在类中的函数之外归档向量

Question

这里有一个相当简单的问题，在类 .cpp 文件中填充函数之外的向量的最佳方法是什么？目前我正在尝试以下不起作用：

std::vector<Player> midfielder(8);

midfielder.at(0) = Midfielder("Default         ",0,"Midfielder");
midfielder.at(1) = Midfielder("David Armitage  ",1,"Midfielder");
midfielder.at(2) = Midfielder("Tom Rockliff    ",2,"Midfielder");
midfielder.at(3) = Midfielder("Gary Ablett     ",3,"Midfielder");
midfielder.at(4) = Midfielder("Dyson Heppel    ",4,"Midfielder");
midfielder.at(5) = Midfielder("Scott Pendlebury",5,"Midfielder");
midfielder.at(6) = Midfielder("Michael Barlow  ",6,"Midfielder");
midfielder.at(7) = Midfielder("Jack Steven     ",7,"Midfielder");

为了提供上下文，“Midfielder”是一个继承自“Player”类的类。

团队管理.h

#ifndef TEAMMANAGEMENT_H
#define TEAMMANAGEMENT_H
#include <vector>
#include "Player.h"
#include "Midfielder.h"
#include <string>

class TeamManagement
{
    public:
        TeamManagement();


        void Display_Players();

};

#endif // TEAMMANAGEMENT_H

团队管理.cpp

#include <iostream>
#include <string>
#include <vector>
#include "Player.h"
#include "Midfielder.h"
#include "TeamManagement.h"
using namespace std;

TeamManagement::TeamManagement()
{

}

std::vector<Player> midfielder(8);
//errors start occurring on line below: 'midfielder' does not name a type
midfielder.at(0) = Midfielder("Default         ",0,"Midfielder");
midfielder.at(1) = Midfielder("David Armitage  ",1,"Midfielder");
midfielder.at(2) = Midfielder("Tom Rockliff    ",2,"Midfielder");
midfielder.at(3) = Midfielder("Gary Ablett     ",3,"Midfielder");
midfielder.at(4) = Midfielder("Dyson Heppel    ",4,"Midfielder");
midfielder.at(5) = Midfielder("Scott Pendlebury",5,"Midfielder");
midfielder.at(6) = Midfielder("Michael Barlow  ",6,"Midfielder");
midfielder.at(7) = Midfielder("Jack Steven     ",7,"Midfielder");
//errors stop occurring here

void TeamManagement::Display_Players(){

cout<<"Position     Name              ID"<<endl;

for (int i=1;i<8;i++)
    {
         cout<<midfielder[i].Player_Details()<<"   "<<midfielder[i].Get_player_id()<<endl;
    }
}

Answer 1

大概默认构造一个Midfielder没有多大意义，所以你可以reserve内存，然后emplace_back变成vector。

std::vector<Player> midfielder {};
midfielder.reserve(8);

midfielder.emplace_back("Default         ",0,"Midfielder");
midfielder.emplace_back("David Armitage  ",1,"Midfielder");
midfielder.emplace_back("Tom Rockliff    ",2,"Midfielder");
midfielder.emplace_back("Gary Ablett     ",3,"Midfielder");
midfielder.emplace_back("Dyson Heppel    ",4,"Midfielder");
midfielder.emplace_back("Scott Pendlebury",5,"Midfielder");
midfielder.emplace_back("Michael Barlow  ",6,"Midfielder");
midfielder.emplace_back("Jack Steven     ",7,"Midfielder");

Answer 2

midfielder.at(0) = Midfielder("Default ",0,"Midfielder"); 是一个声明。您已将该语句和类似语句放在（全局）命名空间范围内。那是你的错误。只有声明可以在命名空间范围内。您必须将语句放在函数中。

错误消息源于不以关键字开头的声明以类型名称开头的事实。由于midfielder 不是关键字，编译器期望它是一个类型名称，但它不是一个，所以你会得到错误。

Answer 3

第一个问题是你不能像在函数之外那样执行赋值。您必须使用构造或初始化。

使用 C++98，您无法在函数之外填充/初始化向量。

使用 C++11/14，您可以使用初始化语法填充一个：

#include <iostream>
#include <vector>

struct Thing {
    int m_i, m_j;
    Thing(int i, int j) : m_i(i), m_j(j) {}
};

std::vector<Thing> things {
    { 1, 2 }, { 2, 3 }
};

int main() {
    std::cout << "things[0].m_j = " << things[0].m_j << '\n';
}

但 std::vector 不喜欢你尝试将“中场”放入 Player 的向量中。让我们使用 SSCCE 来重建您正在造成的损害：

#include <iostream>

struct Base {
    int i;
};

struct Derived : public Base {
    int j;
};

int main() {
    std::cout << "Base size = " << sizeof(Base) << '\n';
    std::cout << "Derived size = " << sizeof(Derived) << '\n';
}

这告诉我们 Base 和 Derived 的大小不同。但是您试图将这两个对象放入同一个容器中，因为它们是相关的。圆钉和方钉是相关的……它们不适合同一个洞，这就是我们现在遇到的问题。

向量会根据您提供的类型在内存中为您的元素创建空间，然后它要求您将准确的类型传递给它以填充这些空间，或者具有到存储类型的转换机制的类型。

如果你想拥有一个不同类型的容器，你需要使用指针，但是你会遇到一个问题，你得到的将是一个指向基本类型的指针，你需要为自己提供一种区分不同玩家类型的方法。

有关 C++98 方法，请参阅 Store derived class objects in base class variables。在现代 C++（11 和 14）中，您应该使用智能指针，例如

std::vector<std::unique_ptr<Base>>

std::vector<std::shared_ptr<Base>>