返回向量成员变量的引用 [英] Return reference to a vector member variable

问题描述

我有一个向量作为类的成员，我想通过一个getVector（）函数返回一个引用，以便以后能够修改它。是不是更好的实现函数getVector（）是const？但是我在下面的代码中有一个错误限定符删除绑定引用类型...。应该修改什么？

I have a vector as member in a class and I want to return a reference to it through a getVector() function, so as to be able to modify it later. Isn’t it better practice the function getVector() to be const? However I got an error "qualifiers dropped in binding reference of type…" in the following code. What should be modified?

class VectorHolder
{
public:
VectorHolder(const std::vector<int>&);
std::vector<int>& getVector() const;

private:
std::vector<int> myVector;

};

std::vector<int> &VectorHolder::getVector() const
{
return myVector;
}

推荐答案

c $ c> const 成员函数，返回类型不能是非const引用。让它 const ：

Since it is a const member function, the return type cannot be non-const reference. Make it const:

const std::vector<int> &VectorHolder::getVector() const
{
   return myVector;
}

现在没关系。

为什么很好？因为在 const 成员函数中，每个成员变成 const ，这样就不能被修改，这意味着 myVector 是函数中的一个 const 向量，这就是为什么你必须使返回类型 const

Why is it fine? Because in a const member function, the every member becomes const in such a way that it cannot be modified, which means myVector is a const vector in the function, that is why you have to make the return type const as well, if it returns the reference.

现在你不能修改 对象。查看您可以做什么以及无法做什么：

Now you cannot modify the same object. See what you can do and what cannot:

 std::vector<int> & a = x.getVector();       //error - at compile time!

 const std::vector<int> & a = x.getVector(); //ok
 a.push_back(10);                            //error - at compile time!

 std::vector<int>  a = x.getVector();        //ok
 a.push_back(10);                            //ok

顺便问一下，我想知道为什么你需要这样的 VectorHolder 。

By the way, I'm wondering why you need such VectorHolder in the first place.

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