在Xcode中寻址C程序

Question

当我在Xcode中的64位Intel中编译下面的代码时,我得到了这个输出.

#include<stdio.h>
#include<limits.h>

int main(void)
{
  /* declare some integer variables */
  long a = LONG_MAX;
  long b = 2L;
  long c = 3L;

  /* declare some floating-point variables */
  double d = 4.0;
  double e = 5.0;
  double f = 6.0;

  printf("A variable of type long occupies %d bytes.", sizeof(long));
  printf("\nHere are the addresses of some variables of type long:");
  printf("\nThe address of a is: %p  The address of b is: %p", &a, &b);
  printf("\nThe address of c is: %p", &c);
  printf("\nThe address of a-b is: %ld\nvalue of a is %ld\nValue of b is %ld\nsize of pointer %ld ", (&a-&c),a,b,sizeof(&a));
  printf("\n\nA variable of type double occupies %d bytes.", sizeof(double));
  printf("\nHere are the addresses of some variables of type double:");
  printf("\nThe address of d is: %p  The address of e is: %p", &d, &e);
  printf("\nThe address of f is: %p\n", &f);

    printf("\n size long - %d", sizeof(a));
  return 0;
}

A variable of type long occupies 8 bytes.

Here are the addresses of some variables of type long:

The address of a is: 0x7fff5fbff880 
The address of b is: 0x7fff5fbff878 
The address of c is: 0x7fff5fbff870 
The address of a-b is: 2

value of a is 9223372036854775807 
Value of b is 2 
size of pointer 8 

A variable of type double occupies 8 bytes.

Here are the addresses of some variables of type double:

The address of d is: 0x7fff5fbff868 
The address of e is: 0x7fff5fbff860 
The address of f is: 0x7fff5fbff858 
size long - 8

对我来说奇怪的是,地址a和之间的区别b仅为2.我希望它为8,这与长的字节数相匹配.有谁知道为什么会这样？

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我在代码中输入了一个拼写错误&a-&c,但这确实与我的问题无关.我的问题是,为什么从变量a的地址到变量b的地址只有2个字节的差异,当long为8个字节长时,我希望看到8的差异？

Answer 1

指针算法基于它指向的类型的大小而不是以字节为单位,这个Pointer Arithmetic的参考很好地涵盖了这个主题,你也有一个错字:

(&a-&c)

你实际上减去c从a.

这也是因为如果指针指向相同的数组指针减法只被定义,见未定义的行为6.5.6/9从C11 draft standard:

[...]当减去两个指针时,两个指针都指向同一个数组对象的元素,或者指向数组对象的最后一个元素的元素; [...]

部分6.5.6/8也是相关的:

[...]如果指针操作数和结果都指向同一个数组对象的元素,或者指向数组对象的最后一个元素,则评估不应产生溢出; 否则,行为未定义.[...]