Fix "b f(std::string)" when current language is C

Fix "b f(std::string)" when current language is C

If you try to set a breakpoint at a function such as "b
f(std::string)", and the current language is C, the breakpoint fails
to be set, like so:

  (gdb) set language c
  break f(std::string)
  Function "f(std::string)" not defined.
  Make breakpoint pending on future shared library load? (y or [n]) n
  (gdb)

The problem is that the code in GDB that expands the std::string
typedef hits this in c-typeprint.c:

      /* If we have "typedef struct foo {. . .} bar;" do we want to
	 print it as "struct foo" or as "bar"?  Pick the latter for
	 C++, because C++ folk tend to expect things like "class5
	 *foo" rather than "struct class5 *foo".  We rather
	 arbitrarily choose to make language_minimal work in a C-like
	 way. */
      if (language == language_c || language == language_minimal)
	{
	  if (type->code () == TYPE_CODE_UNION)
	    gdb_printf (stream, "union ");
	  else if (type->code () == TYPE_CODE_STRUCT)
	    {
	      if (type->is_declared_class ())
		gdb_printf (stream, "class ");
	      else
		gdb_printf (stream, "struct ");
	    }
	  else if (type->code () == TYPE_CODE_ENUM)
	    gdb_printf (stream, "enum ");
	}

I.e., std::string is expanded to "class std::..." instead of just
"std::...", and then the "f(class std::..." symbol doesn't exist.

Fix this by making cp-support.c:inspect_type print the expanded
typedef type using the language of the symbol whose type we're
expanding the typedefs for -- in the example in question, the
"std::string" typedef symbol, which is a C++ symbol.

Use type_print_raw_options as it seems to me that in this scenario we
always want raw types, to match the real symbol names.

Adjust the gdb.cp/break-f-std-string.exp testcase to try setting a
breakpoint at "f(std::string)" in both C and C++.

Change-Id: Ib54fab4cf0fd307bfd55bf1dd5056830096a653b