C++ 获取对 SingleTune 类成员方法的访问权限 获取对不完整类型的成员访问权限

Question

我有一个简单的单例，它拥有另一个类作为它的成员我尝试访问这个成员类的公共(public)方法，我得到:

Member access into incomplete type

我的单例类:

#include "SoundManager.h"
class Singleton : pSoundManager(new SoundManager())
{
  public:
static
  Singleton getInstance()
  {
    if (!instance_)
      {
        instance_ = new Singleton();
      }

    return instance_;
  }
  SoundManager*  getSoundManager( return pSoundManager;);
  private:
    static Singleton* instance_;
    Singleton() {};
    ~Singleton(){};
    SoundManager* pSoundManager;

}

//单例中保留的类

class SoundManager
{
public:
    SoundManager(){};
    ~SoundManager(){};


    unsigned int playEffect() {printf("effect!!");};

private:


};

现在当我尝试做的时候

Singleton::getInstance()->getSoundManager()->playEffect() ;

我得到:

Member access into incomplete type 'SoundManager'

我在这里做错了什么？

Answer 1

这段代码

SoundManager* getSoundManager( return pSoundManager;);

在语法上是错误的，您看到的错误消息可能会产生误导。将此代码替换为

SoundManager* getSoundManager() { return pSoundManager; }

---

我要做的是稍微简化 getInstance() 函数(除了修复一些明显的语法错误之外):

#include "SoundManager.h"

class Singleton {
  public:
static
  Singleton& getInstance() {
    static Singleton instance_;
    return instance_;
  }

  SoundManager& getSoundManager() { return soundManager_; }
  private:
    Singleton() {};
    ~Singleton(){};
    SoundManager soundManager_;    
};

调用playEffect()函数会变成

Singleton::getInstance().getSoundManager().playEffect();

当然。