How to pass a multidimensional array to a function in C and C++

Question

#include<stdio.h>
void print(int *arr[], int s1, int s2) {
    int i, j;
    for(i = 0; i<s1; i++)
        for(j = 0; j<s2; j++)
            printf("%d, ", *((arr+i)+j));
}

int main() {
    int a[4][4] = {{0}};
    print(a,4,4);
}

This works in C, but not in C++.

error:

cannot convert `int (*)[4]' to `int**' for argument `1' to 
`void print(int**, int, int)'

Why does it not work in C++? What change is needed to be made?

Answer 1

This code will not work in neither C nor C++. An array of type int[4][4] is not convertible to a pointer of type int ** (which is what int *arr[] stands for in parameter declaration). If you managed to compile it in C, it is simply because you probably ignored a C compiler warning of basically the same format as the error message you got from C++ compiler. (Sometimes C compilers issue warnings for what is essentially an error.)

So, again, don't make assertions that are not true. This code does not work in C. In order to convert a built-in 2D array into a int ** pointer you can use a technique like this one

http://stackoverflow.com/questions/1584100/converting-multidimensional-arrays-to-pointers-in-c

(See the accepted answer. The problem is exactly the same.)

EDIT: The code appears to work in C because another bug in the printing code is masquerading the effects of the bug in array passing. In order to properly access an element of an int ** pseudo-array, you have to use expression *(*(arr + i) + j), or better a plain arr[i][j] (which is the same thing). You missed the extra * which made it print something that has absolutely nothing to do with the content of your array. Again, initialize your array in main to something else to see that the results you are printing in C have absolutely nothing to do with the your intended content of the array.

If you change the printf statement as shown above, your code will most likely crash because of the array-passing bug I described initially.

One more time: you cannot pass a int[4][4] array as an int ** pseudo-array. This is what the C++ is telling you in the error message. And, I'm sure, this is what your C compiler told you, but you probably ignored it, since it was "just a warning".

Answer 2

#include<stdio.h>
void print(int arr[][4], int s1, int s2) {
    int i, j;
    printf("\n");
    for(i = 0; i<s1; i++) {
        for(j = 0; j<s2; j++) {
            printf("%d, ", *((arr+i)+j));
        }
    }
    printf("\n");
}

int main() {
    int a[4][4] = {{0}};
    print(a,4,4);
}

This will work, where by work I mean compile. @AndreyT explained why your version doesn't work already.

This is how you should pass a 2d array.

For clarity, you can also specify both sizes in the function declaration:

#include<stdio.h>
void print(int arr[4][4], int s1, int s2) {
    int i, j;
    printf("\n");
    for(i = 0; i<s1; i++) {
        for(j = 0; j<s2; j++) {
            printf("%d, ", *((arr+i)+j));
        }
    }
    printf("\n");
}

int main() {
    int a[4][4] = {{0}};
    print(a,4,4);
}

Both will work.

You should also change *((arr+i)+j) to either a[i][j] (preferably) or *(*(arr+i)+j) if your intention is to access the jth element of row i.

Answer 3

#include<cstdio>
template <size_t N, size_t M>
struct DataHolder
{
    int data[N][M];
    DataHolder()
    {
       for(int i=0; i<N; ++i)
           for(int j=0; j<M; ++j)
               data[i][j] = 0;
    }
};

template <size_t N, size_t M>
void print(const DataHolder<N,M>& dataHolder) {
    printf("\n");
    for(int i = 0; i<N; i++) {
        for(int j = 0; j<M; j++) {
            printf("%d, ", dataHolder.data[i][j]);
        }
    }
    printf("\n");
}

int main() {
    DataHolder<4,4> a;
    print(a);
}

Answer 4

Short answer, you may change the program as following

void print(int arr[], int s1, int s2) {
...
printf("%d,", *(a+i + s2*j));
...
print((int*)a,4,4);

This would need a better answer explaining the differences between pointer and pointer arithmetic and arrays in C and C++. I won't launch into that now. Maybe someone else ?

I obviously am not shocked by the same point as other posters in your code. What bother me most in the print function header is that you use a double indirection for an array where you do not intend to change initial pointer back (in fact it can't be done as it is a constant). @|V|lad answer fix this by setting one or two dimensions to a fixed constant, but then passing s1 and s2 become useless.

All depends of what you really want to do. Is print a general purpose array printing function or a specialized one for some array types ?

Answer 5

Here's a version which is both working, but theoretically invalid (see below) C90 and C++98:

#include <stdio.h>

static void print(int *arr, size_t s1, size_t s2)
{
    size_t i, j;
    printf("\n");
    for(i = 0; i < s1; i++) {
        for(j = 0; j < s2; j++) {
            printf("%d, ", arr[i * s2 + j]);
        }
    }
    printf("\n");
}

int main(void) {
    int a[4][4] = {{0}};
    print(a[0], 4, 4);
    return 0;
}

A C++ version using templates (adapted from Notinlist's answer) could look like this:

#include <iostream>
#include <cstring>

using namespace std;

template <size_t N, size_t M>
struct IntMatrix
{
    int data[N][M];
    IntMatrix() { memset(data, 0, sizeof data); }
};

template <size_t N, size_t M>
ostream& operator<<(ostream& out, const IntMatrix<N,M>& m)
{
    out << "\n";
    for(size_t i = 0; i < N; i++) {
        for(size_t j = 0; j < M; j++) {
            out << m.data[i][j] << ", ";
        }
    }
    out << "\n";
    return out;
}

int main()
{
    IntMatrix<4,4> a;
    cout << a;
    return 0;
}

Alternatively, you could use nested STL containers - ie vector< vector<int> > - instead of a plain array.

With C99, you could do

static void print(size_t s1, size_t s2, int arr[s1][s2]) {
    printf("\n");
    for(size_t i = 0; i < s1; i++) {
        for(size_t j = 0; j < s2; j++) {
            printf("%d, ", arr[i][j]);
        }
    }
    printf("\n");
}

and call it as

print(4, 4, a);

---

As Robert pointed out in the comments, the first snippet actually involves undefined behaviour. However, assuming that pointer arithmetics will always result in a pointer even when undefined behaviour is involved (and not blow up your computer), there is only a single possible result because of other restrictions within the standard, ie this is an instance of where the standard leaves something unnecessarily undefined.

As far as I can tell, substituting

print(a[0], 4, 4);

with

union m2f { int multi[4][4]; int flat[16]; } *foo = (union m2f *)&a;
print(foo->flat, 4, 4);

will make it legal C.

Answer 6

You can use int** instead. Its much more flexible:

#include <stdio.h>
#include <stdlib.h>
void print(int **a, int numRows, int numCols )
{
  int row, col ;
  for( int row = 0; row < numRows; row++ )
  {
    for( int col = 0; col < numCols ; col++ )
    {
      printf("%5d, ", a[row][col]);
    }
    puts("");
  }
}

int main()
{
  int numRows = 16 ;
  int numCols = 5 ;
  int **a ;

  // a will be a 2d array with numRows rows and numCols cols

  // allocate an "array of arrays" of int
  a = (int**)malloc( numRows* sizeof(int*) ) ;

  // each entry in the array of arrays of int
  // isn't allocated yet, so allocate it
  for( int row = 0 ; row < numRows ; row++ )
  {
    // Allocate an array of int's, at each
    // entry in the "array of arrays"
    a[row] = (int*)malloc( numCols*sizeof(int) ) ;
  }

  int count = 1 ;
  for( int row = 0 ; row < numRows ; row++ )
  {
    for( int col = 0 ; col < numCols ; col++ )
    {
      a[row][col] = count++ ;
    }
  }

  print( a, numRows, numCols );
}

Another thing which you may be interested in is a structure like D3DMATRIX:

typedef struct _D3DMATRIX {
    union {
        struct {
            float        _11, _12, _13, _14;
            float        _21, _22, _23, _24;
            float        _31, _32, _33, _34;
            float        _41, _42, _43, _44;

        };
        float m[4][4];
    };
} D3DMATRIX;

D3DMATRIX myMatrix ;

The sweet thing about this little tidbit is you can use both myMatrix.m[0][0] (to access the first element), or you can use myMatrix._11 to access that same element as well. The union is the secret.

Answer 7

The problem is, that

int a[4][4];

will actually be stored in a physically continuos memory. So, to access an arbitrary part of your 4x4 array, the function "print" needs to know the dimensions of the array. For example the following little piece of code, will access the same part of the memory in two different ways.

#include <iostream>

void print(int a[][4]){
    for (int i = 0; i <4; i++){
        for (int j = 0; j < 4; j++){
            //accessing as 4x4 array
            std::cout << a[i][j] <<std::endl;        

            //accessing corresponding to the physical layout in memory
            std::cout <<  *(*(a)+ i*4 + j) << std::endl;  

        }
    }
}

int main(){
    int a[4][4];

    //populating the array with the corresponding indices from 0 to 15
    int m = 0;
    for (int i = 0; i<4; i++){
        for (int j= 0; j < 4; j++){
            a[i][j] =  m;
            m++;
        }
    }
    print(a);
}

So the memory layout doesn't change but the way of accessing does. It can be visualized like a checkerboard.

   0  1  2  3
  ----------
0| 1  2  3  4
1| 5  6  7  8
2| 9 10 11 12
3|13 14 15 16

But the real physical memory looks like this.

0*4+0 0*4+1 0*4+2 0*4+3 1*4+0 1*4+1 1*4+2 1*4+3 2*4+1   etc.
-----------------------------------------------------
1      2       3    4     5     6      7     8     9    etc.

In c++ the data of an array is stored row-by-row and the length of a row (in this case 4) is always necessary to get to the proper memory offset for the next row. The first subscript therefore only indicates the amount of storage that is needed when the array is declared, but is no longer necessary to calculate the offset afterwards.

Answer 8

First thing to do is get the types right. If C++'s rules are the same as C's with respect to array types (I'm pretty sure they are), then given the declaration

int a[4][4];

the expression a has type int [4][4], which is implicitly converted ("decays") to a pointer type of int (*)[4] (pointer to 4-element array of int) when passed to print, so you need to change print to

void print(int (*arr)[4], int s1, int s2)
{
  int i, j;        
  for(i = 0; i<s1; i++)        
    for(j = 0; j<s2; j++)        
      printf("%d, ", arr[i][j]);        
}        

The expression arr[i] implicitly dereferences arr, so you don't need to mess with an explicit dereference.

The drawback is that print can only handle Nx4 arrays of int; if you want to handle other array sizes, you'll need to take a different approach.

One thing you can do is instead of passing the array, pass the address of the first element, and have print manually compute the offsets, as so:

int main() {                    
  int a[4][4] = {{0}};                    
  print(&a[0][0],4,4);  // note how a is being passed                  
}  

void print(int *arr, int s1, int s2)  // note that arr is a simple int *
{
  int i, j;
  for (i = 0; i < s1; i++)
    for (j = 0; j < s2; j++)
      printf("%d, ", arr[i * s2 + j]);
}

Answer 9

Aside from using variable-length arrays in C99, you can't really portably write a function to accept a multidimensional array if the sizes of the arrays are not known at compile-time, See Question 6.19 of the C-FAQ. The best way to handle this is to simulate multidimensional arrays using dynamically allocated memory. Question 6.16 does a very good job of explaining the details of doing this.