How to deal with matrices in C using dynamic memory allocation? - c

Question

I was trying to run a simulation in which I need to fill three matrices of size "2 x iterations", this is (iterations=)10^8 columns and 2 rows. I also work with a vector t of size 10^8. Using dynamic memory allocation I wrote the following code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#define T 10000
#define dt 0.0001
#define iterations (T/dt)
/*(more code)*/
int main(){
int i, j;
double *t;
double (*x)[2], (*y)[2], (*c)[2];
t=(double *) malloc((iterations-1)*sizeof(double));
x=(double (*)[2]) malloc((2*(iterations))*sizeof(double));
y=(double (*)[2]) malloc((2*(iterations))*sizeof(double));
c=(double (*)[2]) malloc((2*(iterations))*sizeof(double));
for(i=0; i=1; i++){
x[i][0]=50+500*i;
y[i][0]=300;
c[i][0]=15;
}
for(j=0; j<=iterations-2; j++){
t[j+1]=t[j]+dt;
/*(more code)*/
printf("%G %G %G %G %G %G\n",x[0][j+1],x[1][j+1],y[0][j+1],y[1][j+1],c[0][j+1],c[1][j+1]);
}
return 0;
}
Is the dynamic memory allocation correctly written? I mean, do I really have a vector t of size "iterations" and the three matrices of size "2 x iterations"?
And, if I want to fill each component of the matrices, for example I want a 50 in the position (1,4) of the matrix x, then do I have to write x[1][4]=50? (Like in the first "for".)
The problem is that executing the program I get an error: segmentation fault. Then, using a debugger I get the following:
Program received signal SIGSEGV, Segmentation fault.
x[0][0]=50

Answer 1

A generic way to allocate a matrix:
double **mat_init(int n_rows, int n_cols)
{
double **m;
int i;
m = (double**)malloc(n_rows * sizeof(double*));
for (i = 0; i < n_rows; ++i)
m[i] = (double*)calloc(n_cols, sizeof(double));
return m;
}
void mat_destroy(int n_rows, double **m)
{
int i;
for (i = 0; i < n_rows; ++i) free(m[i]);
free(m);
}
You can also do this:
double **mat_init2(int n_rows, int n_cols)
{
double **m;
int i;
m = (double**)malloc(n_rows * sizeof(double*));
m[0] = (double*)calloc(n_rows * n_cols, sizeof(double));
for (i = 1; i < n_rows; ++i)
m[i] = m[i-1] + n_cols;
return m;
}
void mat_destroy2(double **m)
{
free(m[0]); free(m);
}
For both methods above, you can use matrix[row][col] to access a cell. Sometimes, you may prefer to allocate a single array and use matrix[row*n_cols+col] to access a cell.
BTW, I am sure someone will say "don't use cast", but there are benefits of using cast – that is off-topic.

Answer 2

I mean, do I really have a vector t of size "iterations"
t=(double *) malloc((iterations-1)*sizeof(double));
^^^
Since you subtract one, the answer is no.
.... and the three matrices of size "2 x iterations"?
Well - yes you have three matrices of size "2 x iterations". However, what you have is equivalent to:
double m[iterations][2];
so you have "iterations" rows and 2 columns.
Remember to always check for "out of memory", i.e.
p = malloc(....);
if (p == NULL)
{
printf("out of mem\n");
return -1;
}
So you access it like:
m[0][0]
m[0][1]
m[1][0]
m[1][1]
m[2][0]
m[2][1]
m[3][0]
......