I'm interested in the famous The syntax of C in Backus-Naur Form and studied for a while, what confuse me is that some syntax looks wrong to me but is considered right according to the BNF.
For example, int test {}, what's this? I think this is a ill syntax in C, but the truth is the BNF considered this a function definition:
int -> type_const -> type_spec -> decl_specs
test-> id -> direct_declarator -> declarator
'{' '}' -> compound_stat
decl_specs declarator compound_stat -> function_definition
I tried this with bison, it considered the input int test {} is a right form, but I tried this on a C compiler, it will not compile.
So got questions:
int test {} a right syntax or not?
If it is a right syntax, what is that mean and why compiler do not recognized it?
If it is an ill syntax, can I say the BNF is not rigorous？ And does that mean modern C compiler does not stick with this BNF?
The grammar is necessary but not sufficient to describe a valid C program. For that you need constraints from the standard too. A simpler example of this would be 0++, which follows the syntax of a C expression, but certainly isn't a valid program fragment...
C11 6.9.1p2:
The identifier declared in a function definition (which is the name of the function) shall have a function type, as specified by the declarator portion of the function definition. [162]
The footnote 162 explains that the intent of the constraint is that a typedef cannot be used, i.e. that
typedef int F(void);
F f { /* ... */ }
will not be valid, even though such a typedef could be used for a function declaration, i.e.
F f;
would declare the function
int f(void);
But mere existence of this constraint also proves that the BNF grammar in itself is not sufficient in this case. Hence you are correct in that the grammar would consider such a fragment a function definition.
The BNF form is a precise way to describe the syntax of a language, i.e. what to do precisely to get the parse tree starting from raw input.
For each language you can define infinite many grammars that describe that language. The properties of these grammars that describe the same language can differ a lot.
If you study the grammar of the C language, take care as it is not context free but context sensitive, which means, the decision of choosing a rule or other depends on what there is around that point in input.
Read about the lexer hack to see how to correctly interpret the Backus Naur form of the C grammar.
Related
So from my limited understanding, C has syntax ambiguity as seen in the expression:
T(*b)[4];
Here it is said about this sort of thing:
The well-known "typedef problem" with parsing C is that the standard C grammar is ambiguous unless the lexer distinguishes identifiers bound by typedef and other identifiers as two separate lexical classes. This means that the parser needs to feed scope information to the lexer during parsing. One upshot is that lexing must be done concurrently with parsing.
The problem is it can be interpreted as either multiplication or as a pointer depending on context (I don't 100% understand the details of this since I'm not expert in C, but I get the gist of it and why it's a problem).
typedef a;
b * a; // multiplication
a * b; // b is pointer to type a
What I'm wondering is if you were to parse C with a Parsing Expression Grammar (PEG) such as this C grammar, how does it handle this ambiguity? I assume this grammar is not 100% correct because of this problem, and so am wondering how you would go about fixing it. What does it need to keep track of or do differently to account for this?
The usual way this is handled in a PEG grammar is to use a semantic predicate on a rule such that the rule only matches when the predicate is true, and have the predicate check whether the name in question is a type in the current context or not. In the link you give, there's a rule
typedefName : Identifier
which is the (only) one that needs the semantic predicate to resolve this ambiguity. The predicate simply checks the Identifier in question against the definitions in the current scope. If it is not defined as a type, then it rejects this rule, so the next lower priority one will (try to) match.
I know that C is not a context-free language, a famous example is:
int foo;
typedef int foo;
foo x;
In this case the lexer doesn't know, whether foo in the 3rd line, is an identifier, or typedef.
My question is, is this the only reason that makes C a Context-Sensitive Language?
I mean, if we get rid of typedef, would it become context-free language? Or there are other reasons (examples) that prevent it from being so?
Yes. C can be parsed with a classical lex + yacc combo. The lexer definition and the yacc grammar are freely available at
http://www.quut.com/c/ANSI-C-grammar-l-2011.html
and
http://www.quut.com/c/ANSI-C-grammar-y-2011.html
As you can see from the lex file, it's straightforward except for the context-sensitive check_type() (and comment(), but comment processing technically belongs to the preprocessor), which makes typedef the only source of context-sensitivity there. Since the yacc file doesn't contain any context-sensitivity introducing tricks either, a typedef-less C would be a perfectly context-free language.
No. C cannot be a strict context independent language. For that, you should describe a syntax that doesn't allow to use a nondeclared variable (this is context) in a similar way as what you describe in your question. The language authors always describe syntax using some kind of context free grammar, but just to describe the main syntactic constructs of the language. The case you describe (making a type identifier to fit in a different token class to be able to go in places where it shouldn't) is only an example. If you look for example, the freedom in the order for things like static unsigned long long int variable simplifies the syntax remembering by programmers, but complicates things to the compiler authors.
As per my knowledge and research there are two basic reasons that make C context sensitive language. These are:
Variable is declared before it is used.
Matching the formal and actual parameters of functions or sub-routines.
These two can't be done by PushDown Automata (PDA) but Linear Bounded Automata (LBA) can do thes two.
I think the Question is self sufficient. Is the syntax of C Language completely defined through Context Free Grammars or do we have Language Constructs which may require non-Context Free definitions in the course of parsing?
An example of non CFL construct i thought was the declaration of variables before their use. But in Compilers(Aho Ullman Sethi), it is stated that the C Language does not distinguish between identifiers on the basis of their names. All the identifiers are tokenized as 'id' by the Lexical Analyzer.
If C is not completely defined by CFGs, please can anyone give an example of Non CFL construct in C?
The problem is that you haven't defined "the syntax of C".
If you define it as the language C in the CS sense, meaning the set of all valid C programs, then C – as well as virtually every other language aside from turing tarpits and Lisp – is not context free. The reasons are not related to the problem of interpreting a C program (e.g. deciding whether a * b; is a multiplication or a declaration). Instead, it's simply because context free grammars can't help you decide whether a given string is a valid C program. Even something as simple as int main() { return 0; } needs a more powerful mechanism than context free grammars, as you have to (1) remember the return type and (2) check that whatever occurs after the return matches the return type. a * b; faces a similar problem – you don't need to know whether it's a multiplication, but if it is a multiplication, that must be a valid operation for the types of a and b. I'm not actually sure whether a context sensitive grammar is enough for all of C, as some restrictions on valid C programs are quite subtle, even if you exclude undefined behaviour (some of which may even be undecidable).
Of course, the above notion is hardly useful. Generally, when talking grammars, we're only interested in a pretty good approximation of a valid program: We want a grammar that rules out as many strings which aren't C as possible without undue complexity in the grammar (for example, 1 a or (-)). Everything else is left to later phases of the compiler and called a semantic error or something similar to distinguish it from the first class of errors. These "approximate" grammars are almost always context free grammars (including in C's case), so if you want to call this approximation of the set of valid programs "syntax", C is indeed defined by a context free grammar. Many people do, so you'd be in good company.
The C language, as defined by the language standard, includes the preprocessor. The following is a syntactically correct C program:
#define START int main(
#define MIDDLE ){
START int argc, char** argv MIDDLE return 0; }
It seems to be really tempting to answer this question (which arises a lot) "sure, there is a CFG for C", based on extracting a subset of the grammar in the standard, which grammar in itself is ambiguous and recognizes a superset of the language. That CFG is interesting and even useful, but it is not C.
In fact, the productions in the standard do not even attempt to describe what a syntactically correct source file is. They describe:
The lexical structure of the source file (along with the lexical structure of valid tokens after pre-processing).
The grammar of individual preprocessor directives
A superset of the grammar of the post-processed language, which relies on some other mechanism to distinguish between typedef-name and other uses of identifier, as well as a mechanism to distinguish between constant-expression and other uses of conditional-expression.
There are many who argue that the issues in point 3 are "semantic", rather than "syntactic". However, the nature of C (and even more so its cousin C++) is that it is impossible to disentangle "semantics" from the parsing of a program. For example, the following is a syntactically correct C program:
#define base 7
#if base * 2 < 10
&one ?= two*}}
#endif
int main(void){ return 0; }
So if you really mean "is the syntax of the C language defined by a CFG", the answer must be no. If you meant, "Is there a CFG which defines the syntax of some language which represents strings which are an intermediate product of the translation of a program in the C language," it's possible that the answer is yes, although some would argue that the necessity to make precise what is a constant-expression and a typedef-name make the syntax necessarily context-sensitive, in a way that other languages are not.
Is the syntax of C Language completely defined through Context Free Grammars?
Yes it is. This is the grammar of C in BNF:
http://www.cs.man.ac.uk/~pjj/bnf/c_syntax.bnf
If you don't see other than exactly one symbol on the left hand side of any rule, then the grammar is context free. That is the very definition of context free grammars (Wikipedia):
In formal language theory, a context-free grammar (CFG) is a formal grammar in which every production rule is of the form
V → w
where V is a single nonterminal symbol, and w is a string of terminals and/or nonterminals (w can be empty).
Since ambiguity is mentioned by others, I would like to clarify a bit. Imagine the following grammar:
A -> B x | C x
B -> y
C -> y
This is an ambiguous grammar. However, it is still a context free grammar. These two are completely separate concepts.
Obviously, the semantics analyzer of C is context sensitive. This answer from the duplicate question has further explanations.
There are two things here:
The structure of the language (syntax): this is context free as you do not need to know the surroundings to figure out what is an identifier and what is a function.
The meaning of the program (semantics): this is not context free as you need to know whether an identifier has been declared and with what type when you are referring to it.
If you mean by the "syntax of C" all valid C strings that some C compiler accepts, and after running the pre-processor, but ignoring typing errors, then this is the answer: yes but not unambiguously.
First, you could assume the input program is tokenized according to the C standard. The grammar will describe relations among these tokens and not the bare characters. Such context-free grammars are found in books about C and in implementations that use parser generators. This tokenization is a big assumption because quite some work goes into "lexing" C. So, I would argue that we have not described C with a context-free grammar yet, if we have not used context-free grammars to describe the lexical level. The staging between the tokenizer and the parser combined with the ordering emposed by a scanner generator (prefer keywords, longest match, etc) are a major increase in computational power which is not easily simulated in a context-free grammar.
So, If you do not assume a tokenizer which for example can distinguish type names from variable names using a symbol table, then a context-free grammar is going to be harder. However: the trick here is to accept ambiguity. We can describe the syntax of C including its tokens in a context-free grammar fully. Only the grammar will be ambiguous and produce different interpretations for the same string . For example for A *a; it will have derivations for a multiplication and a pointer declaration both. No problem, the grammar is still describing the C syntax as you requested, just not unambiguously.
Notice that we have assumed having run the pre-processor first as well, I believe your question was not about the code as it looks before pre-processing. Describing that using a context-free grammar would be just madness since syntactic correctness depends on the semantics of expanding user-defined macros. Basically, the programmer is extending the syntax of the C language every time a macro is defined. At CWI we did write context-free grammars for C given a set of known macro definitions to extend the C language and that worked out fine, but that is not a general solution.
What are examples of non - context free languages in C language ? How the following non-CFL exists in C language ?
a) L1 = {wcw|w is {a,b}*}
b) L2 = {a^n b^m c^n d^m| n,m >=1}
The question is clumsily worded, so I'm reading between the lines, here. Still, it's a common homework/study question.
The various ambiguities [1] in the C grammar as normally presented do not render the language non-context-free. (Indeed, they don't even render the grammars non-context-free.) The general rule "if it looks like a declaration, it's a declaration regardless of other possible parses" can probably be codified in a very complex context-free grammar (although it's not 100% obvious that that is true, since CFGs are not closed under intersection or difference), but it's easier to parse with a simpler CFG and then disambiguate according to the declaration rule.
Now, the important point about C (and most programming languages) is that the syntax of the language is quite a bit more complex than the BNF used for explanatory purposes. For example, a C program is not well-formed if a variable is used without being defined. That's a syntax error, but it's not detected by the CFG parser. The grammatical productions needed to define these cases are quite complicated, due to the complicated syntax of the language, but they're going to boil down to requiring that ids appear twice in a valid program. Hence L1 = {wcw|w is {a,b}+} (here w is the identifier, and c is way too complicated to spell out). In practice, checking this requirement is normally done with a symbol table, and the formal language rules, while precise, are not written in a logical formalism. Since L1is not a context-free language, the formalism could not be context-free, but a context-sensitive grammar can recognize L1, so it's not totally impossible. (See, for example, Algol 68.)
The symbol table is also used to decide whether a particular identifier is to be reduced to typedef-name [2]. This is required to resolve a number of ambiguities in the grammar. (It also further restricts the set of strings in the language, because there are some cases where an identifier must be resolved as a typedef-name in order for the program to be valid.)
For another type of context-sensitivity, function calls need to match function declarations in the number of arguments; this sort of requirement is modelled by L2 = {a^n b^m c^n d^m| n,m >=1} where a and c represent the definition and use of some function, and b and d represent the definition and use of a different function. (Again, in a highly-simplified form.)
This second requirement is possibly less clearly a syntactic requirement. Other languages (Python, for example) allow function calls with any number of arguments, and detect a argument/parameter count match as a semantic error only detected at runtime. In the case of C, however, a mismatch is clearly a syntax error.
In short, the set of grammatically valid strings which constitute the C language is a proper subset of the set of strings recognized by the CFG presented in the C language definition; the set of valid parses is a proper subset of the set of derivations generated by the CFG, and the language itself is (a) unambiguous, and (b) not context-free.
Note 1: Most of these are not really ambiguities, because they depend upon how a given identifier is resolved (typedef name, function identifier, declared variable,...).
Note 2: It is not the case that identifier must be resolved as a typedef-name if it happens to be one; that only happens in places where the reduction is possible. It is not a syntax error to use the same identifier for both a type and a variable, even in the same scope. (It's not a good idea, but it's valid.) The following example, adapted from an example in section 6.7.8 of the standard, shows the use of t as both a field name and a typedef:
typedef signed int t;
struct tag {
unsigned t:4; // field named 't' of type unsigned int
const t:5; // unnamed field of type 't' (signed int)
};
These things aren't context-free in C:
foo * bar; // foo multiplied by bar or declaration of bar pointing to foo?
foo(*bar); // foo called with *bar as param or declaration of bar pointing to foo?
foo bar[2] // is bar an array of foo or a pointer to foo?
foo (bar baz) // is foo a function or a pointer to a function?
In the ANSI C grammar for ANTLR v3 ( http://antlr.org/grammar/1153358328744/C.g ), how can init_declarator_list be optional in rule declaration ?
Instead of:
| declaration_specifiers init_declarator_list? ';'
-I would say:
| declaration_specifiers init_declarator_list ';'
What part of the C standard allows statements like:
int;
EDIT:
I just tried, it is allowed! Okay then, why is it allowed?
A wild guess: To make it simpler to write programs that produce machine-generated C.
Probably the ANTLR grammar is directly following that of the C standard grammar. I haven't read the C standard but for C++, the standard says separately that the init_declarator_list can be omitted only when declaring a class or enum type. So the grammar alone only encompasses all the possible forms of declaration while each particular case is further defined using plain language.
As to the case you indicated, int; is disallowed by the rules outside the grammar.
Note that the C/C++ language cannot be completely defined by the grammar alone. Many extra rules must be specified in plain human language.