如何在C ++中将指针Foo **转换为const Foo ** [英] How to pointer-cast Foo** to const Foo** in C++

问题描述

我有

class Fred 
{
public:
  void inspect() const {}; 
  void modify(){};
};

int main()
{
 const Fred x = Fred();
 Fred* p1;
 const Fred** q1 = reinterpret_cast<const Fred**>(&p1);
 *q1 = &x; 
 p1->inspect();
 p1->modify();
}

如何做
const Fred ** q1 =& p1
via pointer-casting？

How would it be possible to do the const Fred** q1 = &p1 via pointer-casting?

（我刚刚读过，可能会这样）

(I have just been reading that this might be possible)

感谢您的回答。 const_cast确实对对象起作用。

Thank you for your answers. The const_cast works indeed for objects

#include <iostream>
#include <stdio.h>
using namespace std;

class Fred 
{
 int a;

public:
Fred(){};
Fred(int a_input)
{
 a = a_input;
};

void inspect() const 
{
 cout << "Inspect called"<< endl;
 cout << "Value is ";
 cout << a << endl;
}; 

void modify()
{
 cout << "Modify called" << endl;
 a++;
};

};

int main()
{
 const Fred x = Fred(7);
 const Fred* q1 = &x;
 Fred* p1 = const_cast<Fred*>(q1); 
 p1->inspect();
 p1->modify();
 p1->inspect();
 x.inspect();
 *p1 = Fred(10);
 p1->inspect();
}

给予

Inspect called
Value is 7
Modify called
Inspect called
Value is 8
Inspect called
Value is 8
Inspect called
Value is 10
Inspect called
Value is 10

但是，对于预定义类型，它不起作用：

However, for predefined types it does not work:

int main()
{
 const double a1 = 1.2;
 const double* b1 = &a1;
 cout << "a1 is " << (*b1) << endl;
 cout << "b1 is " << b1 << endl;
 double* c1 = const_cast<double*>(&a1);
 cout << "b1 is " << b1 << endl;
 cout << "c1 is " << c1 << endl;

 double* d1 = static_cast<double*>(static_cast<void*>(c1));
 cout << "d1 is " << d1 << endl;
 cout<< "*d1 is " << *d1 << endl;

 *d1=7.3;

 cout<< "*d1 is " << *d1 << endl;
 cout<< "*d1 address is "<< d1 << endl;
 cout << "a1 is " << a1 << endl;
 cout << "a1 address is" << &a1 << endl;
 cout<< "*d1 is " << *d1 << endl;
 cout<< "*d1 address is "<< d1 << endl;

 double f1=a1;
 printf("f1 is %f \n", f1);
}

导致：

a1 is 1.2
b1 is 0xffbff208
b1 is 0xffbff208
c1 is 0xffbff208
d1 is 0xffbff208
*d1 is 1.2
*d1 is 7.3
*d1 address is 0xffbff208
a1 is 1.2
a1 address is0xffbff208
*d1 is 7.3
*d1 address is 0xffbff208
f1 is 1.200000 

显然，g ++编译器会优化，它，所以，即使它的值在堆栈上已经改变，它不在乎。

Apparently the g++ compiler optimizes such that it replaces a1 by 1.2 whenever it finds it, so, even if its value on the stack has changed, it does not care.

（在我的例子中，我遇到了直接读取* b1，* c1的问题，所以我不得不做双静态转换 - 重新解释转换没有工作）。

(In my case I had problems with directly reading the *b1, *c1, so I had to do the double static cast - the reinterpret cast did not work).

这是真正改变a1，编译正常的任何方式，因此没有优化没有编译（所以我超过了优化效果）？

Is it any way to really change a1, compiling "normally", therefore not compiling without the optimization (so I overtake the optimization effect)?

推荐答案

您希望 const_cast 。

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