4

我有一个结构

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

我需要为整个结构及其数组的内容分配足够的内存。因此，假设每个数组的长度为 5 ...

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

我现在必须将 array1 和 array2 指向分配的缓冲区中的正确位置：

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

它是否正确？

编辑#1

只是为了消除任何混乱：我试图将内存保持在一个块中，而不是三个。

编辑#2

我不能使用 C99，因为 MSVC 2010 编译器不支持它 (http://stackoverflow.com/questions/6688895/does-microsoft-visual-studio-2010-supports-c99)。

4

3 回答 3

4

您必须分配结构的大小。然后，您必须分配具有各自大小的整数数组。

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);
于 2012-10-14T05:04:48.343 回答
3

按照OP的方法，这应该可以完成工作：

/* 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);
于 2012-10-14T15:12:17.497 回答
1

有一点额外的内存（仅适用于 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;
}
于 2012-10-14T07:38:43.417 回答