Pointers, struct and malloc()

Question

I've made this simple program to test struct, pointers and malloc() of C programming:

#include <stdlib.h>
#include <stdio.h>

 typedef struct l {
    unsigned val;
    struct l * next;
} list;


typedef struct cont {
    list ** mat;
    unsigned riga;
} container;


int main(){

    container * c = (container *)malloc(sizeof(container));
    c->riga = 3;
    c->mat = (list**)malloc(sizeof(list*)*3);

    for(int i = 0; i<3 ;i++){
        c->mat[i] = NULL;
    }

    c->mat[0] = (list *)malloc(sizeof(list));
    c->mat[0]-> val = 4;
    c->mat[0]-> next = NULL;

    printf("val row 0: %d\n", c->mat[0]->val);

    /*************************/

    list * ca = c->mat[1];
    ca = (list *)malloc(sizeof(list));
    ca->val = 2;
    ca->next = NULL;

    printf("val row 1: %d\n", c->mat[1]->val);

    return 0;
}

The problem is that I get a segfault on the 2nd printf() ... shouldn't c->mat[1] and ca be equivalent?

Shouldn't they point to the same allocated memory?

Initially c->mat[1] is NULL so is ca = c->mat[1] = NULL, after ca= malloc() it points to something else ... what about c->mat[i]? And what if I do the viceversa?

Answer 1

ca is a copy of c->mat[1], so when you later change ca to point to the malloc-ed list it is not a copy of c->mat[1] anymore and especially if you modify ca you do not also modify c->mat[1].

If ca were not a pointer but an int you would not expect that, or would you assume that a was 2 in the following example?

int a = 3;
int ca = a;
ca = 2;

Of course, if ca is a copy of c->mat[1] and you change *c then you also change *(c->mat[1]).