c++ - 接受 lambda 作为参数的类方法 - 错误消息

Question

我试图创建这样的方法：

WeaponsDatabase
{
public:

    template<typename Functor> 
    QList<const WeaponData*> getSpecificWeapons(Functor criterium);
};
template<typename Functor>
QList<const WeaponData*> WeaponsDatabase::getSpecificWeapons(Functor criterium)
{
    QList<const WeaponData*> weaponsForCriterium;

    foreach (const WeaponData *weapon, weapons)
    {
        if (criterium(weapon))
        {
            weaponsForCriterium.append(weapon);
        }
    }

    return weaponsForCriterium;
}

似乎没问题，它编译了。然后我尝试在其他地方使用它：

auto criterium = [hardpoint](const WeaponData *weapon)->bool
{
    return weapon->hardpoint == hardpoint;
};

WeaponsDatabase::getInstance().getSpecificWeapons(criterium);

它不起作用，我不明白错误消息。各位大侠能帮我解释一下吗？

In file included from ..\equipmentgroupwidget.cpp:3:0:
..\weaponsdatabase.h:35:57: error: 'QList<const WeaponData*> WeaponsDatabase::getSpecificWeapons(Functor) [with Functor = EquipmentGroupWidget::EquipmentGroupWidget(const QString&, QWidget*)::<lambda(const WeaponData*)>]', declared using local type 'EquipmentGroupWidget::EquipmentGroupWidget(const QString&, QWidget*)::<lambda(const WeaponData*)>', is used but never defined [-fpermissive]
template<typename Functor> QList<const WeaponData*> getSpecificWeapons(Functor criterium);

标签： c++lambda

Answer 1

@WernerHenze 的所有功劳——将实现移到类中解决了这个问题。非常感谢你的伙伴。