我應該總是處理參考返回矢量嗎？

Question

不知道here發生的事情：

#include <iostream> 
#include <vector> 

class Voice 
{ 
public: 
    double mValue = 0.0; 

    Voice() { } 
    Voice(const Voice &copy) { 
    }  
}; 

class VoiceManager 
{ 
public: 
    std::vector<Voice> mVoices; 

    VoiceManager() { 
     mVoices = std::vector<Voice>(numVoices, Voice()); 

     for (int i = 0; i < numVoices; i++) { 
      mVoices[i].mValue = 100.0; 
     }   
    } 

private: 
    int numVoices = 16;  
}; 

int main() 
{  
    VoiceManager voiceManager; 
    Voice voice = voiceManager.mVoices[2]; 
    std::cout << voice.mValue << std::endl; 
} 

我設置mValue每個Voice與運營商[]價值100，但是當我嘗試用相同的[]操作檢索對象，現在看來，這不返回該元素？它打印0

相反，如果我這樣做：

Voice &voice = voiceManager.mVoices[2]; 

我終於可以看到更新後的值。

這裏發生了什麼？

Answer 1

正如在評論中指出的，你的拷貝構造函數是一個）錯誤和b）多餘

class Voice 
{ 
public: 
    double mValue = 0.0; 

    // don't provide default constructor, unless you have another constructor 
    // use compiler-generated copy-constructor 
}; 

class VoiceManager 
{ 
    int numVoices = 16; // move up here so that member-initialization works 
public: 
    std::vector<Voice> mVoices; 

    VoiceManager() 
    : 
     mVoices(numVoices) // member-initialize 
    { 
     for (int i = 0; i < numVoices; i++) { 
      mVoices[i].mValue = 100.0; 
     }   
    } 
}; 

int main() 
{  
    VoiceManager voiceManager; 
    Voice voice = voiceManager.mVoices[2]; 
    std::cout << voice.mValue << std::endl; 
} 

Live Example

請注意，我也沒有在構造未成年人清理了VoiceManager，通過使用成員 - 初始化語法。

至於其他的問題：初始化

Voice voice = voicemanager.mVoice[2]; 

實際調用你的拷貝構造函數

Voice(voicemanager.mVoice[2]); 

（這裏的=跡象並不意味着分配！）