writing to an array of structs - c

Question

I have the following code
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
typedef struct Example
{
uint16_t a;
uint16_t b;
} ExampleStruct;
void derp(struct Example * bar[], uint8_t i)
{
uint8_t c;
for(c = 0; c < i; ++c)
{
bar[c]->a = 1;
bar[c]->b = 2;
}
}
int main()
{
struct Example * foo;
uint8_t i = 3;
foo = malloc(i*sizeof(ExampleStruct));
derp(&foo, i);
free(foo);
return 0;
}
I get segfaults and all debuggers tell me that code stopped working due to
bar[c]->a = 1;
I tried to rearrange this into all of the following
(*bar)[c]->a = 1;
(*bar[c])->a = 1;
bar[c].a = 1;
(*bar)[c].a = 1;
and with no success. What am I doing wrong? I don't understand why is this failing, and I don't understand why the addresses of bar[0], bar[1] and bar[2] are so far away from each other, when each just takes 2 bytes.

Answer 1

There's no need to pass &foo. Keep it simple:
// In a function declaration, it's (almost) always a pointer, not an array.
// "struct Example bar[]" means *exactly* the same thing in this context.
void init(struct Example * bar, int n) {
int i;
for (i = 0; i < n; ++i) {
bar[i].a = 1;
bar[i].b = 2;
}
}
int main() {
int n = 3;
struct Example * foo = malloc(n*sizeof(struct Example));
init(foo, n); // passes the address of the array - &a[0] - to init
printf("The second element is {%u, %u}\n", foo[1].a, foo[1].b);
free(foo);
return 0;
}
output:
The second element is {1, 2}

Answer 2

Some changes were required since you were trying to pass array of objects:
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <inttypes.h>
typedef struct Example
{
uint16_t a;
uint16_t b;
} ExampleStruct;
void derp(struct Example * bar[], uint8_t i)
{
uint8_t c;
for(c = 0; c < i; ++c)
{
bar[c]->a = 1;
bar[c]->b = 2;
}
}
int main()
{
struct Example * foo[3];
uint8_t i = 3, c;
for(i = 0; i < 3; i++)
foo[i] = malloc(sizeof(ExampleStruct));
derp(foo, i);
for(c = 0; c < i; ++c)
{
printf("\n%" PRIu16 " %" PRIu16 ,foo[c]->a,foo[c]->b);
}
for(i = 0; i < 3; i++)
free(foo[i]);
return 0;
}
struct Example * foo; can hold a single pointer to an object of type struct Example. While struct Example * bar[] can hold an array of pointers to objects of type struct Example.
In your original program, this will seg fault when c is greater than 0 since you did not allocate any pointers to an object of type struct Example.
bar[c]->a = 1;
bar[c]->b = 2;
For static objects:
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <inttypes.h>
typedef struct Example
{
uint16_t a;
uint16_t b;
} ExampleStruct;
void derp(struct Example bar[], uint8_t i)
{
uint8_t c;
for(c = 0; c < i; ++c)
{
bar[c].a = 1;
bar[c].b = 2;
}
}
int main()
{
struct Example foo[3];
uint8_t i = 3, c;
derp(foo, i);
for(c = 0; c < i; ++c)
{
printf("\n%" PRIu16 " %" PRIu16 ,foo[c].a,foo[c].b); //accessing in main
}
return 0;
}