传递类的私有方法作为std :: sort的比较运算符 [英] Passing a private method of the class as the compare operator for std::sort
问题描述
我编写一个代码来解决以下问题：给定一组数字 x [0] ， x [1] ，...， x [N-1] ，找到使它们按升序排序的排列。换句话说，我想在{0,2，...，N-1}找到一个排列，例如 i [0] ， i [1] ，...， i [N-1] ，使得 x [i [0 ]] <= x [i [1]] <= ... <= x [i [N-1]] 。
为此，我已经存储了 x 向量和索引向量 i （最初填充 i [j] = j ）作为类的私有成员。我还定义了一个私有方法为
bool MyClass :: compare（size_t s，size_t t）{
return （x [s]
现在，我将调用 std :: sort 如下
std :: sort（i.begin（），i.end比较）;
我希望得到所需的结果。但是代码不编译，我得到以下错误：
错误：没有匹配的函数调用' ：vector< long unsigned int> :: iterator，std :: vector< long unsigned int> :: iterator，< unresolved overloaded function type>）'
我必须正确完成所有的一切，因为 std :: sort 的文档提到我可以传递一个函数作为比较 std :: sort （ http： //www.cplusplus.com/reference/algorithm/sort/ ）
感谢您提供的所有帮助。
您的方法有几个问题。第一个，最明显的是，你不能使用成员函数作为自由函数。要能够调用 compare ，您需要一个类型为 MyClass 的对象和两个整数。内部 std :: sort 的实现将尝试调用一个只有两个整数参数的 free （非成员） >
除此之外，您不能创建指向成员函数的指针，而不显式地获取其地址。 std :: sort（...，compare）; 不会编译成员函数。虽然非成员函数会自动衰减到指向函数的指针，但这里并不是这样。
在C ++ 11中有两种不同的解决方案。最通用的是创建一个用于捕获此参数的lambda：
std :: sort（std :: begin（i），std :: end（i），
[]（int x，int y）{return compare（x，y）;}）; //或者甚至可以在这里实现
一个不同的方法是将对象和成员函数绑定到函数：
std :: sort（std :: begin（i），std :: end（i），
std :: bind（& MyClass :: compare，this，_1，_2））;
在这后一种情况下， std :: bind 函数将创建一个实现 operator（）的对象，它使用两个参数，并调用成员函数 MyClass :: compare this 。
两种方法的语义略有不同，但在这种情况下，可以使用任一个。
I am writing a code to solve the following problem: Given a set of numbers x[0], x[1], ..., x[N-1], find the permutation that makes them sorted in the ascending order. In the other words, I would like to find a permutation on {0,2,...,N-1} such as i[0], i[1], ..., i[N-1] such that x[i[0]] <= x[i[1]] <= ... <= x[i[N-1]].
For this, I have stored the x vector and an index vector i (initially filled with i[j] = j) as private members of a class. I have also defined a private method as
bool MyClass::compare(size_t s, size_t t) {
return (x[s] < x[t]);
}
Now, I would call the std::sort as follows
std::sort(i.begin(), i.end(), compare);
and I expect to get the desired result. But the code does not compile and I get the following error:
error: no matching function for call to ‘sort(std::vector<long unsigned int>::iterator, std::vector<long unsigned int>::iterator, <unresolved overloaded function type>)’
I must have done everything correctly as also the documentation of std::sort mentions that I can pass a function as the compare operator to std::sort (http://www.cplusplus.com/reference/algorithm/sort/)
Thanks for all the helps in advance.
There are a couple of issues with your approach. The first one and most evident is that you cannot use a member function as a free function. To be able to call compare you need an object of type MyClass and two integers. Inside std::sort the implementation is going to try to call a free (non-member) function with only two integer arguments.
Other than that, you cannot create a pointer to a member function without explicitly taking its address. The line std::sort(..., compare); will not compile for a member function. While non-member functions will automatically decay to a pointer to the function that is not the case here.
In C++11 there are two different solutions that you can take. The most generic is creating a lambda that captures the this argument:
std::sort(std::begin(i),std::end(i),
[](int x, int y) { return compare(x,y); }); // or maybe even implement here
A different approach would be binding the object and the member function into a functor:
std::sort(std::begin(i),std::end(i),
std::bind(&MyClass::compare,this,_1,_2));
In this last case, the std::bind function will create an object that implements operator() taking two arguments and will call the member function MyClass::compare on the object pointed by this.
The semantics of both approaches are slightly different, but in this case you can use either one.
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