非静态const成员，不能使用默认赋值运算符 [英] Non-static const member, can't use default assignment operator

问题描述

我正在扩展的程序使用 std :: pair<> 很多。

A program I'm expanding uses std::pair<> a lot.

在我的代码中，编译器抛出一个相当大的一点：

There is a point in my code at which the compiler throws a rather large:

非静态const成员'const Ptr std :: pair， const double *> :: first'不能使用默认赋值运算符

Non-static const member, 'const Ptr std::pair, const double*>::first' can't use default assignment operator

我不确定这是指什么？
Ptr类中缺少哪些方法？

I'm not really sure what this is referring to? Which methods are missing from the Ptr class?

导致此问题的原始调用如下：

The original call that causes this problem is as follows:

vector_of_connections.pushback(pair(Ptr<double,double>,WeightValue*));

std :: Pair< Ptr< double，double> ，WeightValue *> 到向量上，其中 WeightValue * 是来自约3个函数的const变量， Ptr< double，double> 取自在另一个向量上工作的迭代器。

Where it's putting an std::Pair<Ptr<double,double>, WeightValue*> onto a vector, where WeightValue* is a const variable from about 3 functions back, and the Ptr<double,double> is taken from an iterator that works over another vector.

为了将来参考， ; double，double> 是指向 Node 对象的指针。

For future reference, Ptr<double,double> is a pointer to a Node object.

推荐答案

你有这样的情况：

struct sample {
    int const a; // const!

    sample(int a):a(a) { }
};

现在，在某些上下文中使用 sample 是可分配的 - 可能在一个容器（如地图，矢量或其他）。这将失败，因为隐式定义的副本赋值操作符沿着这一行执行某些操作：

Now, you use that in some context that requires sample to be assignable - possible in a container (like a map, vector or something else). This will fail, because the implicitly defined copy assignment operator does something along this line:

// pseudo code, for illustration
a = other.a;

但 a 是const！你必须使它非常量。它不伤害，因为只要你不改变它，它仍然在逻辑上const :)你可以通过引入一个合适的 operator = 修复这个问题，使得编译器不是隐式定义一个。但是这很糟糕，因为你不能更改你的const成员。因此，有一个operator =，但仍然不可分配！ （因为副本和赋值不相同！）：

But a is const!. You have to make it non-const. It doesn't hurt because as long as you don't change it, it's still logically const :) You could fix the problem by introducing a suitable operator= too, making the compiler not define one implicitly. But that's bad because you will not be able to change your const member. Thus, having an operator=, but still not assignable! (because the copy and the assigned value are not identical!):

struct sample {
    int const a; // const!

    sample(int a):a(a) { }

    // bad!
    sample & operator=(sample const&) { }
};

不过，表面上的问题显然在 std :: pair< A，B> 。记住， std :: map 在它包含的键上排序。因此，您无法更改其键，因为这可能会轻易地使地图的状态无效。因此，以下条件成立：

However in your case, the apparent problem apparently lies within std::pair<A, B>. Remember that a std::map is sorted on the keys it contains. Because of that, you cannot change its keys, because that could easily render the state of a map invalid. Because of that, the following holds:

typedef std::map<A, B> map;
map::value_type <=> std::pair<A const, B>

也就是说，它禁止更改其包含的键！所以如果你做

That is, it forbids changing its keys that it contains! So if you do

*mymap.begin() = make_pair(anotherKey, anotherValue);

地图会在您引发错误，因为在地图中存储的一些值对中， :: first 成员具有const限定类型！

The map throws an error at you, because in the pair of some value stored in the map, the ::first member has a const qualified type!

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