yacc.parse(debug=1) does not generate 'copious amounts of debugging during parsing'

haoxiang47 / ply

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yacc.parse(debug=1) does not generate 'copious amounts of debugging during parsing' #6

Closed GoogleCodeExporter closed 8 years ago

GoogleCodeExporter commented 8 years ago

The documentation of PLY promises 'copious amounts of debugging during
parsing' with yacc.parse(debug=1) in the section '5.12 Yacc implementation
notes'.
debug=1 however only prints an additional line when an error is found.

To get the desired output, one should enter a number > 1 (ie
'yacc.parse(debug=2)'.

This probably holds also when re-directing output to file (the next item).

Original issue reported on code.google.com by a.t.hofk...@tue.nl on 22 Jan 2008 at 12:22

GoogleCodeExporter commented 8 years ago

Original comment by dbeaz...@gmail.com on 4 May 2008 at 3:13

Changed state: Fixed