const_cast<type-id> (expression)

Class Type class b{    public:         int m_num;        b (): M_num ({}};   )                                                                                                                                          vOid foo (void)  {    const b* b1 = new b ();     B* b2 = const_cast<B*> (B1);    b2->m_num =  200;    cout << "B1:"  << b1->m_num << endl;// 200    cout << "B2:"  << b2->m_num << endl;// 200    const b b3;    b b4 = const_cast<b &> (b3);    b4.m_num = 300;    cout <<   "B3:"  << b3.m_num << endl;//50    cout <<   "B4:"  << b4.m_num << endl;//300}//************************//

Basic Type Void foo () {    const int a = 100;     int* p1 = const_cast<int*> (&a);     *p1  = 200;    cout << *p1 << endl;//200     cout << a << endl;//100    const int*  p2 = new int (;    int* p3 = const_cast<int*>) (p2);    *p3 = 200;    cout << *p2 < < endl;//200                                                          cout << *p3 << endl;//200}//************************// 

You will find:

A: Can be a basic type or class type;

const a A; Whatever you want to change a.

A const a * p = new A (); After removing the Const property of P, the *p changes.

class a{    public:                                                                                 a () {             m_num=1;         }           int m_num;}; void foo  (void) {    a a;    const a &r  = a;    A a1 = const_cast<A&> (a);     a1.m_num = 200;    cout << a1.m_num << endl;//200    cout  << a.m_num << endl;//1}//****************//

const_cast<type-id> (expression), Type-id can only be a pointer or reference, the other is wrong, the expression can be removed

The Const property or the Volatil property in expression can also increase the Const property or the Volatil property

const int i = 10;

int i1 = const_cast<int> (i)//error

Adding the Const property is the opposite of the Volatil property.

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const_cast<type-id> (expression)