C++ 将数组推入向量
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/4541556/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Pushing an array into a vector
提问by 0x0
I've a 2d array, say A[2][3]={{1,2,3},{4,5,6}};and I want to push it into a 2D vector(vector of vectors). I know you can use two for loopsto push the elements one by on on to the first vector and then push that into the another vector which makes it 2d vector but I was wondering if there is any way in C++ to do this in a single loop. For example I want to do something like this:
我有一个二维数组，比如说A[2][3]={{1,2,3},{4,5,6}};我想把它推入一个二维向量（向量的向量）。我知道您可以使用两个for loops将元素一个接一个地推到第一个向量，然后将其推入另一个向量，使其成为二维向量，但我想知道 C++ 中是否有任何方法可以在单个循环中执行此操作。例如，我想做这样的事情：
myvector.pushback(A[1]+3); // where 3 is the size or number of columns in the array.
I understand this is not a correct code but I put this just for understanding purpose. Thanks
我明白这不是一个正确的代码，但我只是为了理解目的。谢谢
采纳答案by langerra.com
The new C++0x standard defines initializer_listswhich allows you to:
新的 C++0x 标准定义initializer_lists允许您：
vector<vector<int>> myvector = {{1,2,3},{4,5,6}};
gcc 4.3+ and some other compilers have partial C++0x support.
for gcc 4.3+ you could enable c++0x support by adding the flag -std=c++0x
gcc 4.3+ 和其他一些编译器有部分 C++0x 支持。对于 gcc 4.3+，您可以通过添加标志来启用 c++0x 支持-std=c++0x
Its not the best way to have your static data represented like that. However, if your compiler vendor supports C++ tr1 then you could do:
这不是让您的静态数据像这样表示的最佳方式。但是，如果您的编译器供应商支持 C++ tr1，那么您可以执行以下操作：
#include <tr1/array> // or #include <array>
...
typedef vector<vector<int> > vector2d;
vector2d myvector;
// initialize the vectors
myvector.push_back(vector<int>());
myvector.push_back(vector<int>());
typedef std::array<std::array<int, 3>, 2> array2d;
array2d array = {{1,2,3},{4,5,6}};
array2d::const_iterator ai = array.begin(), ae = array.end();
for (vector2d::iterator i = myvector.begin(), e = myvector.end()
; i != e && ai != ae
; i++, a++)
{
// reserve vector space
i->reserve(array.size());
// copy array content to vector
std::copy(ai.begin(), ai->end(), i->begin());
}
回答by crazylammer
You can use vector::assign(pointers to array elements are valid iterators):
您可以使用vector::assign（指向数组元素的指针是有效的迭代器）：
int a[2][3] = {{1, 2, 3}, {4, 5, 6}};
std::vector<std::vector<int> > v(2);
for (size_t i = 0; i < 2; ++i)
v[i].assign(a[i], a[i] + 3);
回答by Simone
This is a little tricky, but you could use template recursion to help you in having the assignment done almost completely at compile-time. I understand that's not exactly what you are looking for, but I think it's worthwhile :-)
这有点棘手，但您可以使用模板递归来帮助您在编译时几乎完全完成分配。我知道这不完全是你要找的，但我认为这是值得的:-)
Here's the code:
这是代码：
#include <vector>
using namespace std;
typedef vector<vector<int> > vector2d;
template<size_t K, size_t M, size_t N>
struct v_copy {
static void copy(vector2d& v, int(&a)[M][N])
{
v[K - 1].assign(a[K - 1], a[K - 1] + N);
v_copy<K - 1, M, N>::copy(v, a);
}
};
template<size_t M, size_t N>
struct v_copy<1, M, N> {
static void copy(vector2d& v, int(&a)[M][N])
{
v[0].assign(a[0], a[0] + N);
}
};
template<size_t M, size_t N>
void copy_2d(vector2d& v, int(&a)[M][N])
{
v_copy<M, M, N>::copy(v, a);
}
int main()
{
int A[2][3] = {{0, 1, 2}, {10, 11, 12}};
vector2d vector(2);
copy_2d(vector, A);
}
it needed a struct because in C++ you can't do partial specialization of functions. BTW , compiling it with gcc version 4.5.0, this code produces the same assembly as
它需要一个结构体，因为在 C++ 中你不能对函数进行部分特化。顺便说一句，用 gcc 4.5.0 版编译它，这段代码产生与
vector[1].assign(A[1], A[1] + 3);
vector[0].assign(A[0], A[0] + 3);
It should not be very hard to have it compile with different types of 2-dimensions arrays.
用不同类型的二维数组编译它应该不是很难。
回答by Vladimir
If you want to push the data into vector of vectors, you have to write something like this:
如果要将数据推入向量向量中，则必须编写如下内容：
vector<int> inner;
vector< vector<int> >outer;
...
outer.pushback(inner);
I think there is no way to do it in a single loop.
我认为没有办法在一个循环中做到这一点。
If you want to use just one vector (something similar like you written), then you can do it in a single loop:
如果你只想使用一个向量（类似于你写的东西），那么你可以在一个循环中完成：
int A[2][3]={{1,2,3},{4,5,6}};
int* p = A[0];
std::vector<int> inner;
std::vector< std::vector<int> >outer;
for(int i = 0; i < 6; ++i)
{
inner.push_back(*p++);
}
回答by Nate Kohl
It's kind of cheating, but you could take advantage of the vector constructorto do one of the loops for you:
这是一种作弊，但您可以利用向量构造函数为您执行以下循环之一：
#include <vector>
int main() {
const int XMAX = 2, YMAX = 3;
int A[XMAX][YMAX] = {{1,2,3}, {4,5,6}};
std::vector<std::vector<int> > v;
for (size_t x = 0; x < XMAX; ++x) {
v.push_back(std::vector<int>(&A[x][0], &A[x][YMAX]));
}
}
回答by Virne
You can resize vectors and then use copy.
您可以调整矢量大小，然后使用复制。
int A[2][3]={{1,2,3},{4,5,6}};
std::vector< std::vector<int> > vec;
vec.resize(2);
for (int i=0; i<2; i++)
{
vec[i].resize(3);
std::copy(A[i], A[i]+3, vec[i].begin());
}
Is it practical? Definetly not.
实用吗？绝对不是。
回答by Ray Hidayat
Hm... I can produce a partial answer but not a full one.
嗯……我可以给出部分答案，但不能给出完整答案。
int elementCount = 6; // I wonder if this can be done somehow with sizeof(A) * sizeof(A[0])
int* end = A + elementCount;
for(int* current = A; current < end; ++current) {
myvector.pushback(*current);
}
回答by Edward Strange
No. The only thing you can do is leverage existing loop functions so that you only have to write one or zero of your own loops.
不可以。您唯一能做的就是利用现有的循环函数，这样您只需编写一个或零个自己的循环。
