常量和非常量运算符重载
我有一个我很困惑的话题，我需要详细说明.它是一个 const 版本和一个非常量版本的运算符重载.
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.
// non-const
double &operator[](int idx) {
if (idx < length && idx >= 0) {
return data[idx];
}
throw BoundsError();
}
我知道这个 lambda 函数接受一个索引并检查其有效性，然后返回类中数组数据的索引.还有一个函数具有相同的主体但函数调用为
I understand that this lambda function, takes an index and checks its validity and then 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
const double &operator[](int idx) const
为什么我们需要两个版本?
Why do we need two versions?
例如，在下面的示例代码中，下面每个实例中使用的是哪个版本?
For example, on the sample code below, which version is used in each instance below?
Array a(3);
a[0] = 2.0;
a[1] = 3.3;
a[2] = a[0] + a[1];
我假设 const 版本仅在 a[2] 上调用，因为我们不想冒险修改 a[0] 或 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].
感谢您的帮助.
推荐答案
当两个版本都可用时，逻辑非常简单:为 const 对象调用 const 版本，为非const 对象调用非const 版本.就是这样.
When both versions are available, the logic is pretty straightforward: const version is called for const objects, non-const version is called for non-const objects. That's all.
在您的代码示例中，a 是一个非 const 对象，这意味着在所有情况下都会调用非 const 版本.const 版本在您的示例中从不调用.
In your code sample a is a non-const object, meaning that the non-const version is called in all cases. The const version is never called in your sample.
拥有两个版本的目的是为非const 对象实现读/写"访问，而为const 对象实现读"访问.对于 const 对象，const 版本的 operator [] 被调用，它返回一个 const double & 引用.您可以通过该 const 引用读取数据，但不能通过它进行写入.
The point of having two versions is to implement "read/write" access for non-const objects and only "read" access for const objects. For const objects const version 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.
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