问题
I was wondering whether the following is undefined behavior
// Case 1:
int *p = 0;
int const *q = *const_cast<int const* const*>(&p);
// Case 2: (I think this is the same)
int *p = 0;
int const *const *pp = &p;
int const *q = *pp;
Is this undefined behavior by reading a int* as if it were a int const*? I think it is undefined behavior, but I previously thought that only adding const in general is safe, so I'm unsure.
回答1:
Qualification-wise, it's fine. With each expression split into a statement:
int *p = 0; // ok
int **addrp = &p; // ok
int const *const *caddrq = addrp; // ok, qualification conv. according to §4.4/4
int const *q = *caddrq; // ok
Note that the rules of const_cast (§5.2.11/3) are identical to those of qualification conversion, but without the requirement of being monotonically increasing in qualification. In your case, because you're only ever adding qualifications the const_cast is unnecessary.
Concerning aliasing, I don't think it's an issue, here, or at least it's not intended to be.
Like you mentioned, there's a new bullet in the C++0x list of allowed access methods (§3.10) that allows similar types ("similar" being types arising from qualification conversions). In C++03 that bullet is missing, but I suspect that the bullet about allowing more cv-qualified access was meant to cover that, but it technically isn't so (that is, the commitee overlooked this).
来源：https://stackoverflow.com/questions/5969867/is-this-const-cast-undefined-behavior