[英]getting an iterator to a generic container inside a generic class: HOW TO
這來自更大的上下文,但是我剝離了很多代碼以簡化問題。 如果您覺得我遺漏了任何東西,請告訴我。
假設您有一個定義為的模板類:
#include <map>
#include <vector>
#include <string>
#include <fstream>
#include <iostream>
#include <iterator>
#include <algorithm>
template <class key, class container, class container_type>
class file_to_map
{
public:
file_to_map()
{
m_file = "";
}
file_to_map(std::string file)
{
m_file = file;
}
~file_to_map()
{
}
std::map<key, container>& get_map()
{
return m_map;
}
void set_file(const std::string file)
{
m_file = file;
}
void insert_into_map(key insert, container_type value)
{
m_map[insert].insert(value);
}
friend std::ostream& operator<< (std::ostream &out, file_to_map<key, container, container_type> &obj)
{
typedef typename std::map<key, container>::const_iterator mapItr;
mapItr mbi = obj.m_map.begin();
mapItr emi = obj.m_map.end();
while (mbi != emi) {
out << " -- " << mbi->first << " -- " << std::endl;
container::iterator cbi;
++mbi;
}
return out;
}
friend std::istream& operator>> (std::ifstream &in, file_to_map<key, container, container_type> &obj)
{
if (in.is_open())
in.close();
if (obj.m_file == "")
return in;
in.open(obj.m_file.c_str(), std::ios::in);
if (in.fail() || in.bad()) {
in.close();
return in;
}
std::vector<key> tmp;
typedef std::istream_iterator<key> string_input;
copy(string_input(in), string_input(), back_inserter(tmp));
typename std::vector<key>::iterator bvi = tmp.begin();
typename std::vector<key>::iterator evi = tmp.end();
while (bvi != evi) {
obj.m_map[*(bvi)] = container();
++bvi;
}
in.close();
return in;
}
private:
std::map<key, container> m_map;
std::string m_file;
};
並在朋友方法“ operator<<
”中
您要打印地圖和通用container
怎么辦呢? 如何獲得適當的迭代器來遍歷通用container
的內容。
我正在嘗試:
container::iterator cbi;
代碼更新
帶有:
friend std::ostream& operator<< (std::ostream &out, file_to_map<key, container, container_type> &obj)
{
typedef typename std::map<key, container>::const_iterator mapItr;
mapItr mbi = obj.m_map.begin();
mapItr emi = obj.m_map.end();
while (mbi != emi) {
out << " -- " << mbi->first << " -- " << std::endl;
typename container::const_iterator cbi = mbi->second.begin();
typename container::const_iterator ebi = mbi->second.end();
std::copy(cbi, mbi, std::ostream_iterator<container_type>(out, "\t\n"));
++mbi;
}
return out;
}
我收到以下編譯器錯誤:
g++ -o file_to_map -Wall ./file_to_map.h ./main.cpp ./code_finder.cpp \
-L/usr/local/boost_1_48_0/stage/lib -lboost_filesystem -lboost_system -I /usr/local/boost_1_48_0/
In file included from ./code_finder.h:4,
from ./code_finder.cpp:1:
./file_to_map.h: In function ‘std::ostream& operator<<(std::ostream&, file_to_map<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::set<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::less<std::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::allocator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > > >, std::basic_string<char, std::char_traits<char>, std::allocator<char> > >&)’:
./code_finder.cpp:36: instantiated from here
./file_to_map.h:62: error: no matching function for call to ‘copy(std::_Rb_tree_const_iterator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > >&, operator<<(std::ostream&, file_to_map<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::set<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::less<std::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::allocator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > > >, std::basic_string<char, std::char_traits<char>, std::allocator<char> > >&)::mapItr&, std::ostream_iterator<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, char, std::char_traits<char> >)’
我想念什么?
使用typename
:
typename container::iterator cbi;
因為iterator
是一個從屬名稱,所以它取決於類模板的類型參數container
。 有趣的是,如果您已在其他地方更正了使用的typename
名,例如在operator<<
:
typedef typename std::map<key, container>::const_iterator mapItr;
由於您正在為地圖使用const_iterator
,這意味着您將使用地圖的const迭代器獲取的容器對象將是const
對象,這又意味着您也需要對容器使用const_iterator
,因為它們是的值地圖。 所以我認為您需要使用:
typename container::const_iterator cbi; //use this instead!
回復您的編輯:
我在這里看到一個錯字:
typename container::const_iterator cbi = mbi->second.begin();
typename container::const_iterator ebi = mbi->second.end();
std::copy(cbi, mbi, std::ostream_iterator<container_type>(out, "\t\n"));
// ^^^ typo
std::copy
的第二個參數應該是ebi
,而不是mbi
。 這就是為什么我通常將這樣的變量命名為begin
和end
,而不是cbi
和ebi
。
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