为什么选择STL <set> 重载运算符&lt;函数必须是const函数吗？

Question

class Test
{

public:

    int v;
    Test(int s)
    {
        v = s;
    }

    bool operator < (const Test & b) const
    {
        return v < b.v;
    }
};

int main()
{
    set <Test> t2;
    return 0;
}

Why must I use a const function when I overload the less operator? If i don't write the "const" there, it won't pass the compiling. So i don't know why MUST I write the "const" there?

Answer 1

The reason is that std::set is a bit special: the values it stores must not change (likewise the key_type in a std::map ). This is because set is a sorted container, and it needs to know how to sort the values; changing the values within the set could result in a broken program, because set::set has no way to know that the values have been changed and therefore would need to be sorted again.

If the comparison function were allowed to modify the values, then it could never be used, because the values in a set are always const.

As an aside, it is true that some modifications of the values are permissible, specifically modifications which do not affect the sort order. So if you have a set with a custom sort function which sorts by one field of a struct, you can modify other fields of the struct because the sort order will remain the same. But std::set has no way to understand which fields are important to the sort order and which are not, so the whole value is const. You can do tricks like making some fields mutable to deal with this, but in most cases it is not necessary (eg you can use a map instead, and put the modifiable parts in the value_type instead).

Answer 2

You are not required to do so. It's generally recommend because one shouldn't modify objects in the process of comparing them. set<Test> makes it a requirement by passing const references to operator< .