std ::对象的对象和const正确性 [英] std::vector of objects and const-correctness

问题描述

请考虑以下内容：

class A {
public:
    const int c; // must not be modified!

    A(int _c)
    :   c(_c)
    {
        // Nothing here
    }

    A(const A& copy)
    : c(copy.c)
    {
        // Nothing here
    }    
};



int main(int argc, char *argv[])
{
    A foo(1337);

    vector<A> vec;
    vec.push_back(foo); // <-- compile error!

    return 0;
}

显然，复制构造函数是不够的。我缺少什么？

Obviously, the copy constructor is not enough. What am I missing?

编辑：
Ofc。我不能在operator =（）方法中改变this-> c，所以我看不到如何使用operator =（）（尽管std :: vector需要）。

Ofc. I cannot change this->c in operator=() method, so I don't see how operator=() would be used (although required by std::vector).

推荐答案

我不知道为什么没有人说的，但正确的答案是放下 const c $ c> A * 在向量中（使用适当的智能指针）。

I'm not sure why nobody said it, but the correct answer is to drop the const, or store A*'s in the vector (using the appropriate smart pointer).

你可以给你的类可怕的语义复制调用UB或不做任何事（因此不是一个副本），但为什么所有这些麻烦在UB和坏代码？通过使 const 得到什么？ （提示：没有。）你的问题是概念性的：如果一个类有一个const成员，该类是const。对象是const，基本上不能被赋值。

You can give your class terrible semantics by having "copy" invoke UB or doing nothing (and therefore not being a copy), but why all this trouble dancing around UB and bad code? What do you get by making that const? (Hint: Nothing.) Your problem is conceptual: If a class has a const member, the class is const. Objects that are const, fundamentally, cannot be assigned.

只要使它成为非常量私有<不可改变。对于用户，这是等价的，const。它允许隐式生成的函数正常工作。

Just make it a non-const private, and expose its value immutably. To users, this is equivalent, const-wise. It allows the implicitly generated functions to work just fine.

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