Below is my code, where I'm passing my array named a to a display function which will just display the array.
I'm getting an invalid type argument of type unary * error on this statement printf("Arr[%d][%d] is %d\n",i,j,*(*(q+i)+j));
However, the commented code prints the values properly. Can you please explain what wrong am I doing? I'm new to C so forgive me if it's a silly mistake/error.
#include <stdio.h>
void display(int *, int, int);
int main()
{
int a[3][4] = {
{1,2,3,4},
{5,6,7,8},
{9,10,11,12}
};
display(a,3,4);
/*
int i,j;
for(i=0; i<3; i++)
{
for(j=0; j<4; j++)
printf("Arr[%d][%d] is %d\n",i,j,*(*(a+i)+j));
}
*/
return 0;
}
void display(int *q, int row, int col)
{
int i,j;
for(i=0; i<row; i++)
{
for(j=0; j<col; j++)
printf("Arr[%d][%d] is %d\n",i,j,*(*(q+i)+j));
}
}
Another approach to solve this would be to explicitly pass the array's address and de-reference it (display()) the desired result however can also be achieved without the additional level of indirection (display1()):
#include <stdio.h>
void display(const size_t row, const size_t col, int (*pa)[row][col]);
void display1(const size_t row, const size_t col, int a[row][col]);
int main(void)
{
int a[3][4] = {
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 }
};
display(3, 4, &a);
display1(3, 4, a);
return 0;
}
void display(const size_t rows, const size_t cols, const int (*pa)[row][col]);
{
for (size_t i = 0; i < rows; ++i)
{
for (size_t j = 0; j < cols; ++j)
{
printf("Arr[%zu][%zu] is %d\n", i, j, (*pa)[i][j]);
}
}
}
void display1(const size_t rows, const size_t cols, const int a[row][col]);
{
for (size_t i = 0; i < rows; ++i)
{
for (size_t j = 0; j < cols; ++j)
{
printf("Arr[%zu][%zu] is %d\n", i, j, a[i][j]);
}
}
}
Update: This needs at least a C99 compatible compiler.
In modern C you should pass the sizes of the dimensions first
void display(size_t row, size_t col, int (*q)[col]);
and you then can use q just as normal as q[i][j] with all index computations done for you.
void display(size_t row, size_t col, int (*q)[col]) {
for(size_t i=0; i<row; i++) {
for(size_t j=0; j<col; j++)
printf("Arr[%zu][%zu] is %d\n", i, j, q[i][j]);
}
}
This needs at leas C99 to work.
In general, to pass 2D arrays to functions you need to specify the array dimensions for all but the first dimension, so in this example you need the [4] for the last dimension but you can optionally omit it for the first dimension, i.e.
void display( int (*q)[4], int row, int col)
And make above declaration as follows
void display(int (*)[], int, int);
Your only problem was with the way you were calling display. You needed to drop the indexes when passing the array and fix the initial definition for display to match:
#include <stdio.h>
void display(int q[][4], int, int);
int main()
{
int a[3][4] = {
{1,2,3,4},
{5,6,7,8},
{9,10,11,12}
};
display (a, 3, 4);
return 0;
}
void display(int q[][4], int row, int col)
{
int i,j;
for(i=0; i<row; i++)
{
for(j=0; j<col; j++)
printf("Arr[%d][%d] is %d\n",i,j,*(*(q+i)+j));
}
}
Output:
Arr[0][0] is 1
Arr[0][1] is 2
Arr[0][2] is 3
Arr[0][3] is 4
Arr[1][0] is 5
Arr[1][1] is 6
Arr[1][2] is 7
Arr[1][3] is 8
Arr[2][0] is 9
Arr[2][1] is 10
Arr[2][2] is 11
Arr[2][3] is 12
Related
I need to do this to persist operations on the matrix as well. Does that mean that it needs to be passed by reference?
Will this suffice?
void operate_on_matrix(char matrix[][20]);
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:
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:
#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.
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:
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.
#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;
}
(See srand() — why call it only once?.)
Easiest Way in Passing A Variable-Length 2D Array
Most clean technique for both C & C++ is: pass 2D array like a 1D array, then use as 2D inside the function.
#include <stdio.h>
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] = { {0, 1, 2}, {3, 4, 5} };
func(2, 3, matrix[0]);
return 0;
}
Internally, no matter how many dimensions an array has, C/C++ always maintains a 1D array. And so, we can pass any multi-dimensional array like this.
I don't know what you mean by "data dont get lost". Here's how you pass a normal 2D array to a function:
void myfunc(int arr[M][N]) { // M is optional, but N is required
..
}
int main() {
int somearr[M][N];
...
myfunc(somearr);
...
}
2D array:
int sum(int array[][COLS], int rows)
{
}
3D array:
int sum(int array[][B][C], int A)
{
}
4D array:
int sum(int array[][B][C][D], int A)
{
}
and nD array:
int sum(int ar[][B][C][D][E][F].....[N], int A)
{
}
I need to do this to persist operations on the matrix as well. Does that mean that it needs to be passed by reference?
Will this suffice?
void operate_on_matrix(char matrix[][20]);
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:
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:
#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.
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:
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.
#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;
}
(See srand() — why call it only once?.)
Easiest Way in Passing A Variable-Length 2D Array
Most clean technique for both C & C++ is: pass 2D array like a 1D array, then use as 2D inside the function.
#include <stdio.h>
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] = { {0, 1, 2}, {3, 4, 5} };
func(2, 3, matrix[0]);
return 0;
}
Internally, no matter how many dimensions an array has, C/C++ always maintains a 1D array. And so, we can pass any multi-dimensional array like this.
I don't know what you mean by "data dont get lost". Here's how you pass a normal 2D array to a function:
void myfunc(int arr[M][N]) { // M is optional, but N is required
..
}
int main() {
int somearr[M][N];
...
myfunc(somearr);
...
}
2D array:
int sum(int array[][COLS], int rows)
{
}
3D array:
int sum(int array[][B][C], int A)
{
}
4D array:
int sum(int array[][B][C][D], int A)
{
}
and nD array:
int sum(int ar[][B][C][D][E][F].....[N], int A)
{
}
I need to do this to persist operations on the matrix as well. Does that mean that it needs to be passed by reference?
Will this suffice?
void operate_on_matrix(char matrix[][20]);
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:
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:
#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.
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:
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.
#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;
}
(See srand() — why call it only once?.)
Easiest Way in Passing A Variable-Length 2D Array
Most clean technique for both C & C++ is: pass 2D array like a 1D array, then use as 2D inside the function.
#include <stdio.h>
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] = { {0, 1, 2}, {3, 4, 5} };
func(2, 3, matrix[0]);
return 0;
}
Internally, no matter how many dimensions an array has, C/C++ always maintains a 1D array. And so, we can pass any multi-dimensional array like this.
I don't know what you mean by "data dont get lost". Here's how you pass a normal 2D array to a function:
void myfunc(int arr[M][N]) { // M is optional, but N is required
..
}
int main() {
int somearr[M][N];
...
myfunc(somearr);
...
}
2D array:
int sum(int array[][COLS], int rows)
{
}
3D array:
int sum(int array[][B][C], int A)
{
}
4D array:
int sum(int array[][B][C][D], int A)
{
}
and nD array:
int sum(int ar[][B][C][D][E][F].....[N], int A)
{
}
I need to do this to persist operations on the matrix as well. Does that mean that it needs to be passed by reference?
Will this suffice?
void operate_on_matrix(char matrix[][20]);
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:
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:
#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.
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:
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.
#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;
}
(See srand() — why call it only once?.)
Easiest Way in Passing A Variable-Length 2D Array
Most clean technique for both C & C++ is: pass 2D array like a 1D array, then use as 2D inside the function.
#include <stdio.h>
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] = { {0, 1, 2}, {3, 4, 5} };
func(2, 3, matrix[0]);
return 0;
}
Internally, no matter how many dimensions an array has, C/C++ always maintains a 1D array. And so, we can pass any multi-dimensional array like this.
I don't know what you mean by "data dont get lost". Here's how you pass a normal 2D array to a function:
void myfunc(int arr[M][N]) { // M is optional, but N is required
..
}
int main() {
int somearr[M][N];
...
myfunc(somearr);
...
}
2D array:
int sum(int array[][COLS], int rows)
{
}
3D array:
int sum(int array[][B][C], int A)
{
}
4D array:
int sum(int array[][B][C][D], int A)
{
}
and nD array:
int sum(int ar[][B][C][D][E][F].....[N], int A)
{
}
I need to do this to persist operations on the matrix as well. Does that mean that it needs to be passed by reference?
Will this suffice?
void operate_on_matrix(char matrix[][20]);
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:
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:
#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.
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:
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.
#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;
}
(See srand() — why call it only once?.)
Easiest Way in Passing A Variable-Length 2D Array
Most clean technique for both C & C++ is: pass 2D array like a 1D array, then use as 2D inside the function.
#include <stdio.h>
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] = { {0, 1, 2}, {3, 4, 5} };
func(2, 3, matrix[0]);
return 0;
}
Internally, no matter how many dimensions an array has, C/C++ always maintains a 1D array. And so, we can pass any multi-dimensional array like this.
I don't know what you mean by "data dont get lost". Here's how you pass a normal 2D array to a function:
void myfunc(int arr[M][N]) { // M is optional, but N is required
..
}
int main() {
int somearr[M][N];
...
myfunc(somearr);
...
}
2D array:
int sum(int array[][COLS], int rows)
{
}
3D array:
int sum(int array[][B][C], int A)
{
}
4D array:
int sum(int array[][B][C][D], int A)
{
}
and nD array:
int sum(int ar[][B][C][D][E][F].....[N], int A)
{
}