• Tim (12/12) May 07 2023 I am releasing DParserGen 0.1.0, which is a parser and lexer
• max haughton (2/14) May 07 2023 Which parts of the D grammar require backtracking?
• Tim (18/19) May 07 2023 Backtracking is used to resolve some ambiguities in the grammar.

Tim <tim.dlang t-online.de> writes:

I am releasing DParserGen 0.1.0, which is a parser and lexer 
generator
for D. It supports LALR and GLR parsers.

The examples include grammars for D and C++. The D grammar uses 
LALR,
but also needs extensions like backtracking. For C++ a GLR parser 
is
used, so ambiguities result in special parse trees.

For more details see the repository:
https://github.com/tim-dlang/dparsergen

It can be used with dub:
https://code.dlang.org/packages/dparsergen

max haughton <maxhaton gmail.com> writes:

On Sunday, 7 May 2023 at 15:16:50 UTC, Tim wrote:
 I am releasing DParserGen 0.1.0, which is a parser and lexer 
 generator
 for D. It supports LALR and GLR parsers.

 The examples include grammars for D and C++. The D grammar uses 
 LALR,
 but also needs extensions like backtracking. For C++ a GLR 
 parser is
 used, so ambiguities result in special parse trees.

 For more details see the repository:
 https://github.com/tim-dlang/dparsergen

 It can be used with dub:
 https://code.dlang.org/packages/dparsergen

Which parts of the D grammar require backtracking?

Tim <tim.dlang t-online.de> writes:

On Sunday, 7 May 2023 at 18:38:09 UTC, max haughton wrote:
 Which parts of the D grammar require backtracking?

Backtracking is used to resolve some ambiguities in the grammar. 
The ambiguities could probably also be resolved with other 
solutions like lookahead.

You can find all nonterminals, which use backtracking, by 
searching for " backtrack" in the grammar file: 
https://github.com/tim-dlang/dparsergen/blob/master/examples/d/grammard.ebnf

One example is distinguishing types and expressions in 
TypeidExpression, TemplateArgument and other nonterminals. The 
parser first tries to parse it as a type and then as an 
expression. The parameter of the template instance 
"Template!(x*[0][0])" could be a type or an expression. As a type 
it would be an array of an array of a pointer to "x". As an 
expression it would be "x" multiplied by "[0][0]", which is 0. 
The parser will always parse this as a type.

For statements it also first tries to parse a declaration and 
then an expression, so "x*y;" is always parsed as a declaration. 
The same ambiguity exists for IfCondition.