C++ 常量和非常量运算符重载
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Const and Non-Const Operator Overloading
提问by James Wilks
I have a topic I'm confused on that I need some elaborating on. It's operator overloading with a const version and a non-const version.
我有一个我很困惑的主题，我需要详细说明。它是具有常量版本和非常量版本的运算符重载。
// non-const
double &operator[](int idx) {
if (idx < length && idx >= 0) {
return data[idx];
}
throw BoundsError();
}
I understand that this function part of a class, takes an index and checks that its logical, returns the index of the array data in the class. There's also a function with the same body but with the function call as
我知道这个函数是类的一部分，需要一个索引并检查它的逻辑，返回类中数组数据的索引。还有一个具有相同主体但函数调用为
const double &operator[](int idx) const
Why do we need two version?
为什么我们需要两个版本？
This sample question also might help elaborate. Which version is used in each instance below?
这个示例问题也可能有助于详细说明。以下每个实例使用哪个版本？
Array a(3);
a[0] = 2.0;
a[1] = 3.3;
a[2] = a[0] + a[1];
My hypothesis that the const version is only called on a[2]because we don't want to risk modifying a[0]or a[1].
我的假设是 const 版本只被调用是a[2]因为我们不想冒险修改a[0]或a[1].
Thanks for any help.
谢谢你的帮助。
回答by AnT
When both versions are available, the logic is pretty straightforward: constversion is called for constobjects, non-constversion is called for non-constobjects. That's all.
当两个版本都可用时，逻辑非常简单：const为const对象const调用版本，为非const对象调用非版本。就这样。
In your code sample ais a non-constobject, meaning that the non-constversion is called in all cases. The constversion is nevercalled in your sample.
在您的代码示例中a是一个非const对象，这意味着const在所有情况下都会调用非版本。您的示例中永远不会调用该const版本。
The point of having two versions is to implement "read/write" access for non-constobjects and only "read" access for constobjects. For constobjects constversion of operator []is called, which returns a const double &reference. You can read data through that const reference, but your can't write through it.
有两个版本的目的是实现非const对象的“读/写”访问和对象的“读”访问const。对于const对象constversion ofoperator []被调用，它返回一个const double &引用。您可以通过该常量引用读取数据，但不能通过它写入。
回答by echo
To supply a code example to complement the answer above:
提供一个代码示例来补充上面的答案：
Array a(3);
a[0] = 2.0; //non-const version called on non-const 'a' object
const Array b(3);
double var = b[1]; //const version called on const 'b' object
const Array c(3);
c[0] = 2.0; //compile error, cannot modify const object
