I have an array of the form (3x3 case as an example)

$$A = [a_{11}, a_{12}, a_{22}, a_{13}, a_{23}, a_{33}] $$

corresponding to the symmetric matrix

$$\begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \end{pmatrix}$$

I want to print out this very matrix (in general of size NxN) given the array A (in general of length N*(N+1)/2) in a nice way. My approach was

void print__symm_matrix_packed(double* arr, int N){
  int idx2 = 0;

  for(int i=0; i<N; i++){
    int idx1 = i;
    for(int j=0; j<N; j++){ 
      if(j < i){
        printf("%f ", arr[idx2 + j]);  
      } else {
        printf("%f ", arr[idx1]);
      idx1 += j+1;
    idx2 += i+1;

Is there room for some elegant improvement?

  • \$\begingroup\$ I think you mean $$ A = [a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}, a_{31}, a_{32}, a_{33}] $$, Am I correct ? \$\endgroup\$ Jun 15 '20 at 18:54
  • \$\begingroup\$ @MiguelAvila No, it is stored column wise back to back \$\endgroup\$
    – P-A
    Jun 15 '20 at 19:33

I think it the following is a good approach:

//use lowercase on non constant values
void print_symmetric_packed_matrix(double* compact_matrix, int n)
    const int entries = N*N;
    for (int i = 0; i < entries;)
        //i++ here avoids i != 0 each iteration
        printf("%f ", compact_matrix[i++]);
        if (i % N == 0) printf("\n");

I hope it helped you.

  • \$\begingroup\$ Thank you, but this does not display the matrix that I mentioned in my question. \$\endgroup\$
    – P-A
    Jun 15 '20 at 19:45

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