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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