[英]C++ iterating a struct
Is it possible to iterate through a struct? 是否可以遍历一个结构?
For example 例如
struct team{
int player1;
int player2;
int player3;
int player4;
...
int player99;
int size = 99;
}
then run a for loop to set or access foo 1-4? 然后运行for循环来设置或访问foo 1-4?
i guess pseudocode would look something like 我猜伪代码看起来像
for(int i = 0; i < size; i++){
player i = (i+1);
}
A more simplified explanation if that doesnt make sense is I Just want to be able to go through each variable without having to hard code player1 = 1; 一个更简单的解释,如果没有意义,我只是希望能够通过每个变量而不必硬编码player1 = 1; player2 =2.
player2 = 2。
One way is to put the players/elements into an array: 一种方法是将player / elements放入一个数组:
struct Team {
static int const size = 99;
int players[size];
};
And then: 接着:
for(int i = 0; i < size; ++i)
int player = players[i];
I would propose to use container instead of many variables, for example you could use std::array
or std::vector
. 我建议使用容器而不是许多变量,例如你可以使用
std::array
或std::vector
。 This way it will be trivial to iterate, much easier to make a copy. 这样,迭代就会变得微不足道,更容易复制。 But also it's better from design point of view: in case you decide to change the number of players it will be much easier to change the container rather than add/remove many fields
但从设计的角度来看也更好:如果您决定更改播放器的数量,更改容器将更容易,而不是添加/删除许多字段
To answer your question as you've asked it, I believe that you can use the pre-compiler macro Pack (the exact phrase depends on your compiler) to guarantee the structure of the memory used to create an instance of your struct. 为了回答你提出的问题,我相信你可以使用预编译器宏包(确切的短语取决于你的编译器)来保证用于创建结构实例的内存结构。 And then you technically could increment a pointer to move through it... if you're mad.
然后你在技术上可以增加一个指针来移动它...如果你生气了。 That would be a very poor way to do and not at all guaranteed to work on different compilers or even different days of the week.
这将是一种非常糟糕的方式,并且完全不能保证在不同的编译器上工作,甚至在一周中的不同日期工作。 No what you want is a data structure to do the job for you;
没有您想要的是一种数据结构来为您完成这项工作。 they come with a 100% cash-back guarantee!
他们提供100%现金返还保证!
The most basic structure to do this with is a fixed size array, eg: 最基本的结构是固定大小的数组,例如:
struct team
{
int players[99]; //an array
int manager;
int coach;
string teamName;
//etc etc
}
Then to access your players 然后访问你的球员
team myTeam;
for(int i(0); i < 99; ++i)
{
myTeam.players[i]; //do whatever
}
The limitation of an array is that you cannot change its size once it's created. 数组的限制是创建数组后就无法更改其大小。 So if you try
所以,如果你尝试
myTeam.players[99]; //accessing invalid memory - the array values are 0 - 98
More advanced 更先进
If you need a data structure that can change size after it's created, eg you might want to add a few more players to your team at some point in the future. 如果您需要一个可以在创建后更改大小的数据结构,例如,您可能希望在将来的某个时刻向您的团队添加更多玩家。 Then you can use a dynamic data structure such as the std::vector or the std::deque or std::list
然后,您可以使用动态数据结构,例如std :: vector或std :: deque或std :: list
You can define pointer to member, like pointer to member function: 您可以定义指向成员的指针,例如指向成员函数的指针:
typedef int team::*member_t;
You can have array of pointers to all your members: 您可以拥有指向所有成员的指针数组:
static member_t member[size];
With this approach defining member function to iterate over all members is easy: 通过这种方法,定义成员函数以遍历所有成员很容易:
template <class F>
void for_each(F f)
{
for (int i = 0; i < size; ++i)
f(this->*member[i]);
}
And with using of preprocessor macro - you can have in one place definition of all members, in other definition of pointer to members - so you will not make any mistake with changing their order. 并且使用预处理器宏 - 您可以在一个地方定义所有成员,在其他指向成员的指针中 - 这样您就不会在更改其顺序时出错。 See full code:
查看完整代码:
struct team {
#define TEAM_MEMBERS(prefix,suffix) \
prefix player1 suffix, \
prefix player2 suffix, \
prefix player3 suffix
int TEAM_MEMBERS(,);
static const int size = 3;
typedef int team::*member_t;
static member_t member[size];
template <class F>
void for_each(F f)
{
for (int i = 0; i < size; ++i)
f(this->*member[i]);
}
};
team::member_t team::member[team::size] = {
TEAM_MEMBERS(&team::,)
};
And some test: 还有一些测试:
#include <iostream>
int main() {
team t = { 0 };
t.for_each([](int m) { std::cout << m << "\n"; }); // prints 0,0,0
int n = 0;
t.for_each([&n](int& m) { m = n++; });
t.for_each([](int m) { std::cout << m << "\n"; }); // prints 0,1,2
t.player2 = 7;
t.for_each([](int m) { std::cout << m << "\n"; }); // prints 0,7,2
}
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