RE: why are we not using const?

["Dave Korn" <dave dot korn at artimi dot com>]

On 14 July 2006 18:22, Kaveh R. Ghazi wrote:

> Now imagine if I change the type of __t to const_tree like so:
> 
> typedef const union tree_node *const_tree;
> #define TREE_CHECK(T, CODE) __extension__ \
> ({  const_tree const __t = (T); \
>     if (TREE_CODE (__t) != (CODE)) \
>       tree_check_failed (__t, __FILE__, __LINE__, __FUNCTION__, (CODE), 0);
>     \ __t; })

> Now the problem is that the statement expression in the TREE_CHECK
> macro uses __t as it's "return value".  The TREE_CHECK macro is used
> in both read and write situations of tree members.  So the "return
> value" of this statement expression cannot be const or all the write
> cases barf.

  So just return T instead of _t.

> Another solution might be to use the macro parameter as the return
> type like so:

  Ah, I see you thought of that!
 
> typedef const union tree_node *const_tree;
> #define TREE_CHECK(T, CODE) __extension__ \
> ({  const_tree const __t = (T); \
>     if (TREE_CODE (__t) != (CODE)) \
>       tree_check_failed (__t, __FILE__, __LINE__, __FUNCTION__, (CODE), 0);
>     \ (T); })
> 
> Then if "T" isn't already const you can write to it.  But this
> evaluates "T" twice so if there are side-effects potentially we get
> hosed.

> Any ideas?

  Use /two/ locals like this?

typedef const union tree_node *const_tree;
#define TREE_CHECK(T, CODE) __extension__ \
({  \
      __typeof__(T) __t = (T); \
      const_tree const __ct = t; \
      if (TREE_CODE (__ct) != (CODE)) \
        tree_check_failed (__ct, __FILE__, __LINE__, __FUNCTION__, (CODE), 0);
      \ __t; })


  So __ct is always const, and __t has the same const-ness as T.

  Hmm.  The docs don't specify whether typeof preserves CV-qualification.  We
could always add a __qualified_typeof__ that did if ordinary __typeof__
doesn't, I suppose.

    cheers,
      DaveK
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