How do I apply a structure offset?

2019-06-25 00:58发布

问题:

I have a structure

typedef struct foo {
    int lengthOfArray1;
    int lengthOfArray2;
    int* array1;
    int* array2;
} foo;

I need to allocate enough memory for the entire structure and its array's contents. So assuming each array had a length of 5...

foo* bar = (foo*)malloc(sizeof(foo) + (sizeof(int) * 5) + (sizeof(int) * 5));

I now have to point array1 and array2 to the correct location in that allocated buffer:

bar->array1 = (int*)(&bar->lengthOfArray2 + sizeof(int));
bar->array2 = (int*)(bar->array1 + lengthOfArray2);

Is this correct?

Edit #1

Just to clear up any confusion: I am trying to keep the memory in one block, and not three.

Edit #2

I cannot use C99 as the MSVC 2010 compiler does not support it (http://stackoverflow.com/questions/6688895/does-microsoft-visual-studio-2010-supports-c99).

回答1:

Following the OP's approach this should do the job:

/* Defining these types allows to change the types without having the need to modify the code. */
typedef int Foo_ArrayElement1_t;
typedef int Foo_ArrayElement2_t;

typedef struct Foo_s {
    size_t lengthOfArray1; /* 'size_t' is the type of choice for array/memory dimensions. */
    size_t lengthOfArray2;
    Foo_ArrayElement1_t * array1;
    Foo_ArrayElement2_t * array2;
} Foo_t;

/*
 * Allocates memory to hold a structure of type Foo_t including size for 
 * 's1' elements referenced by 'array1' and 's2' elements by 'array2'.
 * The arrays' elements are set to 0.
 *
 * Returns a pointer to the freshly allocated memory or NULL if the memory could not 
 * be allocated.
 */
Foo_t * Foo_CreateAndInit(size_t s1, size_t s2)
{
  /* At once allocate all 'Foo_t' (including the memory Foo_t's array pointers shall point to). */
  Foo_t * pfoo = calloc(1,
      sizeof(*pfoo) +
      s1 * sizeof(*(pfoo->array1) + 
      s2 * sizeof(*(pfoo->array2)));
  if (pfoo)
  {
    pfoo->lengthOfArray1 = s1;
    pfoo->lengthOfArray2 = s2;

    /* The first array starts right after foo. */
    pfoo->array1 = (Foo_ArrayElement1_t *) (pfoo + 1); 

    /* The second array starts right after s1 elements of where the first array starts. */
    pfoo->array2 = (Foo_ArrayElement2_t *) (pfoo->array1 + s1); /* That casting here is not 
        necessaryas long as 'Foo_t.array1' and 'Foo_t.array2' point to the same type but makes 
        the code work even if those types were changed to be different. */
  }

  return pfoo;
}

...

Foo_t * foo = Foo_CreateAndInit(5, 5);


回答2:

You have to allocate the sizeof the structure. You then have to allocate the array of ints with their respective sizes.

foo* bar = malloc(sizeof(foo));

/* somewhere in here the array lengths are set then... */

bar->array1 = malloc(sizeof(int) * bar->lengthOfArray1);
bar->array2 = malloc(sizeof(int) * bar->lengthOfArray2);


回答3:

With a little extra memory (only for C99):

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

typedef struct foo {
    int lengthOfArray1;
    int lengthOfArray2;
    int *array1;
    int *array2;
    int array[];
} foo;

int main(void)
{
    foo *bar = malloc(sizeof(foo) + (sizeof(int) * 10));

    bar->array1 = &bar->array[0];
    bar->array2 = &bar->array[5]; /* 5 or lengthOfArray1 */

    return 0;
}