STD ::绑定同类中的静态成员函数

Question

我正在尝试存储一个函数以稍后致电，这是片段。

这很好：

void RandomClass::aFunc( int param1, int param2, double param3, bool isQueued /*= false */ )
{
    /* If some condition happened, store this func for later */
    auto storeFunc = std::bind (&RandomClass::aFunc, this, param1, param2, param3, true);
    CommandList.push( storeFunc );
    /* Do random stuff */
}

但是，如果随机阶级是静态的，所以我相信我应该这样做：

void RandomClass::aFunc( int param1, int param2, double param3, bool isQueued /*= false */ )
{
    /* If some condition happened, store this func for later */
    auto storeFunc = std::bind (&RandomClass::aFunc, param1, param2, param3, true);
    CommandList.push( storeFunc );
    /* Do random stuff */
}

但这不起作用，我得到编译错误

错误c2668：'std :: tr1 :: bind'：对重载功能的含糊不清

任何帮助。

Answer 1

静态成员函数指针的类型看起来像是指向非成员函数的指针：

auto storeFunc = std::bind ( (void(*)(WORD, WORD, double, bool))
                              &CSoundRouteHandlerApp::MakeRoute, 
                              sourcePort, destPort, volume, true );

这是一个简化的示例：

struct Foo
{
  void foo_nonstatic(int, int) {}
  static int foo_static(int, int, int) { return 42;}
};
#include <functional>
int main()
{
  auto f_nonstatic = std::bind((void(Foo::*)(int, int))&Foo::foo_nonstatic, Foo(), 1, 2);
  auto f_static = std::bind((int(*)(int, int, int))&Foo::foo_static, 1, 2, 3);
}