将向量作为数组返回的正确方法是什么

问题描述：

我在这里有两个问题......我正在编写一个函数，它接收一个向量，并且它假设以数组形式返回...假设arrs为'0 1 2'.....

1.这是正确的方法吗？

2.然后现在我想加入returnArr函数中的一个额外元素（例如arrs.push_back（55）），使用我写的代码，对于这个'添加'，我把它放在for循环之前或之后？我问这个，因为我似乎得到了不同的结果

请原谅我使用的简单语言，因为我还是C ++的初学者

我尝试了什么：

I have 2 questions here... I am writing out a function where it takes in a vector and it is suppose to return as array... Suppose arrs are '0 1 2'.....

1. Is this the correct way to do so?

2. Then now I had wanted to add in an additional element ( eg. arrs.push_back(55) ) within the returnArr function, using the code I have wrote, for this 'adding', do I put it before or after the for-loop? I asked this, because I seems to get different result

Pardon the simple language I used as I am still a beginner in C++

What I have tried:

int * returnArr ( vector<int> arrs )
{        
    cout << "[ " ;
    
    for (int i = arrs.size() - 1 ; i >= 0 ; i--)
    {
        cout << i << " ";
    } 

    cout << "]" << endl;
    
    // Results : [ 2 1 0 ]

}

答

请尝试以下方法：

Try the following:

int* returnArr (vector<int>* arrs)
{
    int i;
    // create an array one larger than the number of elements in the vector
    int* pArray = new int[arrs->size() + 1];
    cout << "[ " ;
    for (i = 0; i << (int)arrs->size(); ++i)
    {
        pArray[i] = arrs->at(i);  // copy the vector value to the array
        cout << pArray[i] << ", ";
    }
    pArray[i] = -1;		// put a termination marker at the end of the array
    cout << " ]" << endl;
    
    return pArray; // return the array to the caller
}

试图解决愚蠢的降价处理器搞砸了什么

[/ edit]

[edit]
Trying to fix what the stupid markdown processor has messed up
[/edit]

正确的方法是返回向量并使用其数据 [ ^ ]成员当你需要访问原始指针时。

The correct way is returning the vector and using its data[^] member when you need to access the raw pointer.

有几件事我想发表评论。

1.做不传递vector参数作为传递值。

正确的方法 - （const vector< int>& arrs）

2.你可以就地反转向量，在这种情况下你需要删除第1点的const。

std ::反转（arrs.begin（），arr.end（））;

3.你也可以通过反转将值复制到另一个向量。

这样你就可以简单地返回新创建的向量。

There are several things on which I wish to comment.

1. Do not pass the vector parameter as pass by value.
Right way - (const vector<int>& arrs)

2. You could reverse the vector in-place in which case you need to remove the const in point 1.
std::reverse(arrs.begin(), arr.end());

3. You also could copy values to another vector by reversing it.
This way you can simply return the newly created vector.

vector<int> returnArr(const vector<int>& arrs)
{
    vector<int> arrs2(arrs.size());
    std::copy(arrs.rbegin(), arrs.rend(), arrs2.begin());
    return arrs2;
}

4.这是以相反顺序显示矢量内容的另一种方式。

std :: copy（arrs.rbegin（），arrs.rend（），ostream_iterator< int>（cout，））;

将向量作为数组返回的正确方法是什么

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