const转换为全局var和程序崩溃（C ++） [英] const cast to a global var and program crashed (C++)

问题描述

int main()
{
    const int maxint=100;//The program will crash if this line is put outside the main
    int &msg=const_cast<int&>(maxint);  
    msg=200;  
    cout<<"max:"<<msg<<endl; 
    return 0;
}

如果'const int maxint = 100;'定义函数将运行ok被放置在主函数内部，但是崩溃并弹出一个错误消息，如果放在外面，说访问违反。

The function will run ok if the 'const int maxint=100;' definition is put inside the main function but crash and popup a error message said "Access Violation" if put outside.

有人说这是一种未定义的行为想知道确切的答案，以及我如何可以安全地使用const转换？

Someone says it's some kind of 'undefined behavior', and i want to know the exact answer and how i can use the const cast safely?

推荐答案

它们是正确的，你不允许修改 const 变量的值，这是抛弃 const 的危险：这不是真的 const 。

They are correct, it is undefined behaviour. You're not allowed to modify the value of a const variable, which is the danger of casting away the constness of something: you better know it's not really const.

编译器， maxint 是 const 并且应该永远不会被修改，甚至不必给它的地址。它可以替换 maxint 的所有用法，如果它看上去合适的话。也可能只是把常量放在一部分只读的内存中，正如Matteo Italia指出的，这可能是你发生了什么。这是为什么修改它会产生未定义的行为。

The compiler, seeing that maxint is const and should never be modified, doesn't even have to give it an address. It can just replace all the uses of maxint with 100 if it sees fit. Also it might just put the constant in to a portion of memory that is read-only, as Matteo Italia points out, which is probably what's happening for you. That's why modifying it produces undefined behaviour.

你可以安全地抛弃变量的 const 是如果变量实际上不是 const ，但 const 限定符添加到非 - const 变量，如下所示：

The only way you can safely cast away the constness of a variable is if the variable is not actually const, but the const qualifier was added to a non-const variable, like this:

int f(const int& x) {
    int& nonconst = const_cast<int&>(x);

    ++nonconst;
}

int blah = 523;

f(blah); // this is safe

const int constblah = 123;

f(constblah); // this is undefined behaviour

想想这个例子，它完美编译：

Think about this example, which compiles perfectly:

int f(const int& x) {
    int& nonconst = const_cast<int&>(x);

    ++nonconst;
}

int main() {
    f(4); // incrementing a number literal???
}

你可以看到如何使用 const_cast 是非常危险的，因为没有办法实际告诉变量是否原来 const 或不。你应该避免在可能的情况下使用 const_cast （通过不接受 const 参数的函数）。

You can see how using const_cast is pretty dangerous because there's no way to actually tell whether a variable is originally const or not. You should avoid using const_cast when possible (with functions, by just not accepting const parameters).

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