如何将作为参数传递的长数组复制到std :: list [英] How to copy a long array passed as argument to a std::list

问题描述

Hello Everyone，



我正在尝试将一个long数组复制到std :: list或std：vector。



我有以下代码示例：

Hello Everyone,

I am trying to copy an array of long into a std::list or std:vector.

I have the following code sample:

#include "stdafx.h"
#include <iostream>
#include <list>
#include <iterator>

using namespace std;

class Channels
{
public:
void Configure(long channels[])
{
_channels <= channels; //..... copy channels to _channels
}

private:
   std::initializer_list<long> _channels{};
}

如果我走在正确的轨道上，请指点我。 br />


非常感谢您提前。

祝你好运。

MiQi



我尝试了什么：



我尝试过以下但不确定这是否正确：

Could you please point me if I am on the right track.

Thank you very much in advance.
Best regards.
MiQi

What I have tried:

I have tried the following but not sure if this is the correct way:

#include "stdafx.h"
#include <iostream>
#include <list>
#include <iterator>

using namespace std;

class Channels
{
public:
void Configure(std::initializer_list<long> chans, int size)
{
   _channels = std::initializer_list<long>{ chans };
}

private:
   std::initializer_list<long> _channels{};
}

void main()
{
   long channels[] {1, 3 , 5};
   Channels channel;
   Channel.Configure(channels);
}

示例代码编译愉快。

The sample code compiles happily.

推荐答案

一个简单的旧学校解决方案。

1.将数组的大小与数组一起传递;否则你不知道它的大小。

2.只需将它分配给一个成员变量。



这应该可以解决问题：

How about a simple old school solution.
1. Pass the size of the array along with the array; otherwise you do not know its size.
2. Just assign it to a member variable.

This should do the trick:

class Channels
{
public:
void Configure(long channels[], size_t size)
{
    _channels.assign(channels, channels+size);
}

private:
   std::vector _channels;
}

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