|
||||
|
Section 18:
|
[18.17] Why am I getting an error converting a Foo** → Foo const**?
Because converting Foo** → Foo const** would be invalid and dangerous. C++ allows the (safe) conversion Foo* → Foo const*, but gives an error if you try to implicitly convert Foo** → Foo const**. The rationale for why that error is a good thing is given below. But first, here is the most common solution: simply change Foo const** to Foo const* const*:
class Foo { /* ... */ };
void f(Foo const** p);
void g(Foo const* const* p);
int main()
{
Foo** p = /*...*/;
...
f(p); // ERROR: it's illegal and immoral to convert Foo** to Foo const**
g(p); // OK: it's legal and moral to convert Foo** to Foo const* const*
...
}
The reason the conversion from Foo** → Foo const**
is dangerous is that it would let you silently and accidentally modify a
const Foo object without a cast:
class Foo {
public:
void modify(); // make some modification to the this object
};
int main()
{
const Foo x;
Foo* p;
Foo const** q = &p; // q now points to p; this is (fortunately!) an error
*q = &x; // p now points to x
p->modify(); // Ouch: modifies a const Foo!!
...
}
If the q = &p line were legal, q would be pointing at
p. The next line, *q = &x, changes p itself (since
*q is p) to point at x. That would be a bad thing,
since we would have lost the const qualifier: p is a
Foo* but x is a const Foo. The p->modify()
line exploits p's ability to modify its referent, which is the real
problem, since we ended up modifying a const Foo.
By way of analogy, if you hide a criminal under a lawful disguise, he can then exploit the trust given to that disguise. That's bad. Thankfully C++ prevents you from doing this: the line q = &p is flagged by the C++ compiler as a compile-time error. Reminder: please do not pointer-cast your way around that compile-time error message. Just Say No! (Note: there is a conceptual similarity between this and the prohibition against converting Derived** to Base**.) |
|||