std::lower_bound和std::upper_bound有什么区别？

Question

下面的代码不同时使用MSVC2017和GCC11编译：

#include <deque>
#include <chrono>
#include <algorithm>

using Clock = std::chrono::system_clock;
using TimePoint = std::chrono::time_point<Clock>;

struct PricePoint
{
    TimePoint dt;
    double buy;
    double sell;
};

inline bool operator < (const TimePoint & dt, const PricePoint & a)
{
    return a.dt < dt;
}

int main()
{
    std::deque<PricePoint> priceSequence;
    const auto begin = std::lower_bound(priceSequence.begin(), priceSequence.end(), Clock::now());
    return 0;
}

但是，如果我将std::lower_bound替换为std::upper_bound，它将开始编译。有什么关系？

Answer 1

错误:与“operator<”不匹配

这种错误表明某些模板代码试图使用未定义的操作符。

lower_bound和upper_bound做了相反的比较。< (const TimePoint & dt, const PricePoint & a)适用于upper_bound，但是lower_bound需要定义如下：

inline bool operator < (const PricePoint & a, const TimePoint & dt)
{
    return dt < a.dt;
}