如何在 C/C++ 中获取多维数组的列？
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How to get column of a multidimensional array in C/C++?
提问by Qbik
int matrix[9][9],*p;
p=matrix[0];
this works and gives first row of matrix, but how to get first column of matrixI've tried p=matrix[][0];? Also I don't understand why below code gets compiler error ?
这有效并给出了第一行matrix，但是如何获得matrix我尝试过的第一列p=matrix[][0];？另外我不明白为什么下面的代码会出现编译器错误？
int matrix[9][9],p[9]; // it looks really ugly, byt why it doesn't work ?
p=matrix[0]; // compiler gives "invalid array assigment"
is it because multidimensional arrays are arrays of arrays - and we should interpret matrix[i][j]as j-th element of i-th nested array ?
是不是因为多维数组是数组的数组 - 我们应该解释matrix[i][j]为第 i 个嵌套数组的第 j 个元素？
回答by meyumer
In C/C++, multidimensional arrays are actually stored as one dimensional arrays (in the memory). Your 2D matrix is stored as a one dimensional array with row-first ordering. That is why getting a column out of it is not easy, and not provided by default. There is no contiguous array in the memory that you can get a pointer to which represents a column of a multidimensional array.See below:
在 C/C++ 中，多维数组实际上存储为一维数组（在内存中）。您的二维矩阵存储为行优先排序的一维数组。这就是为什么从中取出一列并不容易，并且默认情况下不提供。内存中没有连续的数组，您可以获得表示多维数组列的指针。见下文：
When you do p=matrix[0], you are just getting the pointer to the first element matrix[0][0], and that makes you think that you got the pointer to first row. Actually, it is a pointer to the entire contiguous array that holds matrix, as follows:
当你这样做时p=matrix[0]，你只是得到了指向第一个元素的指针matrix[0][0]，这让你认为你得到了指向第一行的指针。实际上，它是指向包含 的整个连续数组的指针matrix，如下所示：
matrix[0][0]
matrix[0][1]
matrix[0][2]
.
.
matrix[1][0]
matrix[1][1]
matrix[1][2]
.
.
matrix[8][0]
matrix[8][1]
matrix[8][2]
.
.
matrix[8][8]
As seen above, the elements of any given column are separated by other elements in the corresponding rows.
如上所示，任何给定列的元素都被相应行中的其他元素分隔。
So, as a side note, with pointer p, you can walk through the entire 81 elements of your matrix if you wanted to.
因此，作为旁注，使用 pointer p，如果您愿意，您可以遍历矩阵的全部 81 个元素。
回答by meyumer
You can get the first column using a loop like
您可以使用类似的循环获取第一列
for(int i = 0; i < 9; i++)
{
printf("Element %d: %d", i, matrix[i][0]);
}
I think the assignment doesn't work properly because you're trying to assign something's that's not an address to a pointer.
我认为分配不能正常工作，因为您试图将不是地址的东西分配给指针。
(Sorry this is c code)
（对不起，这是c代码）
回答by AxelOmega
There is no difference between specifying matrix[81]or matrix[9][9]
指定matrix[81]或matrix[9][9]
matrix[r][c]simply means the same as matrix[9*r+c]
matrix[r][c]只是意味着相同 matrix[9*r+c]
There are other containers better suited fort multidimensional arrays like boost::multi_array
还有其他容器更适合多维数组，例如 boost::multi_array
http://www.boost.org/doc/libs/1_53_0/libs/multi_array/doc/index.html
http://www.boost.org/doc/libs/1_53_0/libs/multi_array/doc/index.html
Think of the bare array just like allocating a contiguous piece of memory. You, the programmer then has to handle this piece of memory yourself. The bare name of the array, e.g. matrixis a pointer to the first element of this allocated piece of memory. Then *(matrix+1)is the same as matrix[0][1]or matrix[1].
把裸数组想象成分配一块连续的内存。你，程序员然后必须自己处理这块内存。数组的裸名，例如，matrix是指向该分配内存的第一个元素的指针。then*(matrix+1)与matrix[0][1]or相同matrix[1]。
回答by zzk
p is an array of int, matrix[0] is a pointer..
p 是一个 int 数组，matrix[0] 是一个指针..
回答by Neil
matrixitself is the nearest thing you can get to a column of the array, inasmuch as (matrix + 1)[0][0]is the same as matrix[1][0].
matrix本身是最接近数组列的对象，因为(matrix + 1)[0][0]与matrix[1][0].
回答by Thomas Matthews
If you want your matrix to contiguous locations, declare it as a one dimensional array and perform the row and column calculations yourself:
如果您希望矩阵位于连续位置，请将其声明为一维数组并自行执行行和列计算：
int contiguous_matrix[81];
int& location(int row, int column)
{
return contiguous_matrix[row * 9 + column];
}
You can also iterate over each column of a row:
您还可以遍历一行的每一列：
typedef void (*Function_Pointer)(int&);
void Column_Iteration(Function_Pointer p_func, int row)
{
row = row * MAXIMUM_COLUMNS;
for (unsigned int column = 0; column < 9; ++column)
{
p_func(contiguous_matrix[row + column]);
}
}
回答by Georgi
For static declared arrays you can access them like continuous 1D array, p = matrix[0]will give you the 1st column of 1st row. Then the 1D array can be accessed like p[i], *(p+i), or p[current_raw * raw_size + current_column).
对于静态声明的数组，您可以像访问连续一维数组一样访问它们，p = matrix[0]将为您提供第一行的第一列。然后1D阵列可以像访问p[i]，*(p+i)或p[current_raw * raw_size + current_column)。
The things are getting tricky if a 2D array is represented with **pas it will be interpreted as an array of pointers to 1D arrays.
如果用 2D 数组表示，事情会变得棘手，**p因为它将被解释为指向 1D 数组的指针数组。
