Question

为什么我会参与此代码：

void foo ( const int ** );

int main() {
    int ** v = new int * [10];
    foo(v);

    return 0;
}

此错误：

invalid conversion from ‘int**’ to ‘const int**’ [-fpermissive]|

我认为可以从非const转换为const。

Answer 1

这是因为您尝试转换int** to const int**

int ** v = new int * [10]; // v is int**
foo(v); //but foo takes const int**

int **是：“指向整数的指针”。
const int **是：“指向常量整数的指针”。

使用const是一份合同，你不能通过两个引用的间接来满足这个合同。

来自标准：

const char c = 'c';
char* pc;
const char** pcc = &pc;   // not allowed (thankfully!)
                ^^^ here the bundit is hidden under const: "I will not modify"
*pcc = &c;                // *pcc is "pointer to const" right? so this is allowed...
*pc = 'C';                // would allow to modify a const object, *pc is char right?

因此可以修改const char 始终，只需使用上面的过程即可。

还有：

char *s1 = 0;
const char *s2 = s1; // OK...
char *a[MAX]; // aka char **
const char * const*ps = a; // no error!

从以下链接中引用：

通过类比的方式，如果你在合法的伪装下隐藏罪犯，他然后可以利用对伪装的信任。那很糟糕。

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

与此相关的

也是无效的转化Derived** → Base**。如果转换Derived** → Base**是合法的，则可以取消引用Base**（产生Base*），并且可以使Base *指向不同派生类的对象，可能会导致严重问题。了解原因：

class Vehicle {
public:
  virtual ~Vehicle() { }
  virtual void startEngine() = 0;
};

class Car : public Vehicle {
public:
  virtual void startEngine();
  virtual void openGasCap();
};

class NuclearSubmarine : public Vehicle {
public:
  virtual void startEngine();
  virtual void fireNuclearMissle();
};

int main()
{
  Car   car;
  Car*  carPtr = &car;
  Car** carPtrPtr = &carPtr;
  Vehicle** vehiclePtrPtr = carPtrPtr;  // This is an error in C++
  NuclearSubmarine  sub;
  NuclearSubmarine* subPtr = &sub;
  *vehiclePtrPtr = subPtr;
  // This last line would have caused carPtr to point to sub !
  carPtr->openGasCap();  // This might call fireNuclearMissle()!
  ...
}

http://www.parashift.com/c++-faq-lite/derivedptrptr-to-baseptrptr.html

考虑：

class Vehicle {
public:
  virtual ~Vehicle() { }
  virtual void startEngine() = 0;
};
class Car : public Vehicle {
public:
  virtual void startEngine(){printf("Car engine brummm\n");}
  virtual void openGasCap(){printf("Car: open gas cap\n");}
    virtual void openGasCap2(){printf("Car: open gas cap2\n");}
      virtual void openGasCap3(){printf("Car: open gas cap3\n");}
            virtual void openGasCap4(){printf("Car: open gas cap4\n");}
}; 
class NuclearSubmarine : public Vehicle {
public:
    int i;
  virtual void startEngine(){printf("Nuclear submarine engine brummm\n");}
    virtual void fireNuclearMissle3(){printf("Nuclear submarine: fire the missle3!\n");}
    virtual void fireNuclearMissle(){printf("Nuclear submarine: fire the missle!\n");}
  virtual void fireNuclearMissle2(){printf("Nuclear submarine: fire the missle2!\n");}
};   
int main(){
  Car   car; Car*  carPtr = &car;
  Car** carPtrPtr = &carPtr;
  //Vehicle** vehiclePtrPtr = carPtrPtr;  // This is an error in C++, But:
  Vehicle** vehiclePtrPtr = reinterpret_cast<Vehicle**>(carPtrPtr);
  NuclearSubmarine  sub; NuclearSubmarine* subPtr = &sub;
  *vehiclePtrPtr = subPtr; // carPtr points to sub !
  carPtr->openGasCap();  // Nuclear submarine: fire the missle3!
  carPtr->openGasCap2();  // Nuclear submarine: fire the missle!
  carPtr->openGasCap3();  // Nuclear submarine: fire the missle2!
  //carPtr->openGasCap4();  // SEG FAULT 
}

Answer 2

如果你在cv资格认证的第一个差异中添加了所有级别的const，那么你只能在类似的指针类型之间的转换中添加const限定。

因此，您可以将int**转换为int const* const*，但不能转换为int const* *。如果允许在中间级别省略添加const，则可以执行以下操作：

const int c = 29;
int *pi;
const int** ppci = &pi; // only adding const, right
*ppci = &c;
*pi = 0; // changing c ?! but no const_cast in sight

Answer 3

你被C ++对指针的令人困惑的解析规则误导了。看这种方式可能更清楚：

typedef const int * ptr_to_const_int;
void foo( ptr_to_const_int *);
int main() {
    int ** v = new int * [10];
    foo(v);

    return 0;
}

foo（）的参数列表承诺的是，你将向它传递一个指针（指向常量的东西）。但是新的int * [10]意味着“指向（指向非常量事物）”。

所以如果foo定义如下：

foo( const int **p )
{
  (*p); //<-- this is actually of type const int *
}

虽然我认为你期待

foo( const int **p )
{
  (*p); //<-- you're expecting this to be of type int *
  p = 0; //<-- and this to throw a compiler error because p is const
}

但并不是说你宣称自己是不变的，而是它指向的东西。

无论如何只需在这种情况下使用typedef，一切都清晰可读。

从int 转换为const int

3 个答案:

从int **转换为const int **

3 个答案:

从int 转换为const int