C++ - std::vector foreach - 没有重载函数的实例

Question

I can't figure this one; I recently started to use std::for_each but this one is a pain to solve

I keep receiving an error for the code below :

**no instance of overloaded function "std::vector<_Ty, _Alloc>::push_back [with _Ty=CharacterObject *, _Alloc=std::allocator<CharacterObject *>]" matches the argument list**

code:

class Stuff
{
};  

std::vector< Stuff* > list;
list.push_back(new Stuff());
list.push_back(new Stuff());
list.push_back(new Stuff());

std::vector< Stuff* > stuffs;

std::for_each(list.begin(), list.end(), [this,level, stuff](Stuff &stuff) 
{
    ... // some conditions here
    stuffs.push_back(stuff);  << error
    stuffs.push_back(&stuff);  << also throws an error
});

I tried many different combinations, but it's hard to figure anything with STL template errors

Any help greatly appreciated of course

sheers

[edit: there, works like acharm and no need for bad faith I consider extremely obnoxious]

for (std::vector< CharacterObject* >::iterator it = spawnedCharactersObjects.begin(); it != spawnedCharactersObjects.end(); ++it)
{
    CharacterObject* characterObject = *it;
    if (characterObject->getLevel() == level)
    {
        characterObjects.push_back(characterObject);
    }
}

Answer 1

This is how it would be done correct

#include <vector>
#include <algorithm>

class Stuff
{
};  

int main() {
    std::vector< Stuff* > list;
    list.push_back(new Stuff());
    list.push_back(new Stuff());
    list.push_back(new Stuff());

    std::vector< Stuff* > stuffs;

    std::for_each(list.begin(), list.end(), [&stuffs] (Stuff *stuff) {
        stuffs.push_back(stuff);
    });
    return 0;
}

std::for_each is a template function of kind

std::for_each(Container<Type>::iterator, Container<Type>::iterator, void(Type))

(That's not correct but it shows the relation between the iterator and the unary function). It expects as third parameter a function with one parameter of type Stuff * since list is a container containing elements of that type. Both types have to be same. Also you have to capture stuffs as reference if you want to change it.