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问题描述

我正在通过编写所有默认函数来实现像向量一样的 stl.并且有一个问题,我不明白为什么它为简单类型调用 ragne 版本的assign 并且不默认.下面是实现代码:向量.h

void assign(size_t count, const T& value){//默认版本无效分配(size_t 计数,const T& 值){if(this->_size allocator.deallocate(this->arr, this->_capacity);this->arr = this->allocator.allocate(count);this->_capacity = count;}for(size_t i = 0; i < count; ++i)this->arr[i] = 值;这-> _size = 计数;}模板void assign(InputIt first, InputIt last){//范围版本size_t count = std::distance(first,last);if(this->_size allocator.deallocate(this->arr, this->_capacity);this->arr = this->allocator.allocate(count);this->_capacity = count;}for(size_t i = 0;第一个!=最后一个;i++)this->arr[i] = *first++;这-> _size = 计数;}

主要代码:

 Vectorvec;vec.assign(5,10);

输出:

/MyVector/MyVector.h: 在实例化‘void Vector::assign(InputIt, InputIt) [with InputIt = int;T = 整数;分配器 = std::allocator]’:../MyVector/main.cpp:52:24:从这里需要../MyVector/MyVector.h:99:45: 错误:没有匹配的函数调用距离(int&,int&)"size_t count = std::distance(first,last);~~~~~~~~~~~~~^~~~~~~~~~~~~在/usr/include/c++/7/bits/stl_algobase.h:66:0 包含的文件中,来自/usr/include/c++/7/bits/char_traits.h:39,来自/usr/include/c++/7/ios:40,来自/usr/include/c++/7/ostream:38,来自/usr/include/c++/7/iostream:39,从 ../MyVector/main.cpp:1:/usr/include/c++/7/bits/stl_iterator_base_funcs.h:138:5:注意:候选:模板constexpr 类型名 std::iterator_traits<_Iterator>::difference_type std::distance(_InputIterator, _InputIterator)距离(_InputIterator __first,_InputIterator __last)^~~~~~~~/usr/include/c++/7/bits/stl_iterator_base_funcs.h:138:5: 注意:模板参数推导/替换失败:/usr/include/c++/7/bits/stl_iterator_base_funcs.h: 替换‘模板<class _InputIterator>constexpr typename std::iterator_traits<_Iterator>::difference_type std::distance(_InputIterator, _InputIterator) [with _InputIterator = int]':../MyVector/MyVector.h:99:45: 来自‘void Vector::assign(InputIt, InputIt) [with InputIt = int;T = 整数;分配器 = std::allocator]’../MyVector/main.cpp:52:24:从这里需要/usr/include/c++/7/bits/stl_iterator_base_funcs.h:138:5: 错误:在‘struct std::iterator_traits<int>’中没有名为‘difference_type’的类型在 ../MyVector/main.cpp:2:0 包含的文件中:../MyVector/MyVector.h: 在实例化‘void Vector::assign(InputIt, InputIt) [with InputIt = int;T = 整数;分配器 = std::allocator]’:../MyVector/main.cpp:52:24:从这里需要../MyVector/MyVector.h:107:36: 错误:一元‘*’的无效类型参数(有‘int’)this->arr[i] = *first++;^~~~~~~~Makefile:725: 目标main.o"的配方失败make: *** [main.o] 错误 1

我使用的是 C++17

解决方案

范围版本更适合 vec.assign(5, 10);InputIt = int代码>.您应该以某种方式禁用不代表输入迭代器的模板参数的重载.

我们来看看stdlibc++ 实现:

template>无效分配(首先输入,最后输入){M_assign_dispatch(第一个,最后一个);}

where RequireInputIter

template使用 RequireInputIter = typename enable_if::value>::type;

换句话说,对于推导的类型InputItiterator_traits::iterator_category 类型应该可以转换为input_iterator_tag.否则,由于 SFINAE.

在 C++17 中,RequireInputIter 可以通过 _t_v 助手来简化:

template使用 RequireInputIter = enable_if_t

另请注意,输入迭代器可用于遍历范围仅一次.在您调用 std::distance(first, last) 之后,除非 InputIt 至少是一个前向迭代器,否则所有后续遍历该范围的尝试都是未定义的行为.对于输入迭代器,您无法确定要预分配多少空间.

这就是 assign 在内部使用标签调度技术的原因.经过一些简化,它看起来像这样:

template>无效分配(首先输入,最后输入){M_assign_aux(第一个,最后一个,typename iterator_traits::iterator_category{});}

有两个 M_assign_aux 重载

templatevoid M_assign_aux(先输入,最后输入,std::input_iterator_tag);模板void M_assign_aux(先转发,最后转发,std::forward_iterator_tag);

完成任务.第一个将仅用于输入迭代器,第二个用于前向迭代器及其派生的迭代器,即双向和随机访问迭代器.

I'm implementing stl like vector with writing all default functions. And there is a problem that I don't understand why It calls ragne version of assign for simple types and doesn't default.Here is the implementation code:Vector.h

void assign(size_t count, const T& value){ // Default version
                void assign(size_t count, const T& value){
                if(this->_size < count){
                   this->allocator.deallocate(this->arr, this->_capacity);
                   this->arr = this->allocator.allocate(count);

                   this->_capacity = count;
                }
                for(size_t i = 0; i < count; ++i)
                    this->arr[i] = value;

               this->_size = count;
            }

template<class InputIt>
            void assign(InputIt first, InputIt last){ // Range version
                size_t count = std::distance(first,last);
                if(this->_size < count){
                   this->allocator.deallocate(this->arr, this->_capacity);
                   this->arr = this->allocator.allocate(count);

                   this->_capacity = count;
                }
                for(size_t i = 0; first != last; i++)
                    this->arr[i] = *first++;

                this->_size = count;
            }

Main code:

 Vector<int> vec;
 vec.assign(5,10);

Output:


/MyVector/MyVector.h: In instantiation of ‘void Vector<T, Allocator>::assign(InputIt, InputIt) [with InputIt = int; T = int; Allocator = std::allocator]’:
../MyVector/main.cpp:52:24:   required from here
../MyVector/MyVector.h:99:45: error: no matching function for call to ‘distance(int&, int&)’
                 size_t count = std::distance(first,last);
                                ~~~~~~~~~~~~~^~~~~~~~~~~~
In file included from /usr/include/c++/7/bits/stl_algobase.h:66:0,
                 from /usr/include/c++/7/bits/char_traits.h:39,
                 from /usr/include/c++/7/ios:40,
                 from /usr/include/c++/7/ostream:38,
                 from /usr/include/c++/7/iostream:39,
                 from ../MyVector/main.cpp:1:
/usr/include/c++/7/bits/stl_iterator_base_funcs.h:138:5: note: candidate: template<class _InputIterator> constexpr typename std::iterator_traits<_Iterator>::difference_type std::distance(_InputIterator, _InputIterator)
     distance(_InputIterator __first, _InputIterator __last)
     ^~~~~~~~
/usr/include/c++/7/bits/stl_iterator_base_funcs.h:138:5: note:   template argument deduction/substitution failed:
/usr/include/c++/7/bits/stl_iterator_base_funcs.h: In substitution of ‘template<class _InputIterator> constexpr typename std::iterator_traits<_Iterator>::difference_type std::distance(_InputIterator, _InputIterator) [with _InputIterator = int]’:
../MyVector/MyVector.h:99:45:   required from ‘void Vector<T, Allocator>::assign(InputIt, InputIt) [with InputIt = int; T = int; Allocator = std::allocator]’
../MyVector/main.cpp:52:24:   required from here
/usr/include/c++/7/bits/stl_iterator_base_funcs.h:138:5: error: no type named ‘difference_type’ in ‘struct std::iterator_traits<int>’
In file included from ../MyVector/main.cpp:2:0:
../MyVector/MyVector.h: In instantiation of ‘void Vector<T, Allocator>::assign(InputIt, InputIt) [with InputIt = int; T = int; Allocator = std::allocator]’:
../MyVector/main.cpp:52:24:   required from here
../MyVector/MyVector.h:107:36: error: invalid type argument of unary ‘*’ (have ‘int’)
                     this->arr[i] = *first++;
                                    ^~~~~~~~
Makefile:725: recipe for target 'main.o' failed
make: *** [main.o] Error 1

I'm using C++17

解决方案

The range version is a better match for vec.assign(5, 10); with InputIt = int. You should somehow disable that overload for a template parameter that doesn't represent an input iterator.

Let's take a look at stdlibc++ implementation:

template<typename InputIt, typename = std::RequireInputIter<InputIt>>
void assign(InputIt first, InputIt last) {
    M_assign_dispatch(first, last);
}

where RequireInputIter is

template<typename InputIt>
using RequireInputIter = typename enable_if<is_convertible<typename
    iterator_traits<InputIt>::iterator_category, input_iterator_tag>::value>::type;

In other words, for a deduced type InputIt, iterator_traits<InputIt>::iterator_category type should be convertible into input_iterator_tag. Otherwise, that assign overload is silently excluded from the overload resolution set thanks to SFINAE.

In C++17, RequireInputIter can be simplified with _t and _v helpers:

template<typename InputIt>
using RequireInputIter = enable_if_t<is_convertible_v<typename
    iterator_traits<InputIt>::iterator_category, input_iterator_tag>>;


Also note that input iterators can be used to traverse a range only once. After you call std::distance(first, last), all subsequent attempts to traverse the range are undefined behaviour unless InputIt is at least a forward iterator. For input iterators you can't determine how much space to preallocate.

That's why assign uses tag dispatch technique internally. With some simplifications it looks like this:

template<typename InputIt, typename = std::RequireInputIter<InputIt>>
void assign(InputIt first, InputIt last) {
    M_assign_aux(first, last,
        typename iterator_traits<InputIt>::iterator_category{});
}

There are two M_assign_aux overloads

template<typename InputIt>
void M_assign_aux(InputIt first, InputIt last, std::input_iterator_tag);

template<typename ForwardIt>
void M_assign_aux(ForwardIt first, ForwardIt last, std::forward_iterator_tag);

to do the assignment. The first one will be used for input iterators only, and the second one - for forward iterators and those derived from it, i.e. bidirectional and random access ones.

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07-30 16:34