std::transform中的二元操作符，带unique_ptr的vector对象

Binary operator in std::transform with vector of unique_ptr

本文关键字：ptr vector 对象 unique transform 二元操作符 std      更新时间：2023-10-16

当应用于unique_ptr的向量时，我对标准库转换感到困惑。我定义了一个二元函函数addScalar，它接受对unique_ptr的2个const引用，并返回对unique_ptr的const引用，以避免复制(这是unique_ptr所禁止的)。

然后我尝试在std::transform中使用它，但是unique_ptr似乎根本不可能进行二进制操作，尽管我采取了所有预防措施来避免unique_ptr复制…

有没有人知道如何使用std::transform与std::unique_ptr ?或者我必须用for循环遍历向量并"手动"执行加法?我也想知道我是否可以在我的函子中使用unique_ptr<const Scalar>。

这是我的班级:

#include "space.h"
#include "scalar.h"
#include <vector>
#include <algorithm>
#include <memory>
using std::vector;
using std::ostream;
using std::unique_ptr;
class addScalar
{
   public:
      unique_ptr<Scalar> const& operator()(unique_ptr<Scalar> const& scal1, unique_ptr<Scalar> const& scal2)
      {
         *scal1 += *scal2;
         return scal1;
      };
};
class Tensor4D
{
   public:
      Tensor4D(Space& space_in, int ncomp);
      Tensor4D(const Tensor4D& tens);
      Tensor4D& operator=(const Tensor4D& tens);
      size_t size() const {return comp.size();};
      ~Tensor4D();
   protected:
      Space* const space;
      vector<unique_ptr<Scalar>> comp;
   public:
      Tensor4D& operator+=(const Tensor4D& tens);
};

，这里是操作符+=的实现:

Tensor4D& Tensor4D::operator+=(const Tensor4D& tens)
{
   assert(comp.size() == tens.comp.size());
   transform(tens.comp.begin(), tens.comp.end(), comp.begin(), tens.comp.begin(), addScalar());
   return *this;
}

我得到以下丑陋的编译器错误:

/usr/include/c++/4.8/bits/stl_algo.h: In instantiation of ‘_OIter std::transform(_IIter1, _IIter1, _IIter2, _OIter, _BinaryOperation) [with _IIter1 = __gnu_cxx::__normal_iterator<const std::unique_ptr<Scalar>*, std::vector<std::unique_ptr<Scalar> > >; _IIter2 = __gnu_cxx::__normal_iterator<std::unique_ptr<Scalar>*, std::vector<std::unique_ptr<Scalar> > >; _OIter = __gnu_cxx::__normal_iterator<const std::unique_ptr<Scalar>*, std::vector<std::unique_ptr<Scalar> > >; _BinaryOperation = addScalar]’:
tensor4D.C:44:94:   required from here
/usr/include/c++/4.8/bits/stl_algo.h:4965:12: error: no match for ‘operator=’ (operand types are ‘const std::unique_ptr<Scalar>’ and ‘const std::unique_ptr<Scalar>’)
  *__result = __binary_op(*__first1, *__first2);
        ^
/usr/include/c++/4.8/bits/stl_algo.h:4965:12: note: candidates are:
In file included from /usr/include/c++/4.8/memory:81:0,
             from /home/gmartinon/Kadath/C++/Include/scalar.h:27,
             from tensor4D.h:5,
             from tensor4D.C:1:
/usr/include/c++/4.8/bits/unique_ptr.h:190:7: note: std::unique_ptr<_Tp, _Dp>& std::unique_ptr<_Tp, _Dp>::operator=(std::unique_ptr<_Tp, _Dp>&&) [with _Tp = Scalar; _Dp = std::default_delete<Scalar>]
   operator=(unique_ptr&& __u) noexcept
   ^
/usr/include/c++/4.8/bits/unique_ptr.h:190:7: note:   no known conversion for argument 1 from ‘const std::unique_ptr<Scalar>’ to ‘std::unique_ptr<Scalar>&&’
/usr/include/c++/4.8/bits/unique_ptr.h:203:2: note: template<class _Up, class _Ep> typename std::enable_if<std::__and_<std::is_convertible<typename std::unique_ptr<_Up, _Ep>::pointer, typename std::unique_ptr<_Tp, _Dp>::_Pointer::type>, std::__not_<std::is_array<_Up> > >::value, std::unique_ptr<_Tp, _Dp>&>::type std::unique_ptr<_Tp, _Dp>::operator=(std::unique_ptr<_Up, _Ep>&&) [with _Up = _Up; _Ep = _Ep; _Tp = Scalar; _Dp = std::default_delete<Scalar>]
  operator=(unique_ptr<_Up, _Ep>&& __u) noexcept
  ^
/usr/include/c++/4.8/bits/unique_ptr.h:203:2: note:   template argument deduction/substitution failed:
In file included from /usr/include/c++/4.8/algorithm:62:0,
             from tensor4D.h:8,
             from tensor4D.C:1:
/usr/include/c++/4.8/bits/stl_algo.h:4965:12: note:   types ‘std::unique_ptr<_Tp, _Dp>’ and ‘const std::unique_ptr<Scalar>’ have incompatible cv-qualifiers
  *__result = __binary_op(*__first1, *__first2);
        ^
In file included from /usr/include/c++/4.8/memory:81:0,
             from /home/gmartinon/Kadath/C++/Include/scalar.h:27,
             from tensor4D.h:5,
             from tensor4D.C:1:
/usr/include/c++/4.8/bits/unique_ptr.h:211:7: note: std::unique_ptr<_Tp, _Dp>& std::unique_ptr<_Tp, _Dp>::operator=(std::nullptr_t) [with _Tp = Scalar; _Dp = std::default_delete<Scalar>; std::nullptr_t = std::nullptr_t]
       operator=(nullptr_t) noexcept
       ^
/usr/include/c++/4.8/bits/unique_ptr.h:211:7: note:   no known conversion for argument 1 from ‘const std::unique_ptr<Scalar>’ to ‘std::nullptr_t’
/usr/include/c++/4.8/bits/unique_ptr.h:274:19: note: std::unique_ptr<_Tp, _Dp>& std::unique_ptr<_Tp, _Dp>::operator=(const std::unique_ptr<_Tp, _Dp>&) [with _Tp = Scalar; _Dp = std::default_delete<Scalar>] <near match>
   unique_ptr& operator=(const unique_ptr&) = delete;
               ^
/usr/include/c++/4.8/bits/unique_ptr.h:274:19: note:   no known conversion for implicit ‘this’ parameter from ‘const std::unique_ptr<Scalar>*’ to ‘std::unique_ptr<Scalar>*’

addScalar的返回类型将被分配给unique_ptr<Scalar>，因此它不能返回const引用，因为unique_ptr没有复制赋值。因此，您必须按值返回来调用move赋值。

为了避免构建新的Scalar，您可以使用std:move_iterator移动到addScalar，然后移动assign以覆盖移动的from值:

class addScalar 
{
public:
    unique_ptr<Scalar> operator()(unique_ptr<Scalar> scal1,
                                  unique_ptr<Scalar> const& scal2) {
      *scal1 += *scal2;
      return scal1;
    };
};
Tensor4D& Tensor4D::operator+=(const Tensor4D& tens)
{
   assert(comp.size() == tens.comp.size());
   transform(make_move_iterator(comp.begin()), make_move_iterator(comp.end()),
               tens.comp.begin(), comp.begin(), addScalar());
   return *this;
}

安德烈说得很好，不清楚这是否严格按照标准允许。我把这个问题留给语言律师吧。

现场演示。

std::transform的c++标准规定:

binary_op不能使任何迭代器失效，包括end

最好的方法是实现您自己的转换函数，以满足特定的需求。

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