%type'd non-terminal produces a `has no declared type' error

[Alexander Sorockin]

Hi,

Has anyone besides me noticed that $n with n zero or negative always  
triggers a "$n ... has no declared type" error when a %union is used?  
Cosider this grammar, which is a slightly modified example from the manual:

%union {
        int val;
}

%token <val> NUM

%start foo

%%

foo
        : expr bar '+' expr
        | expr bar '-' expr

bar:    /* empty */
	{ previous_expr = $0; } /* <-- "test.y:16.27-28: $0 of `bar' has no  
declared type" */
        ;

expr
        : NUM;

%%

The error persists even if you insert "%type <val> expr". I know this can  
be worked around by using "$<val>0", but I think this error shouldn't  
appear at all.

In case anyone is interested, here's the piece of code that issues the  
error.

File scan-gram.l:

  ...

static inline bool
handle_action_dollar (char *text, location loc)
{
  const char *type_name = NULL;

  ...

  /* Get the type name if explicit. */
  if (*cp == '<')
    {
      type_name = ++cp;
      while (*cp != '>')
        ++cp;
      *cp = '\0';
      ++cp;
    }

  if (*cp == '$')
    {
        ...
    }
  else
    {
      long int num = strtol (cp, NULL, 10);

      if (1 - INT_MAX + rule_length <= num && num <= rule_length)
        {
          int n = num;
          ...
          if (!type_name && 0 < n)
            //!!!
            //!!! Because n <= 0 in our case, type_name is never retrieved.
            //!!!
            type_name = symbol_list_n_type_name_get (current_rule, loc, n);
          //!!!
          //!!! Thus, unless the type name is referred to explicitly, zero
          //!!! or negative semantic contexts can't coexist with %union's.
          //!!!
          if (!type_name && typed)
            complain_at (loc, _("$%d of `%s' has no declared type"),
                         n, current_rule->sym->tag);
          ...
        }
        ...
    }

That was an extract from version 2.3 code, but this piece doesn't seem to  
have changed much in version 2.4.1.

Best,
Sandy