Verbose parser

[robinwils]

Hi,

Is there a way to make the parser verbose ? I have a problem where the parser generated is segfaulting in result of a stack overflow but I can't really see what is going on in GDB. Any tips would be appreciated.

[PhilippeSigaud]

IIRC, you can evoke any rule by calling YourGramme.ruleX( ... ). That bypasses the call to decimateTree and should give you a more complete parse tree. You could use it to call the very first rule of your gramme. But that's still not logging per se.

Or else, you could add a logging action after your rules (or some selected, low level rules). Here is a small program demonstrating this:

import std.stdio;
//import std.typecons;
//import std.typetuple;

import pegged.grammar;

PT log(PT)(PT p)
{
    writeln("Logging: ", p);
    return p;
}

mixin(grammar("
   MyGram:
      A <{log} 'a' B
      B <{log}  C*
      C <{log} 'b'
"));

void main() {

  // Non-decimated tree:
  writeln(MyGram.A("abbbb_"));
}

Another possibility would be to use the fail pre-defined rule, which makes the rule fail. So in way it can be used as a breakpoint.

[robinwils]

Thanks for the tips, I wasn't using the Semantic Actions properly apparently !

[PhilippeSigaud]

Good! Keep me informed if you have any other difficulty.

On Tue, Mar 17, 2015 at 5:57 PM, Robin WILS notifications@github.com wrote:

Thanks for the tips, I wasn't using the Semantic Actions properly apparently !

— Reply to this email directly or view it on GitHub https://github.com/PhilippeSigaud/Pegged/issues/150#issuecomment-82478471 .