On 18.12.2013 11:11:10, Richard Braun wrote: > The expression is ((a == NULL) || a->something), and I agree it is > equivalent to !((a != NULL) && !a->something). And again, both the > && and || operators are guaranteed to be evaluated left-to-right and > *yield* without evaluating the second operand if the first compares > or not to 0, depending on the operator. > > So, let's take the seconds form since that's what you've used, > without the negation for simplicity : ((a != NULL) && !a->something) > > If a isn't NULL, then it returns !a->something. If a is NULL, then > (a != NULL) compares equal to 0, and && returns 0 before evaluating > !a->something. So no, there can't be a null pointer dereference here. > > And this really is basic C, so please double check your changes.
Yes, vou're right. I got confused. But then something else is happening here. When I write the assertion this way: assert((entry == IE_NULL) || ((entry != IE_NULL) ? IE_BITS_TYPE (entry->ie_bits) : TRUE)); GCC stops complaing about uninitialized variable.
