c++ - sort() 和 stable_sort() 中比较器要求的区别

Question

如果我将 lambda 作为引用参数传递给排序算法，它就可以正常工作。但是当我将相同的 lambda 传递给 stable_sort 时，它没有编译，我必须声明参数 const？为什么会这样？

// sort algorithm example
#include <iostream>     // std::cout
#include <algorithm>    // std::sort
#include <vector>       // std::vector

using namespace std;

void print(std::vector<int> &vec)
{
    for (auto i : vec)
    {
        cout << i << " ";
    }
    cout << endl;
}

int main() {
    std::vector<int> myvector = { 32, 71, 12, 45, 26, 80, 53, 33 };

    sort(myvector.begin(), myvector.end(), [](int &i, int &j)//working
    {
        return i < j;
    });

    stable_sort(myvector.begin(), myvector.end(), [](int &i, int &j)//not compiling
    {
        return i < j;
    });

    stable_sort(myvector.begin(), myvector.end(), [](const int &i, const int &j)//compiling
    {
        return i < j;
    });

    print(myvector);

    return 0;
}

Answer 1

[Bug libstdc++/82891] stable_sort() won't compile with function object that takes parameters by non-const reference

There is now an LWG issue submission request waiting to become published for this. I'll return when it has been added.

GCC 和 LLVM 都存在问题，但它们等待请求^{[ 3 ]} 到 Library Working Group改变标准。