Why am I getting an error converting a Foo → Foo const?

最新推荐文章于 2019-09-29 08:59:32 发布

转载最新推荐文章于 2019-09-29 08:59:32 发布 · 511 阅读

C/C++ fundamentals 专栏收录该内容

53 篇文章

订阅专栏

本文详细解释了在 C++ 中为什么不能将 Foo** 隐式转换为 Fooconst**，以及如何通过改变类型定义来避免潜在的运行时错误。通过具体的代码示例展示了不当转换可能引发的问题，并提出了安全的解决方案。

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**.)

From:http://www.parashift.com/c++-faq/constptrptr-conversion.html

It is correct that a double ** cannot be implicitly converted to a const double **. It can be converted to a const double * const *, though.

Imagine this scenario:

const double cd = 7.0;
double d = 4.0;
double *pd = &d;
double **ppd = &pd;
const double **ppCd = ppd;  //this is illegal, but if it were possible:
*ppCd = &cd;  //now *ppCd, which is also *ppd, which is pd, points to cd
*pd = 3.14; // pd now points to cd and thus modifies a const value!

So, if your function does not intend to modify any of the pointers involved, change it to take a const double * const *. If it intends to do modifications, you must decide whether all the modifications it does are safe and thus const_cast can be used, or whether you really need to pass in a const double **.

From: http://stackoverflow.com/questions/19910296/how-to-convert-pointer-to-pointer-type-to-const