问题
I would like to create a function with a signature like this:
// Set found to be an iterator to the location of key in map or end()
// if not found.
bool lookup(const Key &key,
const std::map<Key, Value> &map,
std::map<Key, Value>::const_iterator &found);
But I would like to also call it in cases where the map and iterator are not const so that I can modify the found value:
const Key key;
std::map<Key, Value> map;
std::map<Key, Value>::iterator found;
if (lookup(key, map, found)) {
found->second.modifingNonConstFunction()
}
But I do not believe I can pass a std::map<Key, Value>::iterator
object to a function expecting a reference to a std::map<Key, Value>::const_iterator
since they are different types, whereas I normally could if the const
was part of C++ declaration of the type like this and I could promote the non-const type to a const type:
void someFunction(const int &arg);
int notConstArg = 0;
someFunction(nonConstArg);
Other than by using templates to provide two definitions for lookup()
, one as shown with const
arguments 2 and 3 and another with non-const arguments 2 and 3, is there a better way in C++ to accomplish this more akin to how const int &
can be passed a non-const int
in the example above. In other words, can I just have a single function and not two?
回答1:
No, I don't think you can do this without overloads/template magic.
The compiler is protecting you from the following scenario:
typedef vector<int> T;
const T v; // Can't touch me
void foo(T::const_iterator &it) {
it = v.begin(); // v.begin() really is a const_iterator
}
int main() {
T::iterator it;
foo(it);
*it = 5; // Uh-oh, you touched me!
}
回答2:
If the function is simple or you don't mind binary bloat, just make every parameter a template parameter.
template <typename Key, typename T, typename Iter>
bool lookup(Key const& key,
T& map,
Iter &found)
{
return (found=map.find(key))!=map.end();
}
int main()
{
std::map<std::string, int> m; m["hello"] = 42;
std::map<std::string, int> const cm(m.begin(), m.end());
std::map<std::string, int>::iterator it;
std::map<std::string, int>::const_iterator cit;
std::cout << std::boolalpha << lookup("hello", m, it) << '\n'; // Key isn't even std::string
std::cout << std::boolalpha << lookup("hello", m, cit) << '\n';
//std::cout << std::boolalpha << lookup("hello", cm, it) << '\n'; // error
std::cout << std::boolalpha << lookup("hello", cm, cit) << '\n';
}
This works since T
can be both, map
and const map
so T&
is map&
or const map&
.
来源:https://stackoverflow.com/questions/14571221/how-to-provide-const-interface-with-iterators-to-a-collection