C++ 如何找出C中分配在堆上的数组的大小

我有以下C代码:

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

#define ARRAYSIZE(a) (sizeof(a))/(sizeof(a[0]))
#define ARRAYSIZEWITHPTR(a) (&(a)[1] - a)

int main (void){

  int array[5] = {1,2,3,4,5};
  int* pArray = array;
  int* p = calloc(ARRAYSIZE(array), sizeof(int));
  
  assert(NULL != p);
  
  printf("size of array: %ld\n", ARRAYSIZE(array));
  printf("size of pArray: %ld\n", ARRAYSIZE(pArray));
  printf("size of p: %ld\n", ARRAYSIZE(p));
  printf("size of pArray: %ld\n", ARRAYSIZEWITHPTR(pArray));
  printf("size of p: %ld\n", ARRAYSIZEWITHPTR(p));

  free(p);

return 0;
}

当我编译这个程序时，我从GCC那里收到了以下警告消息:

test.c: In function ‘main’:
test.c:5:33: warning: division ‘sizeof (int *) / sizeof (int)’ does not compute the number of array elements [-Wsizeof-pointer-div]
    5 | #define ARRAYSIZE(a) (sizeof(a))/(sizeof(a[0]))
      |                                 ^
test.c:15:35: note: in expansion of macro ‘ARRAYSIZE’
   15 |   printf("size of pArray: %ld\n", ARRAYSIZE(pArray));
      |                                   ^~~~~~~~~
test.c:11:8: note: first ‘sizeof’ operand was declared here
   11 |   int* pArray = array;
      |        ^~~~~~
test.c:5:33: warning: division ‘sizeof (int *) / sizeof (int)’ does not compute the number of array elements [-Wsizeof-pointer-div]
    5 | #define ARRAYSIZE(a) (sizeof(a))/(sizeof(a[0]))
      |                                 ^
test.c:16:30: note: in expansion of macro ‘ARRAYSIZE’
   16 |   printf("size of p: %ld\n", ARRAYSIZE(p));
      |                              ^~~~~~~~~
test.c:12:8: note: first ‘sizeof’ operand was declared here
   12 |   int* p = calloc(ARRAYSIZE(array), sizeof(int));
      |        ^

以下是程序输出:

size of array: 5
size of pArray: 2
size of p: 2
size of pArray: 1
size of p: 1
 

从程序输出中可以明显看出，只有在以数组名称作为参数调用宏时，我才能获得正确的值.然而，数组名称不是指向数组的第一个元素的指针吗？如果是这样的话，使用数组名称调用宏(这样会产生正确的结果)与指向堆上的数组的指针(不会产生正确的结果)之间有什么本质的区别？

从源代码中可以看出，我try 了两种宏方法来获取在堆上创建的数组的大小，但都没有成功.

那么下一个问题是，在C编程中使用calloc()在堆上这样创建的数组的大小是不是可能的？

提亚