[英]C++ nested map no matching member function const member
I have a nested map, something like我有一个嵌套的地图,比如
map = {
key : {
innerKey: innerVal
}
}
I'm trying to search for innerVal
from member function marked as const
.我正在尝试从标记为
const
成员函数中搜索innerVal
。 I'm using at()
as described here C++ map access discards qualifiers (const) which gets me to map pointed by key
.我正在使用
at()
,如此处所述C++ 映射访问丢弃限定符 (const) ,它使我能够映射由key
指向的映射。 But, when I try to use at()
on nested map, I get an error:但是,当我尝试在嵌套地图上使用
at()
时,出现错误:
error: no matching member function for call to 'at'
Workaround: I can use a iterator and linearly search on the nested map, which works perfectly.解决方法:我可以使用迭代器并在嵌套地图上线性搜索,这非常有效。 How can I use functions such as
at()
or find()
to search in nested map.如何使用
at()
或find()
等函数在嵌套地图中进行搜索。
TLDR:域名注册地址:
private std::map<int, std::map<int, int> > privateStore;
int search(int key1, int key2) const {
return privateStore.at(key1).at(key2); //works when I remove `const` from function signature
}
Edit: it works for above simplified code, try this , and try removing const
keyword from line 20.编辑:它适用于上述简化代码,试试这个,并尝试从第 20 行中删除
const
关键字。
#include <iostream>
#include <map>
#include <thread>
template <typename T>
class Foo
{
public:
Foo()
{
std::cout << "init";
}
void set(T val)
{
privateStore[std::this_thread::get_id()][this] = val;
}
T search(std::thread::id key1) const
{
std::map<Foo<T>*, T> retVal = privateStore.at(key1); //works when I remove `const` from function signature
return retVal.at(this);
}
private:
static std::map<std::thread::id, std::map<Foo<T>*, T> > privateStore;
};
template<typename T> std::map<std::thread::id, std::map<Foo<T>*, T> > Foo<T>::privateStore = {};
int main()
{
Foo<int> a;
a.set(12);
std::cout << a.search(std::this_thread::get_id());
}
Declare your inner map's key to be a pointer to const
object.将内部映射的键声明为指向
const
对象的指针。 Otherwise when you pass this
in a const function you pass Foo<T> const*
instead of Foo<T>*
and you can't convert that implicitly.否则,当您在 const 函数中传递
this
时,您将传递Foo<T> const*
而不是Foo<T>*
并且您无法隐式转换它。
So所以
static std::map<std::thread::id, std::map<Foo<T> *, T> > privateStore;
to到
static std::map<std::thread::id, std::map<Foo<T> const*, T> > privateStore;
And the same in the definition.和定义相同。
live example of your example - fixed.你的例子的活生生的例子 - 固定。
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