C++循环std :: vector >
问题描述:
如何循环这个?C++循环std :: vector <std :: map <std :: string,std :: string>>
我已经尝试过:
//----- code
std::vector<std::map<std::string, std::string> >::iterator it;
for (it = users.begin(); it != users.end(); it++) {
std::cout << *it << std::endl; // this is the only part i changed according to the codes below
}
//----- error
error: initializing argument 1 of ‘std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&&, const _Tp&) [with _CharT = char; _Traits = std::char_traits<char>; _Tp = std::map<std::basic_string<char>, std::basic_string<char> >]’
//----- code
std::cout << *it["username"] << std::endl;
//----- error
note: template argument deduction/substitution failed:
note: ‘std::map<std::basic_string<char>, std::basic_string<char> >’ is not derived from ‘const std::complex<_Tp>’
//----- code
std::cout << *it->second << std::endl; // also tried with parenthesis - second()
//----- error
error: ‘class std::map<std::basic_string<char>, std::basic_string<char> >’ has no member named ‘second’
//----- code
for(const auto& curr : it) std::cout << curr.first() << " = " << curr.second() << std::endl;
//----- error
error: unable to deduce ‘const auto&’ from ‘<expression error>’
,最后
//----- code
std::map<std::string, std::string>::iterator curr, end;
for(curr = it.begin(), end = it.end(); curr != end; ++curr) {
std::cout << curr->first << " = " << curr->second << std::endl;
}
//----- error
‘std::vector<std::map<std::basic_string<char>, std::basic_string<char> > >::iterator’ has no member named ‘begin‘ & ‘end’
我希望我给出一个明确的细节..然后上面是下面的代码是错误..而目前我的脑子一片空白。
和抱歉这个..
我已经让这种类型的工作:std::map<int, std::map<std::string, std::string> >
和IM尝试使用向量作为一个选项。
答
您的迭代代码是正确的;问题是你的输出语句。您的代码是这样做的:
std::cout << *it << std::endl;
在这种情况下,*it
指std::map<string,string>
和std::cout
不知道如何输出的地图。也许你想是这样的:
std::cout << (*it)["username"] << std::endl;
确保使用()s左右*it
否则,你将有运算符优先级的问题。
答
std::vector<std::map<std::string, std::string> >::iterator it;
for (it = users.begin(); it != users.end(); it++) {
std::cout << *it << std::endl;
当users
不.empty()
,上面<<
运营商尝试串流一个std::map<std::string, std::string>
对象,但标准库不提供过载流映射:如何将它知道你按键之间想要的分隔符价值观和元素之间?
我建议你打破下来的问题是这样的:
std::vector<std::map<std::string, std::string> >::iterator it;
for (it = users.begin(); it != users.end(); it++)
{
std::map<std::string, std::string>& m = *it;
for (std::map<std::string, std::string>::iterator mit = m.begin();
mit != m.end(); ++mit)
std::cout << mit->first << '=' << mit->second << '\n';
std::cout << "again, username is " << m["username"] << '\n';
}
这可以简化在C++ 11:
for (auto& m : users)
for (auto& kv : m)
std::cout << kv.first << '=' << kv.second << '\n';
THANK YOU VERY MUCH ...我想你所有的答案,这一切工作.. – xeroblast 2015-02-12 06:28:29