How to modify pointer to array of structs inside another struct by reference

Question

I am trying to modify a struct that is a pointer inside of another struct within a function and then have the change reflect back outside of the function.

I have simplified my code below to the exact issue I am having. I understand that inside my function, as currently constructed, I am modifying the value of x, rather than the utilizing the address of x to modify, but I can't seem to wrap my head around how to properly modify each particle in the array. I have tried

b[0].p->x++;

inside the for loop in the test function, but that only changes p[0] from 1 to 5.

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

typedef struct {
        int x;
} particle_t;

typedef struct {
        int num_part;
        particle_t* p;
} bins;

void test (bins *b) {
        for (int i = 0; i < 4; i++) {
                b[0].p[i].x++;
        }
}

int main (int argc, char **argv) {
        particle_t *particles = (particle_t*) malloc (4 * sizeof(particle_t));
        particles[0].x = 1;
        particles[1].x = 2;
        particles[2].x = 3;
        particles[3].x = 4;

        bins *b = (bins*) malloc (4 * sizeof(bins));
        b[0].num_part = 1;
        b[0].p = (particle_t*) malloc (4 * sizeof(particle_t));

        for (int i = 0; i < 4; i++) {
                b[0].p[i] = particles[i];
        }

        test(b);

        printf("%d %d %d %d\n",particles[0].x,particles[1].x,particles[2].x,particles[3].x);

        return 0;
}

After running the test function, I expected that printing out particles[i].x would yield 2, 3, 4, and 5, but I am still receiving 1, 2, 3, and 4.

"After running the test function, I expected that" - the program terminates after running the test function... — Crowman
– Crowman, Commented Apr 18, 2019 at 0:04
yes, I updated again, I am want the changes done on the b[0].p array to be reflected in the particles array — jrod091
– jrod091, Commented Apr 18, 2019 at 0:09
OT: regarding: int main (int argc, char **argv) To avoid the compiler outputting two warning messages about unused parameters, When the parameters to main() are not used, then use the signature: int main( void ) — user3629249
– user3629249, Commented Apr 18, 2019 at 16:33

marcelodmmenezes · Accepted Answer · 2019-04-18 00:11:55Z

2

The problem is that you are allocating a new array for b[0].p and copying the particles array to b[0].p array. Your function changes the memory in b[0].p and not the memory pointed by particles.

Instead of malloc just make your b[0].p array point to particles.

Summing up, change

b[0].p = (particle_t*) malloc (4 * sizeof(particle_t));

to

b[0].p = particles;

answered Apr 18, 2019 at 0:11

marcelodmmenezes

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4 Comments

jrod091 Over a year ago

Thanks! Would you mind helping me take this one step further? How would I make this work if I wanted to split the 4 particles I have in this example among 2 bins, as an example?

marcelodmmenezes Over a year ago

You could, for instance, make the first bin point to the beginning of particles{ b[0].p = particles; } and the second point to the other half { b[1].p = particles + 2; }. But I don't see why you are doing this kind of stuff. Seems to me that there is no need for the particles array. Just allocate each bin with its own particles...

marcelodmmenezes Over a year ago

If you insist in having everyone sharing the same memory at least make sure that each bin only access its share of the particles array (Make sure the for iterate only for each bin size (num_part)).

jrod091 Over a year ago

Thanks again! The reason I have the 2 different arrays is because in the grander scheme of my program, the particles are set with position, velocity, and acceleration and then interact with each other, but only if they are close enough to each other within a predefined space. The bins collect the particles so that I can limit the number of particles each particle checks to see if an interaction should occur. And yes, I am iterating over num_parts, I had simplified the code here to highlight where I was getting stuck.

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