如何强制 C++ 使用编译器库定义的运算符 `<`？

Question

I'm trying to compare strings, where the strings are composed of numerical values only.

I defined my own operator < to be

bool operator<(const string &s1, const string &s2)
{
    if(s1.size() != s2.size()) return s1.size() < s2.size();
    return s1 < s2; // I don't want this to use mine
}

In the last return statement, I want the < used to be the one defined by the C++ library. How can I force the program to do this?

Answer 1

Although I would not recommend overloading operator< for standard library classes, you could achieve what you ask for by calling the standard library implementation explicitly, eg:

bool operator<(const std::string& s1, const std::string& s2)
{
    if (s1.size() != s2.size()) {
        return s1.size() < s2.size();
    }
    return std::operator<(s1, s2);
}

UPDATE : As pointed out in the comments, std::operator<(s1, s2) would not work in C++20. I guess using std::less()(s1, s2) would do the trick then, but the way it looks, and the reason why it works (different namespaces), both suggest that this is not the way to go, and creating a custom class is a better solution for your problem.

Answer 2

Others already answered the question directly. I rather suggest you to not do it;). As mentioned in a comment: Better don't write an operator< for std::string . Its not your type.

You have different options if you want to make:

std::string a,b;
bool b = a < b;

use your custom comparison. Strings that don't compare like std::string aren't std::string s, so it would be natural to have a wrapper. Either make the wrapper own the string, or just let it keep a reference:

struct my_string {
     const std::string& data;
     bool operator<(const my_string& other) {
         return ...;
     }
};

So that you can write:

std::string a,b;
bool b = my_string{a} < my_string{b};