I have a 2 dimensional array dynamically allocated in my C code, in my function main. I need to pass this 2D array to a function. Since the columns and rows of the array are run time variables, I know that one way to pass it is :
-Pass the rows and column variables and the pointer to that [0][0] element of the array
myfunc(&arr[0][0],rows,cols)
then in the called function, access it as a 'flattened out' 1D array like:
ptr[i*cols+j]
But I don't want to do it that way, because that would mean a lot of change in code, since earlier, the 2D array passed to this function was statically allocated with its dimensions known at compile time.
So, how can I pass a 2D array to a function and still be able to use it as a 2D array with 2 indexes like the following?
arr[i][j].
Any help will be appreciated.
See the code below. After passing the 2d array base location as a double pointer to myfunc()
, you can then access any particular element in the array by index, with s[i][j]
.
#include <stdio.h>
#include <stdlib.h>
void myfunc(int ** s, int row, int col)
{
for(int i=0; i<row; i++) {
for(int j=0; j<col; j++)
printf("%d ", s[i][j]);
printf("\n");
}
}
int main(void)
{
int row=10, col=10;
int ** c = (int**)malloc(sizeof(int*)*row);
for(int i=0; i<row; i++)
*(c+i) = (int*)malloc(sizeof(int)*col);
for(int i=0; i<row; i++)
for(int j=0; j<col; j++)
c[i][j]=i*j;
myfunc(c,row,col);
for (i=0; i<row; i++) {
free(c[i]);
}
free(c);
return 0;
}
If your compiler supports C99 variable-length-arrays (eg. GCC) then you can declare a function like so:
int foo(int cols, int rows, int a[][cols])
{
/* ... */
}
You would also use a pointer to a VLA type in the calling code:
int (*a)[cols] = calloc(rows, sizeof *a);
/* ... */
foo(cols, rows, a);
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