指向c ++中const int的指针

Question

在C ++中，我有下一个代码

int main() {
    int i = 1;
    cout<<"i = "<<i<<endl; //prints "i = 1"

    int *iPtr = &i;
    cout<<"*iPtr = "<<*iPtr<<endl; //prints "*iPtr = 1"

    (*iPtr) = 12; //changing value through pointer

    cout<<"i = "<<i<<endl; //prints "i = 12"
    cout<<"*iPtr = "<<*iPtr<<endl; //prints "*iPtr = 12"

    system("pause");
    return 0;
}

现在与常数整数 i]相同的代码

int main() {
    const int i = 1;
    cout<<"i = "<<i<<endl; //prints "i = 1"

    int *iPtr = (int*)&i; //here I am usint a type conversion
    cout<<"*iPtr = "<<*iPtr<<endl; //prints "*iPtr = 1"

    (*iPtr) = 12; //changing value through pointer

    cout<<"i = "<<i<<endl; //prints "i = 1"
    cout<<"*iPtr = "<<*iPtr<<endl; //prints "*iPtr = 12"

    system("pause");
    return 0;
}
如您所见，在第二种情况下使用常量整数，* iPtr和const i有两个不同的值，但是指针* iPtr显示为常量i。请告诉我第二种情况会发生什么，为什么？

在c ++中，我有下一个代码int main（）{int i = 1; cout <

Answer 1

您的第二个代码具有未定义的行为

Answer 2

您正在尝试删除变量的const限定词。在C ++中，您应该使用const_cast来做到这一点。