这无法在GCC编译器4.1.2/RedHat的5:错误:不对应的 '操作[]' 在...... <near match>
#include <string>
#include <vector>
#include <map>
class Toto {
public:
typedef std::string SegmentName;
};
class Titi {
public:
typedef Toto::SegmentName SegmentName; // import this type in our name space
typedef std::vector<SegmentName> SegmentNameList;
SegmentNameList segmentNames_;
typedef std::map<SegmentName, int> SegmentTypeContainer;
SegmentTypeContainer segmentTypes_;
int getNthSegmentType(unsigned int i) const {
int result = -1;
if(i < segmentNames_.size())
{
SegmentName name = segmentNames_[i];
result = segmentTypes_[ name ];
}
return result;
}
};
的错误是:
error: no match for 'operator[]' in '(...)segmentTypes_[name]'
/usr/lib/gcc/x86_64-redhat-linux/4.1.2/../../../../include/c++/4.1.2/bits/stl_map.h:340:
note: candidates are: _Tp& std::map<_Key, _Tp, _Compare, _Alloc>::operator[](const _Key&)
[with _Key = std::basic_string<char, std::char_traits<char>, std::allocator<char> >, _Tp = int, _Compare = std::less<std::basic_string<char, std::char_traits<char>, std::allocator<char> > >, _Alloc = std::allocator<std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, int> >]
为什么?地图相当简单。我想这与typedefs有关,但是出了什么问题?
即使我删除所有typedefs
并使用std::string
无处不在,问题仍然存在......我滥用地图吗?
或者'at'在C++中11。 +1 – 2013-03-27 14:07:52
@LuchianGrigore我很确定'at'在C++之前就存在了11 – Saage 2013-03-27 14:09:27
@Saage只是作为一个扩展。现在它是标准的。 – juanchopanza 2013-03-27 14:09:52