Passing an array of structs where one of the fields is another array, to a function

Question

This is my struct.

    typedef struct {
        uint8_t *data;
        size_t size;
    } buffer;

I initialize it like this in my main function.

buffer *array;
array = calloc(255, sizeof(*array));
//check for error
for (int i = 0; i < 255; i++) {
    array[i].data = malloc(sizeof(*array->data));
    //check for error
}

Then I pass it to a function.

ArrayFunction(&array);

This is how it's defined

void ArrayFunction(buffer **array) {
    // Doing something to determine somesize
    for (int i = 0; i < 255; i++) {
        array[i]->size = 100;
        array[i]->data = calloc(100, sizeof(uint8_t));
        //check for error
        //Setting the data to something useful
    }
}

When looking at my program through a debugger, my initialization in the main function seems fine, memory is valid. Once it enters the ArrayFunction, memory past the 0th element is invalid or uninitialized which causes a segmentation fault at the point of trying to calloc. It does work for the 0th and so does the setting data step.

Answer 1

You're trying too hard. You don't need double pointers.

Explanations in the comments starting with //**

You probably just want this:

#include <stdint.h>
#include <stdlib.h>

typedef struct {
  uint8_t* data;
  size_t size;
} buffer;

void ArrayFunction(buffer *arr) {  //** just pass the buffer (t's a pointer)
  const int somesize = 11;         //** just made up some value

  for (int i = 0; i < 255; i++) {
    arr[i].size = somesize;
    arr[i].data = calloc(somesize, sizeof(uint8_t));
        
  }
}

int main(int argc, char* argv[])
{
  buffer* array;
  array = calloc(255, sizeof(*array));
  //check for error
  for (int i = 0; i < 255; i++) {
    array[i].data = malloc(sizeof(*array->data));
    //check for error
  }

  ArrayFunction(array);             //** just pass the pointer
}