Question

I have a struct that looks like this:

typedef struct _my_struct {
    float first_vector[SOME_NUM][OTHER_NUM];
    float second_vector[SOME_NUM][OTHER_NUM];
    int some_val;
} my_struct;

I’d like to do:

my_struct * thing = (my_struct *)malloc(sizeof(my_struct));

But when I do so and then try to access anything in the vectors, I get a seg fault.

If I instead declare the vectors in the struct as:

typedef struct _my_struct {
    float **first_vector;
    float **second_vector;
    int some_val;
} my_struct;

and then allocate with:

my_struct->first_vector = (float**)malloc(sizeof(float*) * SOME_NUM);
my_struct->second_vector = (float**)malloc(sizeof(float*) * SOME_NUM);
int i;
for (i = 0; i < SOME_NUM; i++){ 
    my_struct->first_vector[i] = (float*)malloc(sizeof(float) * OTHER_NUM);
    my_struct->second_vector[i] = (float*)malloc(sizeof(float) * OTHER_NUM);
} 

Then everything works fine.

Since first and second vector are fixed in size and known at compile time, it seems strange that I have to individually allocate memory for them. Is there a way to declare them in the struct so that I can just malloc a new struct without also allocating memory for each vector separately?

Answer 1

What you have should be fine.. aside from the funky typedef at the top and odd declaration of my_struct * struct = malloc... this small test case works for me:

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

typedef struct {
    float first[10][20];
    float second[10][20];
    int val;
} my_struct;

int main(void) {
    printf("Sizeof mystruct %d\n", sizeof(my_struct));

    my_struct* str = malloc(sizeof(my_struct));
    if(!str) {
        printf("Memory allocation error!");
        exit(1);
    }

    str->first[0][0] = 1;
    str->second[1][19] = 15;

    printf("values %f %f\n", str->first[0][0], str->second[1][19]);

    free(str);

    return 0;

}

Output is:

Sizeof mystruct 1604
values 1.000000 15.000000

Additionally, you should not cast the return value of malloc. It can hide important compiler warnings.