Is it possible to pass pointers to a struct in C?

Question

I am calling from main the function init_latent_variables and passing pointer to a struct sample of type SAMPLE

int main(){
    SAMPLE sample;
    sample = read_struct_examples();
    init_latent_variables(&sample);
    return 0;
}

SAMPLE read_struct_examples() {
    SAMPLE sample;        
    sample.examples = (EXAMPLE *) malloc(1*sizeof(EXAMPLE));
    if(!sample.examples) die("Memory error.");
    return(sample); 
}

Following is function definition. The memory allocation works fine in this function, but the original variable in main remains unchanged.

void init_latent_variables(SAMPLE *sample) {
    sample->examples[0].h.h_is = (int *) malloc(5*sizeof(int));
    if(!sample->examples[0].h.h_is) die("Memory error.");       
}

Following are the struct definitions:

typedef struct latent_var {
  int *h_is;
} LATENT_VAR;

typedef struct example {
  LATENT_VAR h;
} EXAMPLE;

typedef struct sample {
  EXAMPLE *examples;
} SAMPLE;

Am I right in passing a pointer to struct. Is it possible to do so in C?

UPDATE : Not sure what was wrong earlier. Clean and recompile seems to work. Thanks and apologies for wasting your time. Already flagged the question so that moderators could delete it.

Answer 1

You're dereferenceing sample->examples before you allocate it:

void init_latent_variables(SAMPLE *sample) {
    sample->examples[0].h.h_is = (int *) malloc(5*sizeof(int));
    if(!sample->examples[0].h.h_is) die("Memory error.");       
}

this should be :

void init_latent_variables(SAMPLE *sample, int num_examples) {
    sample->examples = (EXAMPLE *) malloc(sizeof EXAMPLE * num_examples);
    if(!sample->examples) die("Memory error.");   
    sample->examples[0].h.h_is = (int *) malloc(5*sizeof(int));
    if(!sample->examples[0].h.h_is) die("Memory error.");       
}

Consider using calloc() instead of malloc, or use memset to clear your structs to 0 before using them.

Answer 2

首先，您必须分配示例SAMPLE结构的成员……据我所知，它必须是一个对象数组……而且，当然，在C中没有引用，只有“值”或“指针”。