c++ - 在 map 上使用 find_if 按值查找

Question

我有一个有 map 的类(class)。我需要通过搜索特定值而不是键来在 map 中找到迭代器。使用成员函数谓词 IsValueFound，正在尝试这个。

class A
{
public:
  void findVal();
private:
 int state;
 map<int, int> exmap;
 bool IsValueFound(pair<int key, int val> itr)
 {
   return state == itr.second;
 }
};

void A::findVal
{
  itr = find_if(exmap.begin, exmap.end, mem_fun1_ref(&A::IsValueFound));
}

我遇到编译错误。我不确定这些函数适配器的语法是什么。请帮忙。

编辑:抱歉。请忽略 finf_if stmt 之外的编译错误。我需要先更正 find_if stmt。此外，代码没有提升 :(

Answer 1

将对象 A 转换为仿函数稍微容易一些:
但是您的代码还有其他问题(见下文):

#include <map>
#include <memory>
#include <functional>
#include <algorithm>

using namespace std;

class A {
    public:
      void findVal();
    private:
      int state;
      map<int, int> exmap;

      // Changed the function IsValueFound() to operator()
      // This makes the whole object behave like a function.
      // Its a lot easier then getting member functions and binding
      // the this reference.
      bool operator()(map<int,int>::value_type const& itr) const
                    //         ^^^^^^^^^^^^^^^^^
                    //  std::pair<int,int> is not the type held in the map
                    //  so you are not going to bind against it using a pair.
      {
          return state == itr.second;
      }
};

void A::findVal()
{
    // You did not specify a type
    // for the iterator in your code.
    std::map<int,int>::iterator itr1 = find_if(exmap.begin(), exmap.end(), *this);
                                                   //   ^^^^^^       ^^^^^^
                                                   // begin() and end() are methods.

    // Just pass the `this` object as the the third parameter it will act as a function now.
}