Regular expressions are a standard tool used for parsing strings across many languages. However their scope of applicability is limited. Regular expressions can only match a list. There is no way to describe arbitrary deep nested structures using regular expressions. Question: what is a technology/framework as widely used/spread and as standatd as regular expessions are that can match tree structures (produce AST).
Regular expressions describe a finite-state automaton.
Since the late 1960's, the "bread and butter" of parsing (though not necessarily the "state of the art") has been push-down automata generated by parser generators according to "LR" algorithms like LALR(1).
The connection to regular expressions is this: the parsing machine does in fact use rules very similar to regular expressions in order to recognize viable prefixes. The "shift" state transitions among the "core LR(0) items" constitute a finite automaton, and can be described by a regular expression. The recursion is is handled thanks to the semantic action of pushing symbols onto a stack when doing the "shifts", and removing them ("reducing"). Reductions rewrite a portion of the stack, and perform a "goto" to another state. This type of goto, together with the stack, is absent in the regular expression automaton.
Parse Expression Grammars are also related to regular expressions. Regular expressions themselves can be endowed with recursion. Firstly, we can take pieces of regular expressions and give them names, and then construct bigger regular expressions by writing expressions which invoke these names. (Such as feature is found in the lex tool where you can define a named expressions like letters [A-Za-z]+ and refer to it later as {letters}. Now suppose you allow circular references, like letters [A-Za-z]{letters}?. You now have recursion; the only problem is to adjust the model in order to implement it.
Implementations of so-called "regular expressions" in various modern languages and environments in fact support recursion. Perl-compatible regular expressions (PCRE) support it, for instance.
Expressions that feature recursion or backreferencing are not handled by the classic NFA compilation route (possibly converted to a DFA); they cannot be.
How the above letters recursion can be handled is with actual recursion. The ? operator can be implemented as a function which tries to match its respective argument object. If it succeeds, then it consumes whatever it has matched, otherwise it consumes nothing. That is to say, the regular expression can be converted to a syntax tree, and interpreted "as is" rather than compiled to a state machine (or trivially compiled to functions corresponding to the nodes of the tree), and such interpretation can naturally handle recursion. The interpretation then constitutes, effectively, a syntax-directed recursive-descent parser. (Note how I avoided left recursion in defining letters to make that example compatible with this approach).
Example: parenthesis-matching regex:
par-match := ({par-match})|
This gets compiled to a tree:
branch-op <-- "par-match" name points at this node
/ \
catenate-op <empty>
/ \
"(" catenate-op
/ \
{par-match} ")"
This can then converted to a recursive descent parser, or interpreted directly.
Pattern matching starts by invoking the top-level "branch-op". This operator simply tries all of the alternatives. Suppose the input is empty. Then the left alternative will fail: it demands an open parenthesis. So then the right alternative will succeed: empty matches empty. (The operators either "fail" or indicate "success" and consume input.)
But suppose your input is (()). The left catenate-op will in turn invoke its left subtree, which matches and consumes the left parenthesis, leaving ()). It will then invoke its right subtree, another catenate-op. This catenate-op matches its left subtree, which triggers recursion into the top level via the named par-match references. That recursion will match and consume (), leaving ). The catenate-op then invokes its right subtree which matches ). Control returns up to branch-op. (Though the left side of branch-op matched something, branch-op must still try the other alternative; more than one branch can match, and some can match longer than others.)
This is closely related to Parsing Expression Grammars work.
Practically speaking, the recursive definition could be encoded into the regex syntax somehow. Say we invent some new operator like (?name:definition) which means "match definition which is allowed to contain invocations of itself via name. The invocation syntax could be (*name), so that we can write the par-match example as (?par-match:\((*par-match)\)|). The combinations (? and (* are invalid under "classic" regex syntax and so we can use them for extension.
As a final note, regexes correspond to grammars. That is the fundamental connection btween regexes and parsing. That is to say, regexes correspond to a particular subset of grammars describe only regular languages. An example of a grammar which describes a regular language:
S -> A | B
B -> b
A -> A a | c
Although there is A -> A ... recursion, this is still regular, and corresponds to the regex ac*|b, which is just a more compact way to denote the same language. The grammar lets us notate languages that aren't regular and for which we can't write a regex, but as we have seen, we can extend the regex notation and semantics to express some of these things. Regular expressions aren't separate from grammars. The two aren't counterparts, but rather one is a special case or subset of the other.
Parser generators like Yacc, Bison, and derivatives are what you're after. They aren't as widespread as regular expressions because they generate actual C code. There are translations like Jison for example which implement the Yacc/Bison syntax using javascript. I know there are similar tools for other languages.
I get the impression Parsing expression grammar systems are up and coming though.
I've posted this on the D newsgroup some months ago, but for some reason, the answer never really convinced me, so I thought I'd ask it here.
The grammar of D is apparently context-free.
The grammar of C++, however, isn't (even without macros). (Please read this carefully!)
Now granted, I know nothing (officially) about compilers, lexers, and parsers. All I know is from what I've learned on the web.
And here is what (I believe) I have understood regarding context, in not-so-technical lingo:
The grammar of a language is context-free if and only if you can always understand the meaning (though not necessarily the exact behavior) of a given piece of its code without needing to "look" anywhere else.
Or, in even less rigor:
The grammar cannot be context-free if I need I can't tell the type of an expression just by looking at it.
So, for example, C++ fails the context-free test because the meaning of confusing<sizeof(x)>::q < 3 > (2) depends on the value of q.
So far, so good.
Now my question is: Can the same thing be said of D?
In D, hashtables can be created through a Value[Key] declaration, for example
int[string] peoplesAges; // Maps names to ages
Static arrays can be defined in a similar syntax:
int[3] ages; // Array of 3 elements
And templates can be used to make them confusing:
template Test1(T...)
{
alias int[T[0]] Test;
}
template Test2(U...)
{
alias int[U] Test2; // LGTM
}
Test1!(5) foo;
Test1!(int) bar;
Test2!(int) baz; // Guess what? It's invalid code.
This means that I cannot tell the meaning of T[0] or U just by looking at it (i.e. it could be a number, it could be a data type, or it could be a tuple of God-knows-what). I can't even tell if the expression is grammatically valid (since int[U] certainly isn't -- you can't have a hashtable with tuples as keys or values).
Any parsing tree that I attempt to make for Test would fail to make any sense (since it would need to know whether the node contains a data type versus a literal or an identifier) unless it delays the result until the value of T is known (making it context-dependent).
Given this, is D actually context-free, or am I misunderstanding the concept?
Why/why not?
Update:
I just thought I'd comment: It's really interesting to see the answers, since:
Some answers claim that C++ and D can't be context-free
Some answers claim that C++ and D are both context-free
Some answers support the claim that C++ is context-sensitive while D isn't
No one has yet claimed that C++ is context-free while D is context-sensitive :-)
I can't tell if I'm learning or getting more confused, but either way, I'm kind of glad I asked this... thanks for taking the time to answer, everyone!
Being context free is first a property of generative grammars. It means that what a non-terminal can generate will not depend on the context in which the non-terminal appears (in non context-free generative grammar, the very notion of "string generated by a given non-terminal" is in general difficult to define). This doesn't prevent the same string of symbols to be generated by two non-terminals (so for the same strings of symbols to appear in two different contexts with a different meaning) and has nothing to do with type checking.
It is common to extend the context-free definition from grammars to language by stating that a language is context-free if there is at least one context free grammar describing it.
In practice, no programming language is context-free because things like "a variable must be declared before it is used" can't be checked by a context-free grammar (they can be checked by some other kinds of grammars). This isn't bad, in practice the rules to be checked are divided in two: those you want to check with the grammar and those you check in a semantic pass (and this division also allows for better error reporting and recovery, so you sometimes want to accept more in the grammar than what would be possible in order to give your users better diagnostics).
What people mean by stating that C++ isn't context-free is that doing this division isn't possible in a convenient way (with convenient including as criteria "follows nearly the official language description" and "my parser generator tool support that kind of division"; allowing the grammar to be ambiguous and the ambiguity to be resolved by the semantic check is an relatively easy way to do the cut for C++ and follow quite will the C++ standard, but it is inconvenient when you are relying on tools which don't allow ambiguous grammars, when you have such tools, it is convenient).
I don't know enough about D to know if there is or not a convenient cut of the language rules in a context-free grammar with semantic checks, but what you show is far from proving the case there isn't.
The property of being context free is a very formal concept; you can find a definition here. Note that it applies to grammars: a language is said to be context free if there is at least one context free grammar that recognizes it. Note that there may be other grammars, possibly non context free, that recognize the same language.
Basically what it means is that the definition of a language element cannot change according to which elements surround it. By language elements I mean concepts like expressions and identifiers and not specific instances of these concepts inside programs, like a + b or count.
Let's try and build a concrete example. Consider this simple COBOL statement:
01 my-field PICTURE 9.9 VALUE 9.9.
Here I'm defining a field, i.e. a variable, which is dimensioned to hold one integral digit, the decimal point, and one decimal digit, with initial value 9.9 . A very incomplete grammar for this could be:
field-declaration ::= level-number identifier 'PICTURE' expression 'VALUE' expression '.'
expression ::= digit+ ( '.' digit+ )
Unfortunately the valid expressions that can follow PICTURE are not the same valid expressions that can follow VALUE. I could rewrite the second production in my grammar as follows:
'PICTURE' expression ::= digit+ ( '.' digit+ ) | 'A'+ | 'X'+
'VALUE' expression ::= digit+ ( '.' digit+ )
This would make my grammar context-sensitive, because expression would be a different thing according to whether it was found after 'PICTURE' or after 'VALUE'. However, as it has been pointed out, this doesn't say anything about the underlying language. A better alternative would be:
field-declaration ::= level-number identifier 'PICTURE' format 'VALUE' expression '.'
format ::= digit+ ( '.' digit+ ) | 'A'+ | 'X'+
expression ::= digit+ ( '.' digit+ )
which is context-free.
As you can see this is very different from your understanding. Consider:
a = b + c;
There is very little you can say about this statement without looking up the declarations of a,b and c, in any of the languages for which this is a valid statement, however this by itself doesn't imply that any of those languages is not context free. Probably what is confusing you is the fact that context freedom is different from ambiguity. This a simplified version of your C++ example:
a < b > (c)
This is ambiguous in that by looking at it alone you cannot tell whether this is a function template call or a boolean expression. The previous example on the other hand is not ambiguous; From the point of view of grammars it can only be interpreted as:
identifier assignment identifier binary-operator identifier semi-colon
In some cases you can resolve ambiguities by introducing context sensitivity at the grammar level. I don't think this is the case with the ambiguous example above: in this case you cannot eliminate the ambiguity without knowing whether a is a template or not. Note that when such information is not available, for instance when it depends on a specific template specialization, the language provides ways to resolve ambiguities: that is why you sometimes have to use typename to refer to certain types within templates or to use template when you call member function templates.
There are already a lot of good answers, but since you are uninformed about grammars, parsers and compilers etc, let me demonstrate this by an example.
First, the concept of grammars are quite intuitive. Imagine a set of rules:
S -> a T
T -> b G t
T -> Y d
b G -> a Y b
Y -> c
Y -> lambda (nothing)
And imagine you start with S. The capital letters are non-terminals and the small letters are terminals. This means that if you get a sentence of all terminals, you can say the grammar generated that sentence as a "word" in the language. Imagine such substitutions with the above grammar (The phrase between *phrase* is the one being replaced):
*S* -> a *T* -> a *b G* t -> a a *Y* b t -> a a b t
So, I could create aabt with this grammar.
Ok, back to main line.
Let us assume a simple language. You have numbers, two types (int and string) and variables. You can do multiplication on integers and addition on strings but not the other way around.
First thing you need, is a lexer. That is usually a regular grammar (or equal to it, a DFA, or equally a regular expression) that matches the program tokens. It is common to express them in regular expressions. In our example:
(I'm making these syntaxes up)
number: [1-9][0-9]* // One digit from 1 to 9, followed by any number
// of digits from 0-9
variable: [a-zA-Z_][a-zA-Z_0-9]* // You get the idea. First a-z or A-Z or _
// then as many a-z or A-Z or _ or 0-9
// this is similar to C
int: 'i' 'n' 't'
string: 's' 't' 'r' 'i' 'n' 'g'
equal: '='
plus: '+'
multiply: '*'
whitespace: (' ' or '\n' or '\t' or '\r')* // to ignore this type of token
So, now you got a regular grammar, tokenizing your input, but it understands nothing of the structure.
Then you need a parser. The parser, is usually a context free grammar. A context free grammar means, in the grammar you only have single nonterminals on the left side of grammar rules. In the example in the beginning of this answer, the rule
b G -> a Y b
makes the grammar context-sensitive because on the left you have b G and not just G. What does this mean?
Well, when you write a grammar, each of the nonterminals have a meaning. Let's write a context-free grammar for our example (| means or. As if writing many rules in the same line):
program -> statement program | lambda
statement -> declaration | executable
declaration -> int variable | string variable
executable -> variable equal expression
expression -> integer_type | string_type
integer_type -> variable multiply variable |
variable multiply number |
number multiply variable |
number multiply number
string_type -> variable plus variable
Now this grammar can accept this code:
x = 1*y
int x
string y
z = x+y
Grammatically, this code is correct. So, let's get back to what context-free means. As you can see in the example above, when you expand executable, you generate one statement of the form variable = operand operator operand without any consideration which part of code you are at. Whether the very beginning or middle, whether the variables are defined or not, or whether the types match, you don't know and you don't care.
Next, you need semantics. This is were context-sensitive grammars come into play. First, let me tell you that in reality, no one actually writes a context sensitive grammar (because parsing it is too difficult), but rather bit pieces of code that the parser calls when parsing the input (called action routines. Although this is not the only way). Formally, however, you can define all you need. For example, to make sure you define a variable before using it, instead of this
executable -> variable equal expression
you have to have something like:
declaration some_code executable -> declaration some_code variable equal expression
more complex though, to make sure the variable in declaration matches the one being calculated.
Anyway, I just wanted to give you the idea. So, all these things are context-sensitive:
Type checking
Number of arguments to function
default value to function
if member exists in obj in code: obj.member
Almost anything that's not like: missing ; or }
I hope you got an idea what are the differences (If you didn't, I'd be more than happy to explain).
So in summary:
Lexer uses a regular grammar to tokenize input
Parser uses a context-free grammar to make sure the program is in correct structure
Semantic analyzer uses a context-sensitive grammar to do type-checking, parameter matching etc etc
It is not necessarily always like that though. This just shows you how each level needs to get more powerful to be able to do more stuff. However, each of the mentioned compiler levels could in fact be more powerful.
For example, one language that I don't remember, used array subscription and function call both with parentheses and therefore it required the parser to go look up the type (context-sensitive related stuff) of the variable and determine which rule (function_call or array_substitution) to take.
If you design a language with lexer that has regular expressions that overlap, then you would need to also look up the context to determine which type of token you are matching.
To get to your question! With the example you mentioned, it is clear that the c++ grammar is not context-free. The language D, I have absolutely no idea, but you should be able to reason about it now. Think of it this way: In a context free grammar, a nonterminal can expand without taking into consideration anything, BUT the structure of the language. Similar to what you said, it expands, without "looking" anywhere else.
A familiar example would be natural languages. For example in English, you say:
sentence -> subject verb object clause
clause -> .... | lambda
Well, sentence and clause are nonterminals here. With this grammar you can create these sentences:
I go there because I want to
or
I jump you that I is air
As you can see, the second one has the correct structure, but is meaningless. As long as a context free grammar is concerned, the meaning doesn't matter. It just expands verb to whatever verb without "looking" at the rest of the sentence.
So if you think D has to at some point check how something was defined elsewhere, just to say the program is structurally correct, then its grammar is not context-free. If you isolate any part of the code and it still can say that it is structurally correct, then it is context-free.
There is a construct in D's lexer:
string ::= q" Delim1 Chars newline Delim2 "
where Delim1 and Delim2 are matching identifiers, and Chars does not contain newline Delim2.
This construct is context sensitive, therefore D's lexer grammar is context sensitive.
It's been a few years since I've worked with D's grammar much, so I can't remember all the trouble spots off the top of my head, or even if any of them make D's parser grammar context sensitive, but I believe they do not. From recall, I would say D's grammar is context free, not LL(k) for any k, and it has an obnoxious amount of ambiguity.
The grammar cannot be context-free if I need I can't tell the type of
an expression just by looking at it.
No, that's flat out wrong. The grammar cannot be context-free if you can't tell if it is an expression just by looking at it and the parser's current state (am I in a function, in a namespace, etc).
The type of an expression, however, is a semantic meaning, not syntactic, and the parser and the grammar do not give a penny about types or semantic validity or whether or not you can have tuples as values or keys in hashmaps, or if you defined that identifier before using it.
The grammar doesn't care what it means, or if that makes sense. It only cares about what it is.
To answer the question of if a programming language is context free you must first decide where to draw the line between syntax and semantics. As an extreme example, it is illegal in C for a program to use the value of some kinds of integers after they have been allowed to overflow. Clearly this can't be checked at compile time, let alone parse time:
void Fn() {
int i = INT_MAX;
FnThatMightNotReturn(); // halting problem?
i++;
if(Test(i)) printf("Weeee!\n");
}
As a less extreme example that others have pointed out, deceleration before use rules can't be enforced in a context free syntax so if you wish to keep your syntax pass context free, then that must be deferred to the next pass.
As a practical definition, I would start with the question of: Can you correctly and unambiguously determine the parse tree of all correct programs using a context free grammar and, for all incorrect programs (that the language requires be rejected), either reject them as syntactically invalid or produce a parse tree that the later passes can identify as invalid and reject?
Given that the most correct spec for the D syntax is a parser (IIRC an LL parser) I strongly suspect that it is in fact context free by the definition I suggested.
Note: the above says nothing about what grammar the language documentation or a given parser uses, only if a context free grammar exists. Also, the only full documentation on the D language is the source code of the compiler DMD.
These answers are making my head hurt.
First of all, the complications with low level languages and figuring out whether they are context-free or not, is that the language you write in is often processed in many steps.
In C++ (order may be off, but that shouldn't invalidate my point):
it has to process macros and other preprocessor stuffs
it has to interpret templates
it finally interprets your code.
Because the first step can change the context of the second step and the second step can change the context of the third step, the language YOU write in (including all of these steps) is context sensitive.
The reason people will try and defend a language (stating it is context-free) is, because the only exceptions that adds context are the traceable preprocessor statements and template calls. You only have to follow two restricted exceptions to the rules to pretend the language is context-free.
Most languages are context-sensitive overall, but most languages only have these minor exceptions to being context-free.
I had a work for the university which basically said:
"Demonstrates that the non-regular language L={0^n 1^n : n natural} had no infinite regular sublanguages."
I demonstrated this by contradiction. I basically said that there is a language S which is a sublanguage of L and it is a regular language. Since the possible Regular expressions for S are 0*, 1*, (1+0)* and (0o1)*. I check each grammar and demonstrate that none of them are part of the language L.
However, how I could prove that ANY non regular context free language could not contain any regular infinite sublanguages?
I don't want the prove per se, I just want to be pointed in the right direction.
L = {0^n 1^n : n natural} is non-regular context free.
M = 2*3* is infinite regular.
N = L∪M is non-regular context free. N contains M.
For the 0^n 1^n language, it might be valuable to look into the pumping lemma. I think when I learned the pumping lemma it was used on the a^n b^n language (same thing.) Possibly the pumping lemma might help in your proof.
Also you can consider that regular languages are closed under complement, union, intersection, and the kleene star.
That is if L1 and L2 are regular then:
L1 L2 (concatenation) is also regular.
L1 n L2 is regular
L1 U L2 is regular
¬L1 is regular
L1* is regular
It's possible that you could prove that any language that contains an regular infinite sublanguage is regular by using some of these rules.
Your instincts are good. Two things here.
First, almost always when the question takes the form "show that L is not regular/not CF" the answer is going to involve using the pumping lemmas. Similarly, when you get a question like "show there are no X that ..." the easy route is (almost always) going to be a proof by contradiction.
EDIT: false statement, only applies to context free language
Since you just want hints (and thankfully so, since I forgot how to do proofs since college), look at the definition of a regular language and what properties it has. Just from looking there I had enough info to prove the statement.
Is there a way to find out if two arbitrary regular expressions are equivalent? Looks like complex problem to me, but there might be some DFA simplification mechanism or something?
To test equivalence you can compute the minimal DFAs for the expressions and compare them.
Testability of equality is one of the classical properties of regular expressions. (N.B. This doesn't hold if you're really talking about Perl regular expressions or some other technically nonregular superlanguage.)
Turn your REs to generalised finite automata A and B, then construct a new automaton A-B such that the accepting states of A have null transitions to the start states of B, and that the accepting states of B are inverted. This gives you an automaton that accepts all those strings accepted by A, except for all those accepted by B.
Do the same for B-A, and reduce both to pure FAs. If an FA has no accepting states accessible from a start state then it accepts the empty language. If you can show that both A-B and B-A are empty, you've shown that A = B.
Edit Heh, I can't believe no one noticed the gigantic error there -- an intentional one, of course :-p
The automata A-B as described will accept those strings whose first half is accepted by A and whose second half is not accepted by B. Building the desired A-B is a slightly trickier process. I can't think of it off the top of my head, but I do know it's well-defined (and likely involves creating states to the represent the products of accepting states in A and non-accepting states in B).
This really depends on what you mean by regular expressions. As the other posters pointed out, reducing both expressions to their minimal DFA should work, but it only works for the pure regular expressions.
Some of the constructs used in the real world regex libs (backreferences in particular) give them power to express languages that aren't regular, so the DFA algorithm won't work for them. For example the regex : ([a-z]*) \1 matches a double occurence of the same word separated by a space (a a and b b but not b a nor a b). This cannot be recognized by a finite automaton at all.
These two Perlmonks threads discuss this question (specifically, read blokhead's responses):
Comparative satisfiability of regexps
Testing regex equivalence