std::remove_if 中的谓词出错

Question

我基本上有一个std::deque 的对象，我想根据对象的给定成员变量的条件删除其中一些对象，以便我使用谓词，但我有我没有的错误真懂。

出于 STL 原因（共享指针），我正在使用 g++ 和 -std=c++11，但我试图在带有非 c++11 代码的 MVS 的 Windows 上解决这个问题，所以我正在寻找一个非 @987654325 @ 解决方案，没有 lamdas 等

代码是：

#include <iostream> // for std::cout and std::endl
#include <cstdio> // for getchar()
#include <memory> // for std::shared_ptr
#include <deque> // for std::deque
#include <algorithm> // for std::earse and std::remove_if

class A
{
    private:
        int _i;
        double _d;
    public:
        A(int i, double d)
        {
            _i = i;
            _d = d;
        }
        int geti()const
        {
            return _i;
        }
        double getValueOnWhichToCheck()const
        {
            return _d;
        }
};

typedef std::shared_ptr<A> A_shared_ptr;
typedef std::deque<A_shared_ptr> list_type;

void PrintDeque(list_type & dq)
{
    if (0 ==  dq.size())
    {
        std::cout << "Empty deque." << std::endl;
    }
    else
    {
        for (int i = 0 ; i < dq.size() ; ++i)
        {
            std::cout << i+1 << "\t" << dq[i] << std::endl;
        }
    }
}

class B
{
    public:
        double getThreshold() // Non constant for a reason as in real code it isn't
        {
            return 24.987; // comes from a calculation not needed here so I return a constant.
        }
        bool Predicate(A_shared_ptr & a)
        {
            return a->getValueOnWhichToCheck() >= getThreshold();
        }
        void DoStuff()
        {
            A_shared_ptr pT1 = std::make_shared<A>(A(2,      -6.899987));
            A_shared_ptr pT2 = std::make_shared<A>(A(876,    889.878762));
            A_shared_ptr pT3 = std::make_shared<A>(A(-24,    48.98924));
            A_shared_ptr pT4 = std::make_shared<A>(A(78,     -6654.98980));
            A_shared_ptr pT5 = std::make_shared<A>(A(6752,   3.141594209));
            list_type dq = {pT1,pT2,pT3,pT4,pT5};
            PrintDeque(dq);
            bool (B::*PtrToPredicate)(A_shared_ptr &) = &B::Predicate;
            dq.erase(std::remove_if(dq.begin(), dq.end(), PtrToPredicate),dq.end());
            PrintDeque(dq);
        }
};

int main()
{
    B * pB = new B();
    pB->DoStuff();
    getchar();
}

g++ -std=c++11 main.cpp -o main 的输出是：

In file included from /usr/include/c++/5/bits/stl_algobase.h:71:0,
                 from /usr/include/c++/5/bits/char_traits.h:39,
                 from /usr/include/c++/5/ios:40,
                 from /usr/include/c++/5/ostream:38,
                 from /usr/include/c++/5/iostream:39,
                 from main.cpp:1:
/usr/include/c++/5/bits/predefined_ops.h: In instantiation of ‘bool __gnu_cxx::__ops::_Iter_pred<_Predicate>::operator()(_Iterator) [with _Iterator = std::_Deque_iterator<std::shared_ptr<A>, std::shared_ptr<A>&, std::shared_ptr<A>*>; _Predicate = bool (B::*)(std::shared_ptr<A>&)]’:
/usr/include/c++/5/bits/stl_algo.h:866:20:   required from ‘_ForwardIterator std::__remove_if(_ForwardIterator, _ForwardIterator, _Predicate) [with _ForwardIterator = std::_Deque_iterator<std::shared_ptr<A>, std::shared_ptr<A>&, std::shared_ptr<A>*>; _Predicate = __gnu_cxx::__ops::_Iter_pred<bool (B::*)(std::shared_ptr<A>&)>]’
/usr/include/c++/5/bits/stl_algo.h:936:30:   required from ‘_FIter std::remove_if(_FIter, _FIter, _Predicate) [with _FIter = std::_Deque_iterator<std::shared_ptr<A>, std::shared_ptr<A>&, std::shared_ptr<A>*>; _Predicate = bool (B::*)(std::shared_ptr<A>&)]’
main.cpp:67:73:   required from here
/usr/include/c++/5/bits/predefined_ops.h:234:30: error: must use ‘.*’ or ‘->*’ to call pointer-to-member function in ‘((__gnu_cxx::__ops::_Iter_pred<bool (B::*)(std::shared_ptr<A>&)>*)this)->__gnu_cxx::__ops::_Iter_pred<bool (B::*)(std::shared_ptr<A>&)>::_M_pred (...)’, e.g. ‘(... ->* ((__gnu_cxx::__ops::_Iter_pred<bool (B::*)(std::shared_ptr<A>&)>*)this)->__gnu_cxx::__ops::_Iter_pred<bool (B::*)(std::shared_ptr<A>&)>::_M_pred) (...)’
  { return bool(_M_pred(*__it)); }
                              ^

Answer 1

您将无法将非静态成员函数指针与remove_if 一起使用。 remove_if 需要普通函数指针、静态成员函数指针或函数对象。它不能使用成员函数指针的原因是它需要一个类的实例才能调用该函数，而您不能将其包装到调用中

您要么需要创建Predicate static，创建一个函数对象并传递它的一个实例，要么在调用站点中使用一个lambda。

Answer 2

不喜欢函数指针，但使用 lambda 的 this 似乎可以编译...https://ideone.com/0StRcw

        dq.erase(std::remove_if(dq.begin(), dq.end(), [this](const std::shared_ptr<A>& a) {
            return a->getValueOnWhichToCheck() >= getThreshold();
        }),dq.end());

或者没有 lambdas..

        auto predicateToUse = std::bind(&B::Predicate, this, std::placeholders::_1);
        dq.erase(std::remove_if(dq.begin(), dq.end(), predicateToUse), dq.end());

不自动更新：

dq.erase(std::remove_if(dq.begin(), dq.end(), std::bind(&B::Predicate, this, std::placeholders::_1)), dq.end());