指针是表示内存位置的抽象。请注意，这句话并没有说把指针当作内存地址是错误的，它只是说它“通常会导致悲伤”。换句话说，它会让你产生错误的期望。

The most likely source of grief is certainly pointer arithmetic, which is actually one of C's strengths. If a pointer was an address, you'd expect pointer arithmetic to be address arithmetic; but it's not. For example, adding 10 to an address should give you an address that is larger by 10 addressing units; but adding 10 to a pointer increments it by 10 times the size of the kind of object it points to (and not even the actual size, but rounded up to an alignment boundary). With an int * on an ordinary architecture with 32-bit integers, adding 10 to it would increment it by 40 addressing units (bytes). Experienced C programmers are aware of this and put it to all kinds of good uses, but your author is evidently no fan of sloppy metaphors.

There's the additional question of how the contents of the pointer represent the memory location: As many of the answers have explained, an address is not always an int (or long). In some architectures an address is a "segment" plus an offset. A pointer might even contain just the offset into the current segment ("near" pointer), which by itself is not a unique memory address. And the pointer contents might have only an indirect relationship to a memory address as the hardware understands it. But the author of the quote cited doesn't even mention representation, so I think it was conceptual equivalence, rather than representation, that they had in mind.

在这幅图中，

Pointer_p是一个位于0x12345的指针，它指向0x34567的变量variable_v。

C或c++指针与简单内存地址的另一个不同之处是，我在其他答案中没有看到不同的指针类型(尽管考虑到它们的总大小，我可能忽略了它)。但它可能是最重要的一个，因为即使是经验丰富的C/ c++程序员也会被它绊倒:

编译器可能会假设不兼容类型的指针不指向相同的地址，即使它们很明显指向相同的地址，这可能会导致简单的pointer==address模型不可能出现的行为。考虑以下代码(假设sizeof(int) = 2*sizeof(short)):

unsigned int i = 0;
unsigned short* p = (unsigned short*)&i;
p[0]=p[1]=1;

if (i == 2 + (unsigned short)(-1))
{
  // you'd expect this to execute, but it need not
}

if (i == 0)
{
  // you'd expect this not to execute, but it actually may do so
}

注意，char*有一个例外，所以使用char*操作值是可能的(尽管不是很可移植)。

指针是一个保存内存地址的变量，而不是地址本身。但是，您可以解除对指针的引用-并访问内存位置。

例如:

int q = 10; /*say q is at address 0x10203040*/
int *p = &q; /*means let p contain the address of q, which is 0x10203040*/
*p = 20; /*set whatever is at the address pointed by "p" as 20*/

就是这样。就是这么简单。

一个演示我所说内容的程序，其输出如下:

http://ideone.com/rcSUsb

程序:

#include <stdio.h>

int main(int argc, char *argv[])
{
  /* POINTER AS AN ADDRESS */
  int q = 10;
  int *p = &q;

  printf("address of q is %p\n", (void *)&q);
  printf("p contains %p\n", (void *)p);

  p = NULL;
  printf("NULL p now contains %p\n", (void *)p);
  return 0;
}

C标准没有在内部定义指针是什么以及它在内部是如何工作的。这样做的目的是为了不限制平台的数量，在这些平台上，C可以作为编译或解释语言实现。

指针值可以是某种ID或句柄，也可以是几个ID的组合(对x86段和偏移量说你好)，不一定是真正的内存地址。这个ID可以是任何东西，甚至是固定大小的文本字符串。非地址表示可能对C解释器特别有用。

它说“因为它让那些不知道地址是什么的人感到困惑”——而且，这是真的:如果你知道地址是什么，你就不会困惑了。从理论上讲，指针是一个指向另一个变量的变量，实际上保存着一个地址，即它所指向的变量的地址。我不知道为什么要隐瞒这个事实，这又不是什么高深的科学。如果你理解了指针，你就离理解计算机的工作原理更近了一步。去吧!