c - 在 C 中使用 calloc 时，指针地址中存储的是什么？

Question

我很难理解这个程序来说明指针(来自 http://theocacao.com/document.page/234 ):

下面我不明白为什么:

int * currentSlot = memoryBlock

没有使用&memoryBlock。我读了评论，但不明白。什么是 memoryBlock 放在那里而 &memoryBlock 不会？不会都返回指向用 calloc 创建的整数集的指针(假设我了解已经完成的工作)？ calloc 之后的* memoryBlock 中到底是什么？

然后在这里，*currentSlot = rand();，这里的解引用是如何工作的？我认为取消引用会阻止 *currentSlot 将内存地址(引用)的值赋予实际值(不再是引用而是值)。

#include <stdio.h>
#include <stdlib.h> // for calloc and free
#include <time.h>   // for random seeding

main ()
{
  const int count = 10; 
  int * memoryBlock = calloc ( count, sizeof(int) );

  if ( memoryBlock == NULL )
  {
    // we can't assume the memoryBlock pointer is valid.
    // if it's NULL, something's wrong and we just exit
    return 1;
  }

  // currentSlot will hold the current "slot" in the,
  // array allowing us to move forward without losing
  // track of the beginning. Yes, C arrays are primitive
  //
  // Note we don't have to do '&memoryBlock' because
  // we don't want a pointer to a pointer. All we
  // want is a _copy_ of the same memory address

  int * currentSlot = memoryBlock;


  // seed random number so we can generate values
  srand(time(NULL));

  int i;    
  for ( i = 0; i < count; i++ )
  {
    // use the star to set the value at the slot,
    // then advance the pointer to the next slot    
    *currentSlot = rand();
    currentSlot++; 
  }

  // reset the pointer back to the beginning of the
  // memory block (slot 0)
  currentSlot = memoryBlock;

  for ( i = 0; i < count; i++ )
  {
    // use the star to get the value at this slot,
    // then advance the pointer
    printf("Value at slot %i: %i\n", i, *currentSlot);
    currentSlot++;
  }

  // we're all done with this memory block so we
  // can free it  
  free( memoryBlock );
}

感谢您的帮助。

Answer 1

Below I don't understand why:

int * currentSlot = memoryBlock

isn't using &memoryBlock.

因为memoryBlock和currentSlot都是指向int的指针。 &memoryBlock 将是指向 int 的指针的地址，即 int **。

“在”memoryBlock 中的是指向内存块的指针。

Then here, *currentSlot = rand();, how does the dereferencing work here?

这是 C 的规则:当像这样的解引用表达式出现在表达式的左侧时，右侧的值存储在被解引用的指针指向的内存位置。