C++ 如何将向量转换为数组
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How to convert vector to array
提问by ganuke
How do I convert a std::vector<double>to a double array[]?
如何将 a 转换std::vector<double>为 a double array[]？
回答by Michael Mrozek
There's a fairly simple trick to do so, since the spec now guaranteesvectors store their elements contiguously:
有一个相当简单的技巧可以做到这一点，因为规范现在保证向量连续存储它们的元素：
std::vector<double> v;
double* a = &v[0];
回答by GManNickG
What for? You need to clarify: Do you need a pointer to the first element of an array, or an array?
做什么的？您需要澄清：您需要一个指向数组或数组第一个元素的指针吗？
If you're calling an API function that expects the former, you can do do_something(&v[0], v.size()), where vis a vector of doubles. The elements of a vector are contiguous.
如果您正在调用需要前者的 API 函数，您可以这样做do_something(&v[0], v.size())，其中v是doubles的向量。向量的元素是连续的。
Otherwise, you just have to copy each element:
否则，您只需复制每个元素：
double arr[100];
std::copy(v.begin(), v.end(), arr);
Ensure not only thar arris big enough, but that arrgets filled up, or you have uninitialized values.
确保不仅 thararr足够大，而且会arr被填满，或者您有未初始化的值。
回答by gzbwb
For C++11, vector.data()will do the trick.
对于C++11，vector.data()可以解决问题。
回答by user168715
vector<double> thevector;
//...
double *thearray = &thevector[0];
This is guaranteed to work by the standard, however there are some caveats: in particular take care to only use thearraywhile thevectoris in scope.
这是保证该标准的工作，但也有一些注意事项：特别注意只使用thearray，而thevector在范围内。
回答by user168715
Vectors effectively are arrays under the skin. If you have a function:
向量实际上是皮肤下的数组。如果你有一个功能：
void f( double a[]);
you can call it like this:
你可以这样称呼它：
vector <double> v;
v.push_back( 1.23 )
f( &v[0] );
You should not ever need to convert a vector into an actual array instance.
您永远不需要将向量转换为实际的数组实例。
回答by Jayhello
As to std::vector<int> vec, vec to get int*, you can use two method:
至于std::vector<int> vec, vec 要 get int*，您可以使用两种方法：
int* arr = &vec[0];
int* arr = vec.data();
int* arr = &vec[0];
int* arr = vec.data();
If you want to convert any type Tvector to T* array, just replace the above intto T.
如果要将任何类型的T向量转换为T* array，只需将上面的替换int为T。
I will show you why does the above two works, for good understanding?
我将向您展示为什么上述两个有效，以便更好地理解？
std::vectoris a dynamic array essentially.
std::vector本质上是一个动态数组。
Main data member as below:
主要数据成员如下：
template <class T, class Alloc = allocator<T>>
class vector{
public:
typedef T value_type;
typedef T* iterator;
typedef T* pointer;
//.......
private:
pointer start_;
pointer finish_;
pointer end_of_storage_;
public:
vector():start_(0), finish_(0), end_of_storage_(0){}
//......
}
The range (start_, end_of_storage_)is all the array memory the vector allocate;
的range (start_, end_of_storage_)是所有阵列存储器中的向量分配;
The range(start_, finish_)is all the array memory the vector used;
的range(start_, finish_)是所有的阵列中的存储器使用的载体;
The range(finish_, end_of_storage_)is the backup array memory.
的range(finish_, end_of_storage_)是备份阵列存储器。
For example, as to a vector vec. which has {9, 9, 1, 2, 3, 4} is pointer may like the below.
例如，对于向量 vec。其中有 {9, 9, 1, 2, 3, 4} 的指针可能如下所示。
So &vec[0]= start_ (address.) (start_ is equivalent to int* array head)
So &vec[0]= start_ (address.) (start_ 相当于 int* 数组头)
In c++11the data()member function just return start_
在c++11该data()成员函数只返回start_
pointer data()
{
return start_; //(equivalent to `value_type*`, array head)
}
回答by rashedcs
We can do this using data() method. C++11 provides this method.
我们可以使用 data() 方法来做到这一点。C++11 提供了这种方法。
Code Snippet
代码片段
#include<bits/stdc++.h>
using namespace std;
int main()
{
ios::sync_with_stdio(false);
vector<int>v = {7, 8, 9, 10, 11};
int *arr = v.data();
for(int i=0; i<v.size(); i++)
{
cout<<arr[i]<<" ";
}
return 0;
}
回答by Toby
std::vector<double> vec;
double* arr = vec.data();
回答by Toby
If you have a function, then you probably need this:foo(&array[0], array.size());. If you managed to get into a situation where you need an array then you need to refactor, vectors are basically extended arrays, you should always use them.
如果你有一个功能，那么你可能需要这样的：foo(&array[0], array.size());。如果您设法进入需要数组的情况，那么您需要重构，向量基本上是扩展数组，您应该始终使用它们。
回答by Sakthi Vignesh
You can do some what like this
你可以做一些这样的
vector <int> id;
vector <double> v;
if(id.size() > 0)
{
for(int i = 0; i < id.size(); i++)
{
for(int j = 0; j < id.size(); j++)
{
double x = v[i][j];
cout << x << endl;
}
}
}
