返回要在另一个函数中使用的ifstream实例

Question

我分离了ifstream的开口处，这样我就可以遍历它而不需要重新打开它，但是我不确定如何返回它，这样它就可以在另一个函数中使用。实际上，Npc_B_File超出了第二个函数的范围。如何返回ifstream？

void battle_start(char const* P_Name)
{
ifstream Npc_B_File(P_Name);
if(Npc_B_File.fail())
    {
    cout << "could not read file.";
    }

}

 void battle_npc(string npc)
 {

    while(btlcommand != npc)
    {
    Npc_B_File >> btlcommand;
    }
    if(btlcommand == npc_pick_dog)
    Npc_B_File >> btlcommand;

    if(btlcommand == "1" && bat_response == true)
    {
    cout << "You are in" << btlcommand;
    Npc_B_File >> btlcommand;
    bat_response = false;
    }
    if(btlcommand == "2" && bat_response == true)
    {
    cout << "You are in" << btlcommand;
    Npc_B_File >> btlcommand;
    bat_response = false;
    }
    if(btlcommand == "3" && bat_response == true)
    {
    cout << "You are in" << btlcommand;
    Npc_B_File >> btlcommand;
    bat_response = false;
    }
}

Answer 1

选项1

如果您编译器和库支持ifstream的移动语义，您可以简单地使用：

ifstream battle_start(char const* P_Name)
{
    ifstream Npc_B_File(P_Name);
    if(Npc_B_File.fail())
    {
        cout << "could not read file.";
    }
    return Npc_B_File;
}

int main()
{
    ifstream file(battle_start("filename"));
}

选项2

对于没有为ifstream实现移动语义的旧版编译器或库，您可以使用：

void battle_start(char const* P_Name, /*out*/ifstream &Npc_B_File)
{
    Npc_B_File.open(P_Name);
    if(Npc_B_File.fail())
    {
        cout << "could not read file.";
    }
}

int main()
{
    ifstream file;
    battle_start("filename", file);
}

Answer 2

你可能想要这样的东西：

bool battle_start(char const* P_Name, std::ifstream &file) {
    file.open(P_Name);
    if(file.fail()) {
        cout << "could not read file.";
        return false;
     }
     return true;
}

作为不相关的补充，代码如下：

btlcommand == "2" && bat_response == true

...makes你看起来像个笨蛋。只需使用：

btlcommand == "2" && bat_response