Initialize values of dynamic array in a function (C language)

Question

this is my code:

void init_array(int** array) {

  *array = (int*) malloc(3 * sizeof(int));

  /* ???? **(array+2) = 666; */

  return;
}

void init(int* array, int* length) {

  *length = 3;

  *(array+0) = 0;
  *(array+1) = 1;
  *(array+2) = 2;

  return;

}

int main(void) {

  /* Variables */

  int array_length;
  int* array;

  /* Initialize */

  init_array(&array);
  init(array, &array_length);

  free(array);

  return 0;

}

My question is: How can I initialize values of the array in a function init_array().

I have attempted things such as:

**(array+2) = 666;
*(*(array+2)) = 666;
*array[2] = 666;
**array[2] = 666;

When I used pencil and paper I came to result that **(array+2) should work but it gives me a segmentation fault.

I would appreciate your answer because I am confused how pointers in C actually work.

I'm familiar with C++ but I thing you need to use the new operator or what ever to declare an object and then create the values in the specified address. — pandoragami
– pandoragami, Commented May 24, 2017 at 14:52

dbush · Accepted Answer · 2017-05-24 14:58:35Z

3

You have the address of a pointer passed to the function:

array

You want to dereference that to get the pointer:

*array

Then apply the array subscript operator to the result:

(*array)[2]

Or equivalently:

*((*array) + 2)

The parenthesis are required because the array subscript operator [] has higher precedence than the dereference operator *.

Generally speaking, you should use the array subscript operator whenever possible, as it tends to be easier to read.

edited May 24, 2017 at 14:58

answered May 24, 2017 at 14:52

dbush

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