C语言 如何在C中的函数中传递二维数组（矩阵）？
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How to pass 2D array (matrix) in a function in C?
提问by Shweta
I need to do this to persist operations on the matrix as well. Does that mean that it needs to be passed by reference?
我也需要这样做以保持对矩阵的操作。这是否意味着它需要通过引用传递？
Will this suffice?
这足够了吗？
void operate_on_matrix(char matrix[][20]);
void operate_on_matrix(char matrix[][20]);
回答by Bart van Ingen Schenau
C does not really have multi-dimensional arrays, but there are several ways to simulate them. The way to pass such arrays to a function depends on the way used to simulate the multiple dimensions:
C 并没有真正的多维数组，但是有几种方法可以模拟它们。将此类数组传递给函数的方式取决于用于模拟多维的方式：
1) Use an array of arrays. This can only be used if your array bounds are fully determined at compile time, or if your compiler supports VLA's:
1) 使用数组数组。仅当您的数组边界在编译时完全确定，或者您的编译器支持VLA时，才可以使用：
#define ROWS 4
#define COLS 5
void func(int array[ROWS][COLS])
{
int i, j;
for (i=0; i<ROWS; i++)
{
for (j=0; j<COLS; j++)
{
array[i][j] = i*j;
}
}
}
void func_vla(int rows, int cols, int array[rows][cols])
{
int i, j;
for (i=0; i<rows; i++)
{
for (j=0; j<cols; j++)
{
array[i][j] = i*j;
}
}
}
int main()
{
int x[ROWS][COLS];
func(x);
func_vla(ROWS, COLS, x);
}
2) Use a (dynamically allocated) array of pointers to (dynamically allocated) arrays. This is used mostly when the array bounds are not known until runtime.
2）使用（动态分配的）指向（动态分配的）数组的指针数组。这主要用于直到运行时才知道数组边界的情况。
void func(int** array, int rows, int cols)
{
int i, j;
for (i=0; i<rows; i++)
{
for (j=0; j<cols; j++)
{
array[i][j] = i*j;
}
}
}
int main()
{
int rows, cols, i;
int **x;
/* obtain values for rows & cols */
/* allocate the array */
x = malloc(rows * sizeof *x);
for (i=0; i<rows; i++)
{
x[i] = malloc(cols * sizeof *x[i]);
}
/* use the array */
func(x, rows, cols);
/* deallocate the array */
for (i=0; i<rows; i++)
{
free(x[i]);
}
free(x);
}
3) Use a 1-dimensional array and fixup the indices. This can be used with both statically allocated (fixed-size) and dynamically allocated arrays:
3) 使用一维数组并修正索引。这可以用于静态分配（固定大小）和动态分配的数组：
void func(int* array, int rows, int cols)
{
int i, j;
for (i=0; i<rows; i++)
{
for (j=0; j<cols; j++)
{
array[i*cols+j]=i*j;
}
}
}
int main()
{
int rows, cols;
int *x;
/* obtain values for rows & cols */
/* allocate the array */
x = malloc(rows * cols * sizeof *x);
/* use the array */
func(x, rows, cols);
/* deallocate the array */
free(x);
}
4) Use a dynamically allocated VLA. One advantage of this over option 2 is that there is a single memory allocation; another is that less memory is needed because the array of pointers is not required.
4) 使用动态分配的 VLA。与选项 2 相比，它的一个优点是只有一个内存分配；另一个是需要更少的内存，因为不需要指针数组。
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
extern void func_vla(int rows, int cols, int array[rows][cols]);
extern void get_rows_cols(int *rows, int *cols);
extern void dump_array(const char *tag, int rows, int cols, int array[rows][cols]);
void func_vla(int rows, int cols, int array[rows][cols])
{
for (int i = 0; i < rows; i++)
{
for (int j = 0; j < cols; j++)
{
array[i][j] = (i + 1) * (j + 1);
}
}
}
int main(void)
{
int rows, cols;
get_rows_cols(&rows, &cols);
int (*array)[cols] = malloc(rows * cols * sizeof(array[0][0]));
/* error check omitted */
func_vla(rows, cols, array);
dump_array("After initialization", rows, cols, array);
free(array);
return 0;
}
void dump_array(const char *tag, int rows, int cols, int array[rows][cols])
{
printf("%s (%dx%d):\n", tag, rows, cols);
for (int i = 0; i < rows; i++)
{
for (int j = 0; j < cols; j++)
printf("%4d", array[i][j]);
putchar('\n');
}
}
void get_rows_cols(int *rows, int *cols)
{
srand(time(0)); // Only acceptable because it is called once
*rows = 5 + rand() % 10;
*cols = 3 + rand() % 12;
}
回答by casablanca
I don't know what you mean by "data dont get lost". Here's how you pass a normal 2D array to a function:
我不知道您所说的“数据不会丢失”是什么意思。以下是将普通二维数组传递给函数的方法：
void myfunc(int arr[M][N]) { // M is optional, but N is required
..
}
int main() {
int somearr[M][N];
...
myfunc(somearr);
...
}
回答by Minhas Kamal
Easiest Way: Passing A Variable-Length 2D Array
最简单的方法：传递可变长度的二维数组
Most clean technique for both C & C++ is: to pass 2D array like a 1D array, then use as 2D inside the function.
C 和 C++ 最干净的技术是：像一维数组一样传递二维数组，然后在函数内部用作二维数组。
void func(int row, int col, int* matrix){
int i, j;
for(i=0; i<row; i++){
for(j=0; j<col; j++){
printf("%d ", *(matrix + i*col + j)); // or better: printf("%d ", *matrix++);
}
printf("\n");
}
}
int main(){
int matrix[2][3] = { {1, 2, 3}, {7, 8, 9} };
func(2, 3, matrix[0]);
return 0;
}
回答by shinxg
2D array:
二维数组：
int sum(int array[][COLS], int rows)
{
}
3D array:
3D阵列：
int sum(int array[][B][C], int A)
{
}
4D array:
4D阵列：
int sum(int array[][B][C][D], int A)
{
}
and nD array:
和 nD 数组：
int sum(int ar[][B][C][D][E][F].....[N], int A)
{
}
回答by Casper Ghost
#include <iostream>
using namespace std;
void printarray(int *a, int c,int r)
{
for (int i = 0; i < r; i++)
{
for (int j = 0; j < c; j++)
{
cout << "\t" << *(a + i*c + j) << "\t"; // a is a pointer refer to a 2D array
}
cout << endl << "\n\n";
}
}
int main()
{
int array[4][4] =
{{1 ,2 ,3 ,4 },
{12,13,14,5 },
{11,16,15,6 },
{10,9 ,8 ,7 }};
printarray((int*)array,4,4);
// here I use print function but u can use any other useful function like
//setArray((int *) array,4,4);
return 0;
}
