std::remove_if 中的谓词错误

Question

我基本上有一个 std::deque 个对象，我想根据对象的给定成员变量的条件删除其中一些对象，这样我就可以使用谓词，但是我有错误不是很懂

出于 STL 原因（共享指针），我将 g++ 与 -std=c++11 一起使用，但我试图在 windows 上使用非 c++11 的 MVS 来解决这个问题代码，所以我正在寻找一个非 c++11 解决方案，没有 lamdas 等

密码是：

#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

您将无法在 remove_if 中使用非静态成员函数指针。 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因为它需要 class 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，创建函数对象并传递其实例，要么在调用站点中使用 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());

std::remove_if 中的谓词错误

Error with a predicate in std::remove_if

c++

stl

deque