Passing 2-d array to a function in C

Question

I'm trying to make a function that initializes a random 2 dimensional square matrix given its variable name and number of rows and columns (n in this case). This code returns a "Segmetation fault" error and I don't know why. I've tried using & in if-else in the function initM, but then returns "error: lvalue required as left operand of assignment"Anyone knows how to make it work?

void initM (int n, int **matrix){
    for (int i = 0; i < n; i++){
        for (int j = 0; j < n; j++){
            if (i == j) matrix[i][j] = 101;
            else matrix[i][j] = (rand() % 101);
        }
    }
}

void main (int argc, char *argv[]){

    int n = atoi(argv[1]);

    int **matrix = (int **)malloc(n*sizeof(int));
    for(int i = 0; i < n; i++) matrix[i] = (int *)malloc(n*sizeof(int));

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

    for (int i = 0; i < n; i++) free(matrix[i]);
    free(matrix);
}

Answer 1

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

It should be

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

In your architecture it happens that sizeof(int) and sizeof(void*) do not have the same values.

No need to cast the output of malloc, as void* can correctly behave as any other pointer to object.

Answer 2

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

The main problem is that sizeof (int) is not the same as sizeof (int *) on your system, so you're not allocating enough space. The above call can be rewritten as

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

The cast is unnecessary ¹ , and sizeof *matrix == sizeof (int *) ² .

Similarly, you can allocate each matrix[i] as

matrix[i] = malloc( n * sizeof *matrix[i] ); // sizeof *matrix[i] == sizeof (int)

You should always check the result of a malloc , calloc , or realloc call. While failures are rare, they can happen:

int **matrix = malloc( n * sizeof *matrix );
if ( matrix )
{
  for ( i = 0; i < n; i++ )
  {
    matrix[i] = malloc( n * sizeof *matrix[i] );
    if ( matrix[i] )
      // do stuff
  }
}

Final nit - main returns int , not void . Unless your compiler specifically documents void main( int, char ** ) as a valid signature, then using it invokes undefined behavior . Your code may run just fine, but there are (admittedly old or oddball) platforms where typing main as void leads to problems.

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1. ^{Unless you're compiling this code as C++ or working with an ancient pre-C89 compiler.}
2. ^{sizeof is an operator , not a function - parentheses are only necessary if the operand is a type name.}