Re: Re: [C++-11] User defined literals

[3dw4rd at verizon dot net]

> 
> Oct 31, 2011 11:56:59 AM, jason@redhat.com wrote:
> 
> On 10/30/2011 01:13 PM, Ed Smith-Rowland wrote:
> > +  /* Look for a literal operator taking the exact type of numeric argument
> > +     as the literal value.  */
> 
> Is this right?  Do numeric literals only get here with type unsigned 
> long long or long double?

Yes, the preprocessor interprets a numeric literal either as integral or floating point in the usual way.  If an unrecognized suffix is found then the numeric part is interpreted as unsigned long long or as long double as appropriate.  This happens in the lexer.  Anything with a double quoted front part is sent to string literal.  Anything with a single quoted bit is set to char literal.

> > +  while (fns)
> >      {
> >        tree tmpl_args = make_char_string_pack (num_string);
> > -      decl = lookup_template_function (decl, tmpl_args);
> > -      result = finish_call_expr (decl, &args, false, true, tf_none);
> > -      if (result != error_mark_node)
> > +      tree fn = OVL_CURRENT (fns);
> > +      tree argtypes = NULL_TREE;
> > +      if (TREE_CODE (TREE_TYPE (fn)) != LANG_TYPE)
> > + argtypes = TYPE_ARG_TYPES (TREE_TYPE (fn));
> > +      if (argtypes != NULL_TREE
> > +   && same_type_p (TREE_VALUE (argtypes), void_type_node))

> Let's wait to make the pack until we find a template.  Also, you should 
> be able to just check for TEMPLATE_DECL since we won't accept a literal 
> operator template with different parameter types.

> For string and character literals, we can still just build up a call; we 
> only need to walk the overload list here for numeric literals.

I found that if you don't walk the overload list for chars, a char could be routed to the operator taking wchar_t for example.  This is similar to the comment you made for numeric literals - 1.2_foo potentially going to operator"" _foo(long long unsigned).  I think for strings you may be right.

Jason