Sorry, I mend to say: Even if a grammar is not ambiguous, it can require unbound look-ahead to be parsed correctly [1]. The grammar of C is ambiguous. The stat

Sorry, I mend to say: Even if a grammar is not ambiguous, it can require unbound look-ahead to be parsed correctly [1]. The grammar of C is ambiguous. The statement "a * b;" can be both parsed as a variable declaration of the variable 'b' of type pointer to 'a' and as an expression consisting of a multiplication of 'a' and 'b'. It all depends on whether 'a' is a type or not. In most cases it would be a type definition, because why multiply two variables and ignore the result. One trick to deal with this is to give precedence for the type declaration grammar rule over the expression grammar rule. However, this is not something that can be done with many parser generating tools. Yet the first C compiler where single pass compilers with a single look-ahead lexical token probably implemented as a recursive descent parser. It worked, because it kept a (reverse) list of all variable declarations, which allowed it to determine when 'a' was parsed if it was the start of some declaration or the start of a statement based on whether it was defined before as a type or not. [1] https://stackoverflow.com/questions/12971191/grammars-that-require-unlimited-lookahead https://stackoverflow.com/questions/12971191/grammars-that-r...