from c malloc matrix to java array -

i have c code:

double **a; = (double **)malloc( (m+2+1)*sizeof(double *));   for(i=1; i<=m+2; i++)     a[i] = (double *)malloc( (n+1+1)*sizeof(double)); 

and have convert in java code. not sure but, equivalent write in java?:

double[][] = new double[m+2+1][n+1+1]  

what best way convert data structure in java?

==============================================

debugging 2 codes have 2 different behaviors:

m=8 , n=13 

debugging c code, allowed access a[14][407]

   int main() {      int i,m,n;     m = 2;     n = 2;      double **a;     = (double **)malloc( (m+2+1)*sizeof(double *));       for(i=1; i<=m+2; i++)         a[i] = (double *)malloc( (n+1+1)*sizeof(double));      int num_rows = sizeof(a) / sizeof(a[0]);     int num_cols = sizeof(a[0]) / sizeof(a[0][0]);     printf("r %d c %d", num_rows, num_cols); } 

debugging java code a[10][15]

what's difference?

thanks

============================================== @crawford-whynnes

to do: allocate matrix of rows = r , cols = c

1. st solution

   double **a; = (double **)malloc((r)*sizeof(double *)); a[0]=(double*) malloc((r*c)*sizeof(double)); for(i=1; i<r; i++)   a[i]=a[i-1] + n; 

2. nd solution

   double **a; = (double **)malloc( (r)*sizeof(double *)); for(i=0; i<r; i++)   a[i] = (double *)malloc( (c)*sizeof(double)); 

what correct solution? first or second?

in second solution have matrix of rows = r , cols = r*c , not matrix of rows = r , cols = c...it's correct?

thanks

the loop in c should be

for(i=0; i<=m+2; i++) 

otherwise first row left unallocated.

after change has been done, java code same c code. change malloc calloc (for 0 initialization of allocated memory) equivalent of java code.