Question

Possible Duplicate:
How do I use arrays in C++?

One dimensional array as a function parameter:

#include <stdio.h>
#include <string.h>

int func(int a[], int n)
{
    int i;
    for(i = 0; i < n; i++)
        printf("%d ", a[i][j]);
}

int main(void)
{   
    int a[2] = {1,2};
    func(a, 2);
}

it compiles and runs correctly.

But when a two-dimensional array as a function parameter:

#include <stdio.h>
#include <string.h>

int func(int a[][], int n)
{
    int i, j;
    for(i = 0; i < n; i++)
        for(j = 0 ; j < n; j++)
            printf("%d ", a[i][j]);
        printf("\n");
}

int main(void)
{   
    int a[2][2] = {{1,2}, {3,4}};
    func(a, 2);
}

it can not compile correctly. I must modify the code like this:

#include <stdio.h>
#include <string.h>

int func(int a[][2], int n)
{
    int i, j;
    for(i = 0; i < n; i++)
        for(j = 0 ; j < n; j++)
            printf("%d ", a[i][j]);
        printf("\n");
}

int main(void)
{   
    int a[2][2] = {{1,2}, {3,4}};
    func(a, 2);
}

I do not know why? Anybody can explain how it works? Many thanks.

Answer 1

Arrays (both one and multidimensional) in c reside in continuous memory blocks. This means that when you define char a[3], the array is laid out in memory like this (forgive my terrible ascii art skills):

| a[0] | a[1] | a[2] |

For a two-dimensional array char a[2][3], the layout is like this:

| a[0][0] | a[0][1] | a[0][2] | a[1][0] | a[1][1] | a[1][2] |  
                              ^
                              +--- first row ends here

Therefore, when you index into a two-dimensional array a[i][j], the compiler generates code equivalent to this:

*(a + i*3 + j)

Which can be read as “skip i rows and take cell j in that row”. To accomplish this, the compiler must know the length of the row (which is the second dimension). This means that the second dimension is a part of the type definition!

As such when you want pass a 2d array into a function, you must specify the needed dimension for the type definition.