[英]std::map::const_iterator template compilation error
I have a template class that contains a std::map
that stores pointers to T which refuses to compile: 我有一个模板类,其中包含一个
std::map
,它存储指向T的指针,而T拒绝编译:
template <class T>
class Foo
{
public:
// The following line won't compile
std::map<int, T*>::const_iterator begin() const { return items.begin(); }
private:
std::map<int, T*> items;
};
gcc gives me the following error: gcc给我以下错误:
error: type 'std::map<int, T*, std::less<int>, std::allocator<std::pair<const int, T*> > >' is not derived from type 'Foo<T>'
Similarly, the following also refuses to compile: 同样,以下内容也拒绝编译:
typedef std::map<int, T*>::const_iterator ItemIterator;
However, using a map that doesn't contain the template type works OK, eg: 但是,使用不包含模板类型的地图也可以,例如:
template <class T>
class Foo
{
public:
// This is OK
std::map<int, std::string>::const_iterator begin() const { return items.begin(); }
private:
std::map<int, std::string> items;
};
I assume this is related to templates and begs the question - how can I return a const_iterator
to my map? 我认为这与模板有关,并提出了一个问题-如何将
const_iterator
返回到地图?
Use typename
: 使用
typename
:
typename std::map<int, T*>::const_iterator begin() const ...
When this is first passed by the compiler, it doesn't know what T
is. 当编译器第一次传递它时,它不知道
T
是什么。 Thus, it also doesn't know wether const_iterator
is actually a type or not. 因此,它也不知道
const_iterator
是否实际上是一种类型。
Such dependent names (dependent on a template parameter) are assumed to 此类依赖名称 (取决于模板参数)假定为
typename
typename
template
. template
。 您需要输入typename
:
typename std::map<int, T*>::const_iterator begin() const { return items.begin(); }
You need: 你需要:
typename std::map<int, T*>::const_iterator begin() const { return items.begin(); }
Or simpler 或更简单
typedef typename std::map<int, T*>::const_iterator const_iterator;
const_iterator begin() const { return items.begin(); }
This is because const_iterator
is dependent name on T
so you need to tell compiler that it is actually type. 这是因为
const_iterator
是T
依赖名称,因此您需要告诉编译器它实际上是类型。
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