有关Iterators and Reverse Iterators的程序

[faye_2004]

#include <iostream>
#include <deque>
#include <algorithm>
using namespace std;
void print (int elem)
{
cout << elem << ' ';
}
int main()
{
deque<int> coll;
for (int i=1; i<=9; ++i) {
coll.push_back(i);
}

deque<int>::iterator pos1;
pos1 = find (coll.begin(), coll.end(), 2);
deque<int>::iterator pos2;
pos2 = find (coll.begin(), coll.end(),7);
for_each (pos1, pos2, print);
cout << endl;
deque<int>::reverse_iterator rpos1(pos1);
deque<int>::reverse_iterator rpos2(pos2);
for.each (rpos2, rpos1, print);
cout << endl;
}
the output of the program is as follows:
2 3 4 5 6
6 5 4 3 2
为什么输出是6 5 4 3 2，如何将Iterators范围转化为Reverse Iterators

[fangrk]

你不是已经转换了吗？

[zhouwander]

在反向型迭代器中维护着一个随机迭代器，重载了＋＋和－－运算符也就变成反向的了，deque中的迭代器就是随机的。
_Myt& operator++()
{ // preincrement
--current;
return (*this);
}

_Myt operator++(int)
{ // postincrement
_Myt _Tmp = *this;
--current;
return (_Tmp);
}

_Myt& operator--()
{ // predecrement
++current;
return (*this);
}

_Myt operator--(int)
{ // postdecrement
_Myt _Tmp = *this;
++current;
return (_Tmp);
}

[prozax]

不是很对吗 有什么问题?

[faye_2004]

另一程序
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
vector<int> coll;
for (int i=1; i<=9; ++i) {
coll.push_back(i);}
vector<int>::iterator pos;
pos = find (coll.begin(), coll.end(),5);
cout << "pos: " << *pos << endl;
vector<int>::reverse_iterator rpos(pos);
cout << "rpos: " << *rpos <<endl;
}
输出为
pos: 5
rpos: 4为什么？

[UPCCPU]

上面的问题可以看一看STL源代码：
template<class _RI,
class _Ty,
class _Rt = _Ty&,
class _Pt = _Ty *,
class _D = ptrdiff_t>
class reverse_iterator : public _Ranit<_Ty, _D> {
public:
...
_Rt operator*() const
{return (*(current - 1)); }
...
}
这是visual c＋＋6.0中的reverse_iterator模板类的代码片断。

[faye_2004]

#include <iostream>
#include <deque>
#include <algorithm>
using namespace std;
void print (int elem)
{
cout << elem << ' ';
}
int main()
{
deque<int> coll;
for (int i=1; i<=9; ++i) {
coll.push_back(i);
}

deque<int>::iterator pos1;
pos1 = find (coll.begin(), coll.end(), 2);
deque<int>::iterator pos2;
pos2 = find (coll.begin(), coll.end(),7);
for_each (pos1, pos2, print);
cout << endl;
deque<int>::reverse_iterator rpos1(pos1);
deque<int>::reverse_iterator rpos2(pos2);
for.each (rpos2, rpos1, print);
cout << endl;
}
the output of the program is as follows:
2 3 4 5 6
6 5 4 3 2
为什么输出是6 5 4 3 2，如何将Iterators范围转化为Reverse Iterators
另一程序
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
vector<int> coll;
for (int i=1; i<=9; ++i) {
coll.push_back(i);}
vector<int>::iterator pos;
pos = find (coll.begin(), coll.end(),5);
cout << "pos: " << *pos << endl;
vector<int>::reverse_iterator rpos(pos);
cout << "rpos: " << *rpos <<endl;
}
输出为
pos: 5
rpos: 4为什么？

[nlstone]

详见任何一本解释相关内容(逆迭代器)的书,譬如c++ standard library
简单地说,逆迭代器的内容是取其前一个元素的值.譬如说如果一个Reverse Iterator指向第五个元素,那么它的内容(譬如用*来访问,这是重载过的),会去取第四个元素的值.
这是为了使得find一类的算法不需要做任何改变即可应付逆迭代器(否则因为有半开区间的影响,逆迭代器find()的话就需要做更改,要么改find算法,要么改动传入find()的逆迭代器的位置)