c++ pass auto matrix to function

Question

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

i have a problem. i have to use auto darray(matrix).

const int size_m=10;
const int size_n=10;

void process(int *x)
{
  //i can pass array, all is well, i work as with dynamic allocated array
  for(int i=0;i<size_m;scanf("%d",&x[i]),i++);
}

void input_m(int **x) 
/*
mistake is here, 
a cause of my problem that i'm trying to work  with  matrix allocated in stack(auto matrix,local) as with dynamically allocated matrix.
i receive error like this : "cannot convert ‘int [100][100]’ to ‘int**’ in assignment" how should i pass it? 
*/
{
  for(int i=0;i<size_m;i++)
     for(int j=0;j<size_n;j++)
      scanf("%d",&x[i][j]);
}

int main()
{  
  int x[size_m];
  input(x);
  int matr_x[size_m][size_n];
  input_m(matr_x);
  return 0;
}

THANK YOU! it works.. it was so simple, as usual)

    const int sizem=3;
    const int sizen=3;

      void _input(int x[sizem][sizen])
        {

           for(int i=0;i<sizem;i++)
           for(int j=0;j<sizen;x[i][j]=1,j++);
         }

       int main()
         {
            int m=10,n=10;

             int x[sizem][sizen]={{1,2,3},{5,7,4}};

          _input(x);

           for(int i=0;i<sizem;i++)
            { for(int j=0;j<sizen;printf(" %d",x[i][j]),j++);
              puts("");
              }

          puts("");


           return 0;
               }

Especially see [this part](http://stackoverflow.com/questions/4810676/) under "Conversions" in the other question. — Xeo, May 28 '11 at 17:28
i don't see pass, i see declaration,i know it, i want to find out how to pass auto matrix in function — Tebe, May 28 '11 at 17:31
also. note that your usage of auto is rather arcane at best, and especially these days is getting a whole different meaning in C++: [the storage specifier 'auto' is being removed from the standard](http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2006/n1984.pdf) (page 3, std section 7.1.1) — sehe, May 28 '11 at 17:44

David Grayson · Answer 1 · 2011-05-28T18:05:25.803

Your two-dimensional array is incompatible with the function you wrote for fundamental reasons of how memory works in C.

When you write int matrix[size_m][size_n] you are telling the compiler that you want a block of sizeof(int)*size_m*size_n bytes and you intend to store integers in it.

When you write int ** x you are telling the compiler that x is a pointer to a pointer to an integer. If you want to use x as a two-dimensional array, then x should point to not just one pointer, but the first pointer in an array of pointers, i.e. a contiguous region of memory that contains pointers for each row of your matrix. But you don't have any such array of pointers anywhere in the program you posted.

Since you know the dimensions of your matrix at compile time, you can fix this by changing the type of x be int x[size_m][size_n]. Better yet, make a typedef:

typedef int MyMatrix[size_m][size_n];

void input_m(MyMatrix x){ ... }

int main()
{
    MyMatrix matr_x;
    ...
}

+1 for the typedef! Why didn't I think of that... :) – Chris Frederick May 28 '11 at 17:44 — Chris Frederick, May 28 '11 at 17:44

score 1 · Accepted Answer · edited May 23 '17 at 12:18

1

First of all, see the answer to this question. Your code should compile if you change void input_m(int **x) to void input_m(int x[size_m][size_n]) (assuming that both size_m and size_n are constants). Note that, as stated in the question that I linked to, "in general, to pass 2D arrays to functions you need to specify the array dimensions for all but the first dimension."

edited May 23 '17 at 12:18

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answered May 28 '11 at 17:37

Chris Frederick

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c++ pass auto matrix to function

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