Canonicalize conversion operators

Consider a conversion operator such as:

operator foo const* const* ();

There are two small parser problems, highlighted by this test:

(gdb) p operator foo const* const*
There is no field named operatorfoo const* const *

GDB is looking up the symbol "operatorfoo const* const*" -- it is missing a
space between the keyword "operator" and the type name "foo const* const*".

Additionally, this input of the user-defined type needs to be canonicalized
so that different "spellings" of the type are recognized:

(gdb) p operator const foo* const *
There is no field named operator const foo* const *

gdb/ChangeLog:

	* c-exp.y (oper): Canonicalize conversion operators of user-defined
	types.
	Add whitespace to front of type name.

gdb/testsuite/ChangeLog:

	* gdb.cp/cpexprs.cc (base) <operator fluff const* const*>: New
	method.
	(main): Call it.
	* gdb.cp/cpexprs.exp: Add new conversion operator to test matrix.
	Add additional user-defined conversion operator tests.

1 files changed, 8 insertions, 2 deletions

diff --git a/gdb/c-exp.y b/gdb/c-exp.y
index 7c050b4..18c74d8 100644
--- a/gdb/c-exp.y
+++ b/gdb/c-exp.y
@@ -1554,7 +1554,7 @@ oper:	OPERATOR NEW
 	|	OPERATOR '>'
 			{ $$ = operator_stoken (">"); }
 	|	OPERATOR ASSIGN_MODIFY
-			{ const char *op = "unknown";
+			{ const char *op = " unknown";
 			  switch ($2)
 			    {
 			    case BINOP_RSH:
@@ -1630,7 +1630,13 @@ oper:	OPERATOR NEW
 
 			  c_print_type ($2, NULL, &buf, -1, 0,
 					&type_print_raw_options);
-			  $$ = operator_stoken (buf.c_str ());
+
+			  /* This also needs canonicalization.  */
+			  std::string canon
+			    = cp_canonicalize_string (buf.c_str ());
+			  if (canon.empty ())
+			    canon = std::move (buf.string ());
+			  $$ = operator_stoken ((" " + canon).c_str ());
 			}
 	;