const_cast：仅当原始变量为const时，才可以修改以前的const值

如何解决const_cast：仅当原始变量为const时，才可以修改以前的const值

https://stackoverflow.com/a/332086/462608

仅当原始变量为const时，才修改未定义的const值

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如果您使用它来使const脱离对未使用const声明的内容的引用，那将是安全的。

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例如，这在基于const重载成员函数时非常有用。它还可以用于向对象添加const，例如调用成员函数重载。

我无法理解以上引用的含义。我要求您给我一些例子，以实际说明这些引述的含义。

解决方法

关于前两个引号：

void do_not_do_this(const int& cref) {
    const_cast<int&>(cref) = 42;
}

int main() {
    int a = 0;
    // "if you use it to take the const off a reference 
    // to something that wasn't declared with const,it is safe."
    do_not_do_this(a);  // well-defined
        // a is now 42.
    
    // "modifying a formerly const value is only 
    //  undefined if the original variable is const"
    const int b = 0;
    do_not_do_this(a);  // undefined behavoiur
}

关于您的最终报价：

// "This can be useful when overloading member functions based
//  on const,for instance. It can also be used to add const
//  to an object,such as to call a member function overload."
class A {
    const int& get() const
    {
        // ... some common logic for const and
        // non-const overloads.
        return a_;
    }

    int& get() {
        // Since this get() overload is non-const,the object itself
        // is non-const in this scope. Moreover,the a_ member
        // is non-const,and thus casting away the const of the return
        // from the const get() (after 'this' has been casted to
        // const) is safe.
        A const * const c_this = this;
        return const_cast<int&>(c_this->get());
    }
    
private:
    int a_{0}; 
}

,

如何？

#include <iostream>
void foo(const int& ub,const int& ok)
{
    const_cast<int&>(ub) = 0.0; // undefined behaviour - the original object is const
    const_cast<int&>(ok) = 1.0; // this is fine - the original object is not const
}

int main()
{
    const int ub = 1.0;
    int ok = 0.0;
    foo(ub,ok);
    std::cout << ub << " " << ok << std::ends;
}

请注意，常见编译器的输出为1 1：理由是编译器知道ub不能在main中进行更改，因此它将{中的ub替换为1 {1}}通话。

您的第三段暗示了一个非std::cout成员函数的函数体，该成员函数调用const版本是为了避免代码重复。