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djowel_at_[hidden]

-- 
Joel de Guzman
http://www.boost-consulting.com
http://spirit.sf.net
Michael Burbidge <mburbidg_at_[hidden]> wrote:
> I have the following very simple grammar:
> 
> struct expr_grammar : public grammar<expr_grammar>
> {
> template <typename ScannerT>
> struct definition
> {
> definition(expr_grammar const& self)
> {
> variable_ref
> = lexeme_d[ch_p('$') >> symbol]
> ;
> 
> symbol
> = range_p('a', 'A')
> ;
> }
> 
> rule<ScannerT> variable_ref, symbol;
> 
> rule<ScannerT> const& start() const { return variable_ref; }
> };
> };
> 
> and the following very simple function that uses this grammar:
> 
> bool parse_expr(char* theExpr)
> {
> expr_grammar g;
> bool result = parse(theExpr, g, space_p).full;
> }
> 
> This results in a compiler error because no function or template can be
> found for parse. If I remove the lexeme_d directive from the
> variable_ref rule then it compiles just fine. How did lexeme_d change
> the type of the parameters to parse? How can I fix this so that I can
> use lexeme_d to prevent  accepting spaces between any of the characters
> in variable_ref?