, plus the "syntactic" grammar) are collected in Appendix A, with this important note (emphasis added):

This summary of C++ syntax is intended to be an aid to comprehension. It is not an exact statement of the language. In particular, the grammar described here accepts a superset of valid C++ constructs. Disambiguation rules (6.8, 7.1, 10.2) must be applied to distinguish expressions from declarations. Further, access control, ambiguity, and type rules must be used to weed out syntactically valid but meaningless constructs.

Finally, here's the promised program. Line 21 is syntactically correct if and only if the N in IsPrime<N> is prime. Otherwise, typen is an integer, not a template, so typen<1>() is parsed as (typen<1)>() which is syntactically incorrect because () is not a syntactically valid expression.

template<bool V> struct answer { answer(int) {} bool operator()(){return V;}};

template<bool no, bool yes, int f, int p> struct IsPrimeHelper
  : IsPrimeHelper<p % f == 0, f * f >= p, f + 2, p> {};
template<bool yes, int f, int p> struct IsPrimeHelper<true, yes, f, p> { using type = answer<false>; };
template<int f, int p> struct IsPrimeHelper<false, true, f, p> { using type = answer<true>; };

template<int I> using IsPrime = typename IsPrimeHelper<!(I&1), false, 3, I>::type;
template<int I>
struct X { static const int i = I; int a[i]; }; 

template<typename A> struct foo;
template<>struct foo<answer<true>>{
  template<int I> using typen = X<I>;
};
template<> struct foo<answer<false>>{
  static const int typen = 0;
};

int main() {
  auto b = foo<IsPrime<234799>>::typen<1>(); // Syntax error if not prime
  return 0;
}

[1] To put it more technically, every production in a context-sensitive grammar must be of the form:

αAβ → αγβ

where A is a non-terminal and α, β are possibly empty sequences of grammar symbols, and γ is a non-empty sequence. (Grammar symbols may be either terminals or non-terminals).

This can be read as A → γ only in the context [α, β]. In a context-free (Type 2) grammar, α and β must be empty.

It turns out that you can also restrict grammars with the "monotonic" restriction, where every production must be of the form:

α → β where |α| ≥ |β| > 0  (|α| means "the length of α")

It's possible to prove that the set of languages recognized by monotonic grammars is exactly the same as the set of languages recognized by context-sensitive grammars, and it's often the case that it's easier to base proofs on monotonic grammars. Consequently, it's pretty common to see "context-sensitive" used as though it meant "monotonic".

202
Armen Tsirunyan

Maybe I am not from this planet, but it would seem to me that the following should be a syntax error:

int a[] = {1,2,}; //extra comma in the end

But it's not. I was surprised when this code compiled on Visual Studio, but I have learnt not to trust MSVC compiler as far as C++ rules are concerned, so I checked the standard and it is allowed by the standard as well. You can see 8.5.1 for the grammar rules if you don't believe me.

enter image description here

Why is this allowed? This may be a stupid useless question but I want you to understand why I am asking. If it were a sub-case of a general grammar rule, I would understand - they decided not to make the general grammar any more difficult just to disallow a redundant comma at the end of an initializer list. But no, the additional comma is explicitly allowed. For example, it isn't allowed to have a redundant comma in the end of a function-call argument list (when the function takes ...), which is normal.

So, again, is there any particular reason this redundant comma is explicitly allowed?

Answered By: Jon Skeet ( 274)

It makes it easier to generate source code, and also to write code which can be easily extended at a later date. Consider what's required to add an extra entry to:

int a[] = {
   1,
   2,
   3
};

... you have to add the comma to the existing line and add a new line. Compare that with the case where the three already has a comma after it, where you just have to add a line. Likewise if you want to remove a line you can do so without worrying about whether it's the last line or not, and you can reorder lines without fiddling about with commas. Basically it means there's a uniformity in how you treat the lines.

Now think about generating code. Something like (pseudo-code):

output("int a[] = {");
for (int i = 0; i < items.length; i++) {
    output("%s, ", items[i]);
}
output("};");

No need to worry about whether the current item you're writing out is the first or the last. Much simpler.

179
mcherm

I'd like to convert a float to an int in Javascript. Actually, I'd like to know how to do BOTH of the standard convertions: by truncating and by rounding. And efficiently, not via converting to a string and parsing.

Answered By: moonshadow ( 211)
var intvalue = Math.floor( floatvalue );
var intvalue = Math.ceil( floatvalue ); 
var intvalue = Math.round( floatvalue );

Math object reference

144
Irfan

I was running the following PHP code:

<?php 
    </script>
?>

There were no parse errors and the output was "?>" (example).

In similar cases I do get a parse error:

<?php 
    </div>
?>

Parse error: syntax error, unexpected '<' in ...

Why doesn't <?php </script> ?> give the same error?

Answered By: Pekka 웃 ( 226)

This must be because there are various ways of starting a block of PHP code:

  • <? ... ?> (known as short_open_tag)

  • <?php ... ?> (the standard really)

  • <script language="php"> ... </script> (not recommended)

  • <% ... %> (deprecated and removed ASP-style tag after 5.3.0)

Apparently, you can open a PHP block one way, and close it the other. Didn't know that.

So in your code, you opened the block using <? but PHP recognizes </script> as the closer. What happened was:

<?php       <----- START PHP
</script>   <----- END PHP
?>          <----- JUST GARBAGE IN THE HTML