重载下标运算符[]
为什么它的操作需要是类的成员函数，并且最好返回对私有成员的引用？
class X
{
public:
int& operator[] (const size_t);
const int &operator[] (const size_t) const;
private:
static std::vector<int> data;
};
int v[] = {0, 1, 2, 3, 4, 5};
std::vector<int> X::data(v, v+6);
int& X::operator[] (const size_t index)
{
return data[index];
}
const int& X::operator[] (const size_t index) const
{
return data[index];
}
Why does it require to be a member function of a class for its operation and is good to return a reference to private member?
class X
{
public:
int& operator[] (const size_t);
const int &operator[] (const size_t) const;
private:
static std::vector<int> data;
};
int v[] = {0, 1, 2, 3, 4, 5};
std::vector<int> X::data(v, v+6);
int& X::operator[] (const size_t index)
{
return data[index];
}
const int& X::operator[] (const size_t index) const
{
return data[index];
}
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评论（4）
至于为什么需要[]作为成员，您可以阅读这个问题（由您真诚地提出）。好像就是这样，没有真正令人信服的解释。
至于为什么要返回引用呢？
因为您不仅希望提供一种读取数据的方法，还希望提供一种修改数据的方法（对于非常量对象）。如果返回的不是引用（或某个代理）
v[i] = 4；
行不通。
华泰
As to why is it required to have [] as a member, you can read this question (by yours sincerely). Seems it's just the way it is with no really really convincing explanation.
As to why return reference?
Because you want to provide a way not only to read, but also (for non-const objects) to modify the data. If the return weren't a reference (or some proxyr)
v[i] = 4;
wouldn't work.
HTH
根据 13.5.5，它必须是成员函数：
对私有成员的引用是完全可以的并且非常常见。您向类的用户隐藏了详细信息，但仍然提供您需要的功能（修改单个元素的能力）
您的
data变量可能不应该是静态的，除非您真的想共享它在你班级的所有实例中It needs to be a member function according to 13.5.5:
A reference to a private member is completely OK and pretty common. You hide the details from the user of your class, but still provide the functionality you need (ability to modify individual elements)
Your
datavariable likely shoudn't be static though, unless you really want to share it among all instances of your class对于第一个问题，这只是他们决定的方式，即你不能这样做：
T operator[]( const X &, size_t );作为外部函数。
是的，您可以返回对私有成员的引用，如果您允许用户在那里写入，则返回非常量，否则返回非常量。
在您的示例中，虽然数据是静态的，但如果这是它返回的数据源，则没有意义。
For the first question, it is just the way they decided it had to be, i.e. you can't do:
T operator[]( const X &, size_t );as an external function.
And yes, you are fine returning a reference to a private member, non-const if you allow your users to write there, non-const otherwise.
In your example though data is static, which does not make sense if that is the source for what it returns.
调用非成员运算符[] 的语法是什么？任何语法都会很尴尬。
operator[]采用[和]中的一个参数，该参数通常是查找对象所需的索引或某种数据。另外，是的，返回引用是个好主意，即使它是私人成员。这正是 STL 向量所做的事情，以及我能想到的任何其他提供
operator[]的类的作用。建议保持使用。What would the syntax be for calling a non-member
operator[]? Any syntax for that would be awkward.operator[]takes ones parameter within the[and]and that is usually an index or some kind of data necessary to find an object.Also, yes, it is a good idea to return a reference, even if it's a private member. That is exactly what STL vectors do and just about any other class I can think of that I've ever used that provides
operator[]. It would be advised that it the usage is maintained.