allocating storage to a matrix

Question

I'm writing a function that allocates storage to an nxn matrix.

void assign_matrix_storage(double **matrix, int n){ 
    if((matrix = malloc(n * sizeof(double*))) == NULL){
        printf("ERROR: Memory allocation failed\n");
        exit(EXIT_FAILURE);
    }   

    int i;
    for(i = 0; i < n; i++){
        if((matrix[i] = malloc(n * sizeof(double))) == NULL){
            printf("ERROR: Memory allocation failed\n");
            exit(EXIT_FAILURE);
        }   
    }   

    return;
}

However if I run the following code, I get a segfault on the last statement:

double **A;
assign_matrix_storage(A, 2);
A[1][1] = 42;

Why is this?

http://stackoverflow.com/questions/16004668/c-allocating-a-matrix-in-a-function/27366086#27366086 Above you will find a program that I have made with functions allocating and manipulating matrices in any possible way for C (gcc C11/C99). Maybe it will be usefull for you... — 42n4, Dec 08 '14 at 20:25

score 3 · Answer 1 · answered Apr 25 '11 at 03:55

3

It's happening because A isn't getting changed by assign_matrix_storage(). C is pass-by-value, so you're passing in a copy of A. So the change you make to A in the function is lost. The parameter needs to be something like double ***pointerToA and then when you call the function you'd do assign_matrix_storage(&A, 2); And obviously inside assign_matrix_storage() you'll need to properly deference pointerToA one "level".

answered Apr 25 '11 at 03:55

QuantumMechanic

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1

it might be nicer to return a `double**` rather than accept as an argument a `double***`. – Mark Elliot Apr 25 '11 at 03:56
Great point! Have `assign_matrix_storage()` just take `int n` and have it do the allocation and return the `double **`. – QuantumMechanic Apr 25 '11 at 03:58

Jeff Mercado · Accepted Answer · 2011-04-25T04:12:48.890

You've allocated the memory for your matrix (perfectly) but you don't actually assign it to the A variable from the callee. Instead, A ends up still uninitialized and attempting to assign to A[1][1] caused a segfault. To be able to do that, you'd need a pointer to that variable and assign your matrix to that address. So in effect, your function signature and implementation would need to change:

/* take a pointer to a (double **) */
void assign_matrix_storage(double ***matrix, int n){
    /* then all accesses need to dereference first */
    if(((*matrix) = malloc(n * sizeof(double*))) == NULL){
        printf("ERROR: Memory allocation failed\n");
        exit(EXIT_FAILURE);
    }   

    int i;
    for(i = 0; i < n; i++){
        if(((*matrix)[i] = malloc(n * sizeof(double))) == NULL){
            printf("ERROR: Memory allocation failed\n");
            exit(EXIT_FAILURE);
        }   
    }

    return;
}

/* then call */
double **A;
assign_matrix_storage(&A, 2);
A[1][1] = 42;

A better alternative to what you have would be to return the pointer to the new matrix instead and assign that to your variable.

double **assign_matrix_storage(int n) {
    double **matrix;
    /* the rest of your implementation */
    return matrix;
}

double **A;
A = assign_matrix_storage(2);
A[1][1] = 42;

score 1 · Answer 3 · answered Apr 25 '11 at 03:59

Maybe this implementation is useful.

/* Allocate space for a unit-off-set (m times n) matrix b[1..n][1..m] */
/* Note that b_{ij}=b[j][i], so the i'th column is b[i] */
double **matrix(int m, int n){
  int i;
  double **a = (double **)malloc(n*sizeof(double *)); 
  double **b=a-1;  /* make unit off-set */
  for (i=1; i<=n; i++) b[i] = (double *)malloc((m+1)*sizeof(double)); // m+1, not m!
  return(b);
}

allocating storage to a matrix

3 Answers3