C++ FAQ Celebrating Twenty-One Years of the C++ FAQ!!!
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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 {
  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**.)