how can I pass multiarray into function through pointer with c++. I can do this with simple arrays
void Foo(int *arr) { }
int someArr[10];
Foo(someArr);
What about 2-dimensions array?
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3 Answers
You can achieve what you want in this way:
void foo(int *array) { }
int column_size = 5;
int main() {
int array[column_size][2];
foo(&array[0][0]);
return 0;
}
although you should take care on how you read the elements from inside the foo function.
In order to read the array[c][r] element you must do:
int element = *(array + c * column_size + r);
The general get element function is:
int get_element(int *array, int row, int column, int column_size) {
int element = *(array + row * column_size + column);
return element;
}
So, if you have let's say an 2D array like int array[M][N] and you want to get the array[i][j] element you just call the function in this way:
getElement(&array[0][0], i, j, N)
Why does this work?
The reason the above function works can be clarified if you know how 2D arrays are saved in memory. The arrays are saved row-by-row, so let's say you have the following array:
int a[3][3] = {{1, 2, 4}, {5, 6, 7}, {8, 9, 10}};
let's assume an integer is 4 bytes and &a[0][0] corresponds to 0x10 memory address. Then 1 is saved in 0x10 memory address, 2 is saved in 0x14, ..., 7 is saved in 0x24 memory address, ... and 10 is saved in 0x30 memory address (see the following table).
*Memory* Memory address => Value => Pointer pointing at this memory address 0x10 => 1 => &a[0][0] 0x14 => 2 => (&a[0][0] + 1) or (&a[0][1]) 0x18 => 4 => (&a[0][0] + 2) or (&a[0][2]) 0x1c => 5 => (&a[0][0] + 3 * 1 + 0) or (&a[1][0]) 0x20 => 6 => (&a[0][0] + 3 * 1 + 1) or (&a[1][1]) 0x24 => 7 => (&a[0][0] + 3 * 1 + 1) or (&a[1][2]) 0x28 => 8 => (&a[0][0] + 3 * 2 + 0) or (&a[2][0]) 0x2c => 9 => (&a[0][0] + 3 * 2 + 1) or (&a[2][1]) 0x30 => 10 => (&a[0][0] + 3 * 2 + 2) or (&a[2][2])
Now when you have the following pointer:
int *pt = (&a[0][0] + 2);
the pt pointer will be pointing 2 elements after a[0][0]. So pt is pointing at a[0][2]. *pt will be equal to 4.
Now let's say you want to get the a[i][j] element. In order to get this element
you need to move i * COLUMN_SIZE elements away in order to get in the correct row where the element is (each row has COLUMN_SIZE elements) and then you need to add j in order to get in the correct column.
If you want to get the a[2][1] (where COLUMN_SIZE = 3), then 2 * COLUMN_SIZE = 6 + 1 = 7. So in order to get the a[2][1] element you do *(&a[0][0] + 2 * 3 + 1) or *(&a[0][0] + 7).
For some great tutorials on pointers, have a look here: Stanford CS Ed Library.
2 Comments
Jonathan Leffler
But the function can then only refer to the array as a 1D array (vector), not as a correctly sized 2D array.
Max Frai
@jonathan-leffler Yeah, I knew about code you post. But the real question was about passing pointers, so @functional did what I wanted. Thanks.
It is more or less the same - except for the bits that are different, of course.
#include <stdio.h>
enum { DIM = 6 };
static void Bar(int arr[][DIM], int n)
{
int i, j;
for (i = 0; i < n; i++)
{
for (j = 0; j < DIM; j++)
printf(" %2d", arr[i][j]);
putchar('\n');
}
}
int anotherArr[][DIM] =
{
{ 1, 2, 3, 4, 5, 6 },
{ 6, 5, 4, 3, 2, 1 },
{ 0, 0, 0, 0, 0, 0 },
{ -1, -2, -3, -4, -5, -6 },
};
int main(void)
{
Bar(anotherArr, 4);
return(0);
}
answered Jul 31, 2010 at 23:40
Jonathan Leffler
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4 Comments
Max Frai
Hm, why do you use enum for storing array dimension?
sholsapp
Can you do the same thing, but declare the parameter for Bar as a pointer instead of an array?
Jonathan Leffler
@Ockonal: because I speak C better than C++.
Jonathan Leffler
@gnucom: yes, but you lose the benefit of compiler-calculated subscripts. See the answer by @functional.
void Foo(int **arr) { }
int someArr[10][10];
Foo(someArr);
2 Comments
Jonathan Leffler
That will crash horribly because it expects an array of pointers, and a 2D array is not the same as an array of pointers.