Is D's grammar really context-free?

Solution 1:

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.

Solution 2:

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.

Solution 3:

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

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.

Solution 4:

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.

Solution 5:

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.