我们可以使用const_cast修改常量吗？ [英] Can we use `const_cast` to modify a constant variable?

问题描述

int main()
{
    const int ia = 10;

    int *pia = const_cast<int*>(&ia);
    *pia = 5;

    std::cout << &ia << "\t" <<  pia <<endl;
    std::cout <<  ia << "\t" << *pia <<endl;

    return 0;
}

输出为：

0x28fef4       0x28fef4
10             5

* pia 和 ia 具有相同的地址，但它们的值不同。我的目的是使用 const_cast 修改一个常量值，但结果表明它不起作用。

*pia and ia have the same address, but they have different values. My purpose is to use const_cast to modify a constant value, but as the result shows that it does not work.

有人知道为什么吗？

推荐答案

看到 10 的原因为 ia 打印很可能是编译器优化：它看到一个 const 对象，决定它不会改变，并用以下内容替换最后的打印输出：

The reason why you see 10 printed for ia is most likely the compiler optimization: it sees a const object, decides that it's not going to change, and replaces the last printout with this:

cout<< 10 <<"  "<<*ppa<<endl;

换句话说，生成的代码具有 const的值烘焙到二进制文件中。

In other words, the generated code has the value of the const "baked into" the binary.

放弃最初被声明为 const <的对象的常量性。 / code>并写入该对象是未定义的行为：

Casting away the const-ness of an object that has originally been declared as const and writing to that object is undefined behavior:

$ 5.2.11 / 7-注意：取决于类型该对象，通过由const_cast抛出const-qualifier68的指针，左值或指向数据成员的写入操作，可能会产生未定义的行为（7.1.5.1）。

$5.2.11/7 - Note: Depending on the type of the object, a write operation through the pointer, lvalue or pointer to data member resulting from a const_cast that casts away a const-qualifier68) may produce undefined behavior (7.1.5.1).

根据平台的不同， const 对象可能会放置在受保护的内存区域中，而您不能在其中写入内容。解决类型系统中的 const -ness可能有助于您的程序编译，但是您可能会看到随机结果甚至崩溃。

Depending on the platform, const objects may be placed in a protected region of memory, to which you cannot write. Working around the const-ness in the type system may help your program compile, but you may see random results or even crashes.

这篇关于我们可以使用const_cast修改常量吗？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！