Reputation: 3011
I tried to overload the std::find function, in this way:
include <list>
include <string>
include "Marker.h"
namespace Test {
class MarkerContainer {
private:
std::list<Marker*> list;
double end_t;
inline bool exists(std::list<Marker*> *input_list, Marker* check);
public:
MarkerContainer();
MarkerContainer(double end_time);
bool addMarker(Marker* input);
double computeBeatTime(double sample_t);
double computeSampleTime(double beat_t);
void printAll();
};
}
std::list<Ableton::Marker*>::iterator std::find(std::list<Ableton::Marker*>::iterator first, std::list<Ableton::Marker*>::iterator last, Ableton::Marker* value){
for ( ;first!=last; first++) if ( **first==*value ) break;
return first;
}
But I catch this compiler error:
Out-of-line definition of 'find' does not match any declaration in namespace 'std' in /Users/.../MarkerContainer.h
I hope I did some stupid mistake and what I would like to do is simple.. any idea ?
Thanks in advance! Pietro
Upvotes: 0
Views: 1987
Reputation: 153792
You can't overload a function like this: you need to define the overload in an appropriate namespace. You need to open namespace std
at global scope. Specifically with namespace std
I don't think you are allowed to do this. Otherwise this would work.
That said, I think you should look at using std::find_if()
with a suitable predicate instead of using your hand-crafted approach:
auto it = std::find_if(list.begin(), list.end(),
[=](Abelton::Marker* el)->bool { return *el == *check; });
Upvotes: 0
Reputation: 96233
You aren't allowed to overload find
. You are allowed to specialize templates in the std
namespace, but in this case it's far simpler to use find_if
with a dereferencing functor.
// Create DerefEquality comparison (I'll try to come back later and write it).
container_type::iterator iter = find_if(container.begin(), container.end(), DerefEquality(item_to_find));
Upvotes: 3