Print a square matrix using pointers

Question

I've browsed to previously answered questions regarding pointers and matrices, but in these cases the matrices were seen as pointers to pointers. However, I am trying to create a function which read a matrix using a simple pointer and another function which prints it. This is my code, the read functions seems to work properly, but the program crashes at the printing part. If I remove the "*" from the printf statement the program works(i.e. it prints numbers from 4 to 4- I suppose this is alright, since an int is stored on 4 bytes).

void readm(int *p,int n)
{
 p=(int *)malloc(sizeof(int)*n*n);
 for(int i=0;i<n*n;i++)
        scanf("%d",p+i);
}

void printm(int *p,int n)
{
  for(int i=0;i<n;i++)
  {
    for(int j=0;j<n;j++)
        printf("%d ",*(p+(i*n)+j));
      printf("\n");
  }
}

What do you mean the program crashes ? `SEGFAULT` ? Could you also include the code using those functions ? — Olotiar, Jan 12 '14 at 14:31
[Don't cast the return from `malloc`](http://stackoverflow.com/a/605858/440558). — Some programmer dude, Jan 12 '14 at 14:35

score 1 · Accepted Answer · answered Jan 12 '14 at 14:32

In the readm function you have a problem with this line:

p=(int *)malloc(sizeof(int)*n*n);

Here you assign only to your local copy of the pointer. The variable you use when calling readm will not be changed.

You need to pass the pointer "by reference":

void readm(int **p,int n)  /* Note pointer-to-pointer for `p` */
{
    *p=malloc(sizeof(int)*n*n);  /* Note pointer-dereference of `p` */
    for(int i=0;i<n*n;i++)
        scanf("%d",*p+i);  /* Note pointer-dereference of `p` */
}

You then have to call the function using the address-of operator:

int *p;
readm(&p, X);  /* Note use of address-of operator */

Thank you, it works properly now. Initially I thought that given the fact that I am working with a pointer the modifications which it would suffer in the function would be visible in the exterior( I associated this with the fact that when you want the modifications of a parameter to be visible outside the function you pass it using "&",hence working with its adress.) — user2971971, Jan 12 '14 at 14:44

Vlad from Moscow · Answer 2 · 2014-01-12T14:44:40.573

The problem is that the calling code that calls function readm doesn't know that inside the function variable p (defined as a parameter of the function) got a new value. p is a local variable of the function and its life ends after exiting the function.

You should define the function the following way

void readm( int **p, int n )
{
  *p = (int *)malloc( sizeof(int ) * n * n);
  for ( int i=0; i<n*n; i++ ) scanf( "%d", *p+i );
}

and call it as

int *p;

readm( &p, n );

As for function printmthen there is no any need to redeclare it as

void printm( int **p, int n )

because it does not change the pointer. The only thing I would change is adding qualifier const

void printm( const int *p, int n );

score 1 · Answer 3 · answered Jan 12 '14 at 14:35

A pointer to a 1D array is defined as such:

int *p1dArr;

A pointer to a 2D array is defined as such:

int **p2dArr;

You're using a 1D array as though it's a 2D array. This is probably the source of your troubles. Change you function definitions to the following:

void readm(int **p, int row, int col)
{
    p = malloc(row * sizeof(*p));
    for(int r = 0; r < row; r++)
        p[r] = malloc(col * sizeof(**p))

    for(int r = 0; r < row; r++)
        for(int c = 0; c < col; c++)
            scanf("%d", &p[r][c]);
}

void printm(int **p, int row, int col)
{
    for(int r = 0; r < row; r++)
    {
        for(int c = 0; c < col; c++)
            printf("%d ", p[r][c]);
        printf("\n");
    }
}

Print a square matrix using pointers

3 Answers3

Linked