std :: map有什么用呢？

Question

任何人都可以解释我从这个简单的程序使用的输出std::map.请注意,我插入p到地图中,但还q没有说它发现它们两者,但也说地图中只有1个元素!

#include <map>
#include <iostream>

struct screenPoint {
  float x = 0, y = 0;
  screenPoint(float x_, float y_): x{x_}, y{y_}{}
};

bool operator<(const screenPoint& left, const screenPoint& right){
  return left.x<right.x&&left.y<right.y;
}

std::map<screenPoint, float> positions;

int main(int argc, const char * argv[]) {

  auto p = screenPoint(1,2);
  auto q = screenPoint(2,1);
  positions.emplace(p,3);

  auto f = positions.find(p);
  auto g = positions.find(q);

  if (f == positions.end()){
    std::cout << "f not found";
  } else {
    std::cout << "f found";
  }

  std::cout << std::endl;

  if (g == positions.end()){
    std::cout << "g not found";
  } else {
    std::cout << "g found";
  }

  std::cout << std::endl;

  std::cout << "number elements: " << positions.size() << "\n";
  return 0;
}

输出:

f found
g found
number elements: 1

Answer 1

问题在于您在这种情况下定义比较仿函数的方式.这两个元素,p以及q具有相同的x和y,正好相反.你的逻辑检查x一个小于另一个的那个,以及ys.true对于这些输入,这永远无法评估.

试试这个片段:

int main()
{
    auto p = screenPoint(1,2);
    auto q = screenPoint(2,1);

   std::cout << std::boolalpha << (p < q) << " " << (q < p) << std::endl;
}

它会打印出来

false false

所以p不小于q,q也不小于p.就地图而言,这使它们等同.