C++ 如何使用 const_cast？
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How to use const_cast?
提问by Sarah
I have a private variable in my Studentclass defined as:
我的类中有一个私有变量Student定义为：
const int studentNumnber;
I am trying to write a copy constructor for the Studentand I need to cast away the constness to do this. Unfortunately, I don't understand how to use std::const_cast.
我正在尝试为 编写一个复制构造函数Student，我需要抛弃常量性来做到这一点。不幸的是，我不明白如何使用std::const_cast.
This is what I am trying to do in my copy constructor:
这就是我在复制构造函数中尝试执行的操作：
Student(const Student & s)
: Person(p.getName(), p.getEmailAddress(), p.getBirthDate()), school(0), studentNumber(0) {
school = new char[strlen(s.school) + 1];
strcpy_s(school, strlen(s.school) + 1, s.school);
const_cast<int*>(this)->studentNumber = s.studentNumber;
//studentNumber = s.studentNumber);
}
That doesn't work... I am unsure of the syntax.
那不起作用......我不确定语法。
回答by timrau
You are not allowed to const_castvariables that are actually const. This results in undefined behavior. const_castis used to remove the const-ness from references and pointers that ultimately refer to something that is not const.
不允许您使用const_cast实际上是const. 这会导致未定义的行为。const_cast用于从最终引用不是const.
So, this is allowed:
所以，这是允许的：
int i = 0;
const int& ref = i;
const int* ptr = &i;
const_cast<int&>(ref) = 3;
*const_cast<int*>(ptr) = 3;
It's allowed because i, the object being assigned to, is not const. The below is not allowed:
这是允许的，因为i被分配给的对象不是const。以下是不允许的：
const int i = 0;
const int& ref = i;
const int* ptr = &i;
const_cast<int&>(ref) = 3;
*const_cast<int*>(ptr) = 3;
because here iis constand you are modifying it by assigning it a new value. The code will compile, but its behavior is undefined (which can mean anything from "it works just fine" to "the program will crash".)
因为这里i是const并且您正在通过为其分配一个新值来修改它。代码会编译，但它的行为是未定义的（这可能意味着从“它工作得很好”到“程序会崩溃”。）
You should initialize constant data members in the constructor's initializers instead of assigning them in the body of constructors:
您应该在构造函数的初始值设定项中初始化常量数据成员，而不是在构造函数的主体中分配它们：
Student(const Student & s)
: Person(p.getName(), p.getEmailAddress(), p.getBirthDate()),
school(0),
studentNumber(s.studentNumber)
{
// ...
}
回答by Ravi Podugu
In your code you are trying cast this pointer instead of variable. You can try the following:
在您的代码中，您正在尝试转换此指针而不是变量。您可以尝试以下操作：
Student(const Student & s)
: Person(p.getName(), p.getEmailAddress(), p.getBirthDate()), school(0), studentNumber(0) {
school = new char[strlen(s.school) + 1];
strcpy_s(school, strlen(s.school) + 1, s.school);
*const_cast<int*>(&studentNumber) = s.studentNumber;
}
