lambda功能比分离的方法慢

Question

了解lambda在C ++中的开销11 和 c ++ 0x lambda开头。但是，第一个是Fluke，第二个是将其与功能对象进行比较。

我正在使用lambda和分离的函数来检查一个函数的函数（以下代码）的函数的性能（Compiler：G ++；标准：C ++ 17）。使用多个输入检查时，LAMBDA功能在系统上较慢，在运行时大约增加3％。代码附加在下面。

编译器是否有可能优化对分离方法的调用，但对lambda表达式不进行相同的操作？

#include <vector>
#include <iostream>
#include <chrono>

using namespace std;


template <
    class result_t   = std::chrono::milliseconds,
    class clock_t    = std::chrono::steady_clock,
    class duration_t = std::chrono::milliseconds>
auto since(std::chrono::time_point<clock_t, duration_t> const& start)
{
    return std::chrono::duration_cast<result_t>(clock_t::now() - start);
}

int manhattanDistance(int x1, int y1, int x2, int y2) {
    return abs(x1 - x2) + abs(y1 - y2);
}

int fWithoutLambda(int x, int y, vector<vector<int>>& points) {
    int min_d = numeric_limits<int>::max();
    int idx = -1;
    
    for (int i = 0; i < points.size(); i++) {
        auto d = manhattanDistance(x, y, points[i][0], points[i][1]);
        if ((x == points[i][0] || y == points[i][1]) && d < min_d) {
            idx = i;
            min_d = d;
        }
    }
    
    return idx;
}

int fWithLambda(int x, int y, vector<vector<int>>& points) {
    auto manh = [x, y](int x2, int y2) {return abs(x - x2) + abs(y - y2);};
    
    int min_d = numeric_limits<int>::max();
    int idx = -1;
    
    for (int i = 0; i < points.size(); i++) {
        auto d = manh(points[i][0], points[i][1]);
        if ((x == points[i][0] || y == points[i][1]) && d < min_d) {
            idx = i;
            min_d = d;
        }
    }
    
    return idx;
}

int main() {

    size_t repeats = 10;
    int time_no_lambda{0};
    int time_lambda{0};

    for (size_t j = 0; j < repeats; ++j) {
        int n = 1000000;
        vector<vector<int>> v(n);
        int counter = 0;
        for (auto& el : v) {
            v[counter++] = {counter, counter%10};
        }

        int x = 10, y = 5;

        auto start = std::chrono::steady_clock::now();  
        fWithLambda(x, y, v);
        auto tim_0 = since(start).count();
        time_lambda += tim_0;
        fWithoutLambda(x, y, v);
        time_no_lambda += (since(start).count() - tim_0);
    }

    std::cout << "Elapsed(ms), lambda =" << (time_no_lambda/repeats) << std::endl;  
    std::cout << "Elapsed(ms), no lambda =" << (time_lambda/repeats) << std::endl;  
}