Why is std::basic_string::operator[] a const method if it's also a non-const method?

问题

http://cplusplus.com/reference/string/basic_string/operator[]

I understand that it's advantageous to have a second version which returns const to prevent warnings when a const result is required and to mitigate casting but if the function already provides a non-const method (method-- not result) then what is the point of declaring the const-result method const?

回答1:

You need to understand, that the second (const) version not only returns a different result but is also marked itself as const (this is the second const at the end of the declaration):

const_reference operator[] (size_type pos) const;

These are two different things: The const return value by itself would not be needed in the presence of a non-const method (since the non-const return value can always be casted tò the const version).

But not having a const-version of the operator would mean, that you couldn't use it on const string objects.

The const result is just a result of the constness of the operator itself: if you have a const string and use the operator to get a reference to a single character, clearly this reference must also be const (if not, you could change single characters within a const string).

回答2:

Assume you have

const std::string str{"test"};
std::cout << str[2];

If you don't have a const member function, the code above will fail, as the this pointer passed implicitly to operator[] is const.

回答3:

If you have a const std::basic_string you cannot call (non-const) std::basic_string::operator[], as it is, well, not marked const.

回答4:

If there wasn't a const version of the function, the function could not be invoked on a const instance. For example:

void func (const std::string &h)
{
    if (h[0] == "hello")
        ...
}

How would this work if there wasn't a const version of operator[]?

回答5:

The const method returns a different type, it returns a const-qualified reference. This means if the const method is used, the result of the method is const-qualified (and therefore is “immutable”). The non-const-qualified method returns a regular reference, which is “mutable”.

You can't have just the const method or else you can't modify individual characters in non-const-qualified std::basic_string objects. And you can't have just the non-const method or you won't be able to call the method on const-qualifed std::basic_string objects.

来源：https://stackoverflow.com/questions/34983475/why-is-stdbasic-stringoperator-a-const-method-if-its-also-a-non-const-met