c++ - 为什么要在重载函数中使用 const_cast？

Question

C++ Primer 一书的第 6.4 章陈述如下:

In § 4.11.3 (p. 163) we noted that const_casts are more useful in the context of overloaded functions. As one example, recall our shorterString function from § 6.3.2 (p. 224):

const string &shorterString(const string &s1, const string &s2)
{
    return s1.size() <= s2.size() ? s1 : s2;
}

This function takes and returns references to const string. We can call the function on a pair of nonconst string arguments, but we'll get a reference to a const string as the result. We might want to have a version of shorterString that, when given nonconst arguments, would yield a plain reference. We can write this version of our function using a const_cast:

string &shorterString(string &s1, string &s2)
{
    auto &r = shorterString(const_cast<const string &>(s1), const_cast<const string &>(s2));
    return const_cast<string&>(r);
}

This version calls the const version of shorterString by casting its arguments to references to const. That function returns a reference to a const string, which we know is bound to one of our original, nonconst arguments. Therefore, we know it is safe to cast that string back to a plain string& in the return.

我的问题:与下面的代码相比，这种方法有什么好处？性能不会因为函数调用而受到影响吗？或者这只是一个坏例子？

string &shorterString(string &s1, string &s2) {
    return s1.size() <= s2.size() ? s1 : s2;
}

Answer 1

此示例具有误导性，因为在这种情况下，初始函数是如此简短和简单，与您提出的第二个完整定义相比，该方法几乎没有优势。
想法是你想要两个版本的函数，一个有 const 参数和返回，一个有引用参数和返回。这不是一个很好的例子，因为重载函数和初始函数一样长。
如果您假设初始 shorterString函数相当复杂，重载的函数仍然只需几行即可将引用参数转换为 const 并将返回 const 转换为引用。通过采用这种方法，您可以有两种不同的参数/返回行为，并且可以适本地重用，而不是重复代码。