【发布时间】:2014-03-21 14:33:30
【问题描述】:
我有一个包含不同类型向量的结构。我想创建一个迭代器,它将迭代每个向量的每个元素,并根据特定规则,使用 * 运算符提供特定值。我已经为迭代器的迭代器创建了一个玩具示例:
/*--------------------------------------------------------------------*/
/* Copyright (C) Inria 2014
This program is free software: you can redistribute it and/or modify
it under the terms of the GNU General Public License as published by
the Free Software Foundation, either version 3 of the License, or
(at your option) any later version.
This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
GNU General Public License for more details.
You should have received a copy of the GNU General Public License
along with this program. If not, see <http://www.gnu.org/licenses/>.
*/
/*
* Project: MixtComp
* Created on: Feb 20, 2014
* Author: Vincent KUBICKI <vincent.kubicki@inria.fr>
**/
#include <iostream>
#include <iterator>
#include <vector>
class MyIterator : public std::iterator<std::input_iterator_tag, int>
{
typedef std::vector<int>::const_iterator it ;
typedef std::vector<it> vecit;
typedef std::vector<it>::iterator itit ;
public:
MyIterator(it itA ,
it itB ,
it endA,
it endB)
{
vec_it_.push_back(itA);
vec_it_.push_back(itB);
vec_end_.push_back(endA);
vec_end_.push_back(endB);
it_vec_it_ = vec_it_.begin();
it_vec_end_ = vec_end_.begin();
if (itA == endA)
{
++it_vec_it_;
++it_vec_end_;
}
}
MyIterator(const MyIterator& mit) :
vec_it_(mit.vec_it_),
vec_end_(mit.vec_end_),
it_vec_it_(mit.it_vec_it_),
it_vec_end_(mit.it_vec_end_)
{}
MyIterator& operator++()
{
if (*it_vec_it_ != *it_vec_end_)
{
++(*it_vec_it_);
if (*it_vec_it_ == *it_vec_end_)
{
++it_vec_it_;
++it_vec_end_;
}
}
return *this;
}
MyIterator operator++(int)
{
MyIterator tmp(*this);
operator++();
return tmp;
}
bool operator==(const MyIterator& rhs)
{
return (*it_vec_it_ == *(rhs.it_vec_it_));
}
bool operator!=(const MyIterator& rhs)
{
return (*it_vec_it_ != *(rhs.it_vec_it_));
}
const int operator*()
{
return *(*it_vec_it_);
}
private:
vecit vec_it_; // vector of current iterators
vecit vec_end_; // vector of end iterators
itit it_vec_it_; // iterator on vec_it_
itit it_vec_end_; // iterator on vec_end_
};
int main () {
std::vector<int> dataA {1, 3, 4, 9};
std::vector<int> dataB {11, 34, 43};
MyIterator from(dataA.begin(),
dataB.begin(),
dataA.end(),
dataB.end());
MyIterator until(dataA.end(),
dataB.end(),
dataA.end(),
dataB.end());
for (MyIterator it = from; it != until; ++it)
{
std::cout << *it << std::endl;
}
return 0;
}
但是,在这个实现中,dataA 和 dataB 都是相同类型的向量。假设我想要不同类型的向量,例如 std::vector<std::pair<int, int>> 和 std::vector<std::pair<int, std::vector<int>>>。
是否有一种明智(且简单)的方法来实现这一目标?
我的第一个想法是用 Base 和 Derived 迭代器的组合替换 it 类型。 itit 将是 std::vector<Base> 上的迭代器。为了比较Base 迭代器,我会使用void 指针来比较它们各自指向的对象的地址。 operator*() 将在 Derived 类中定义,并且都将提供相同类型的结果,例如 int。
【问题讨论】:
-
你有一个所有类型派生自一个公共基类吗?否则,您将丢失有关类型的信息。然后, boost::any 可能是一个选项
-
你也可以看看
boost::variant。 -
感谢您提供的信息。
标签: c++ vector stl iterator standard-library