FastDelegate
指的是http://www.codeproject.com/KB/cpp/FastDelegate.aspx,但我认为它不相关。
我有以下代码,但出现错误。
#include <FastDelegate.h>
using namespace fastdelegate;
template <typename T>
T Getter() {}
template <typename T>
void Setter(T) {}
template <typename T>
class Prop
{
public:
typedef FastDelegate0<T> Getter;
typedef FastDelegate1<T> Setter;
Prop(Getter getter, Setter setter) :
m_Getter(getter), m_Setter(setter)
{
}
private:
Getter m_Getter;
Setter m_Setter;
};
template <typename T>
inline Prop<T>* MakeProp(FastDelegate0<T> getter, FastDelegate1<T> setter)
{
return new Prop<T>(getter, setter);
}
static int Target = 0;
int main()
{
FastDelegate0<int> fdGetter(Getter<int>);
Prop<int>* c = MakeProp(fdGetter, Setter<int>);
// ^^^^ error: no matching function for call to 'MakeProp'
}
如果将
main()
更改为:int main()
{
FastDelegate0<int> fdGetter(Getter<int>);
FastDelegate1<int> fdSetter(Setter<int>);
Prop<int>* c = MakeProp(fdGetter, fdSetter); // It works.
}
要么:
int main()
{
FastDelegate0<int> fdGetter(Getter<int>);
Prop<int>* c = MakeProp<int>(fdGetter, Setter<int>); // It works, too.
}
我认为,
MakeProp()
应该从fdGetter(即T
,而不是自动调用int
的构造函数)中获取FastDelegate1<int>
。但这不是。为什么?附言我想将getter和setter保存在
Prop
中,欢迎对此方法提出任何建议。在函数中传递参数的过程中,复制FastDelegate *的实例可能不好。 最佳答案
你有没有尝试过
Prop<int>* c = MakeProp(FastDelegate0<int>(Getter<int>), FastDelegate1<int>(Setter<int>));
?
Setter<int>
不能转换为FastDelegate1<T>
!