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.

Come to think about it, I think it's a matter of semantics. I don't think the author is right, since the C standard refers to a pointer as holding an address to the referenced object as others have already mentioned here. However, address!=memory address. An address can be really anything as per C standard although it will eventually lead to a memory address, the pointer itself can be an id, an offset + selector (x86), really anything as long as it can describe (after mapping) any memory address in the addressable space.

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)。

很难确切地说出这些书的作者到底是什么意思。指针是否包含地址取决于如何定义地址和如何定义指针。

从所有的回答来看，有些人认为(1)地址必须是整数，(2)指针不需要是虚的，因为规范中没有这么说。根据这些假设，显然指针不一定包含地址。

然而，我们看到，虽然(2)可能是真的，(1)可能不一定是真的。根据@ corn秸秆的答案，&被称为操作符的地址，这是怎么回事?这是否意味着规范的作者希望指针包含地址?

我们可以说，指针包含一个地址，但地址不一定是整数?也许吧。

我认为所有这些都是废话连篇的迂腐语义学。实际上，这是毫无价值的。你能想到一个编译器以这样的方式生成代码，指针的值不是一个地址吗?如果有，是什么?我也是这么想的……

我认为这本书的作者(第一个摘录声称指针不一定只是地址)可能指的是指针自带固有类型信息这一事实。

例如,

 int x;
 int* y = &x;
 char* z = &x;

y和z都是指针，但y+1和z+1是不同的。如果它们是内存地址，难道这些表达式不会给你相同的值吗?

在这里，把指针当作地址的想法通常会导致悲伤。之所以会出现bug，是因为人们将指针视为地址，而这通常会导致不幸。

55555可能不是指针，尽管它可能是一个地址，但(int*)55555是一个指针。55555+1 = 55556，但是(int*)55555+1是55559(在sizeof(int)方面的+/-差异)。

把指针看作地址是一种近似。像所有的近似值一样，它有时足够有用，但也不准确，这意味着依赖它会带来麻烦。

指针就像一个地址，它指出在哪里可以找到一个对象。这种类比的一个直接限制是，并非所有指针都实际包含地址。NULL是一个指针，它不是地址。指针变量的内容实际上可以是以下三种类型之一:

对象的地址，可以被解引用(如果p包含x的地址，则表达式*p与x的值相同); 一个空指针，null是一个例子; 无效内容，不指向对象(如果p不持有有效值，则*p可以做任何事情(“未定义行为”)，导致程序崩溃是相当常见的可能性)。

此外，更准确的说法是，一个指针(如果有效且非空)包含一个地址:指针指出在哪里可以找到一个对象，但还有更多与之相关的信息。

In particular, a pointer has a type. On most platforms, the type of the pointer has no influence at runtime, but it has an influence that goes beyond the type at compile time. If p is a pointer to int (int *p;), then p + 1 points to an integer which is sizeof(int) bytes after p (assuming p + 1 is still a valid pointer). If q is a pointer to char that points to the same address as p (char *q = p;), then q + 1 is not the same address as p + 1. If you think of pointer as addresses, it is not very intuitive that the “next address” is different for different pointers to the same location.

It is possible in some environments to have multiple pointer values with different representations (different bit patterns in memory) that point to the same location in memory. You can think of these as different pointers holding the same address, or as different addresses for the same location — the metaphor isn't clear in this case. The == operator always tells you whether the two operands are pointing to the same location, so on these environments you can have p == q even though p and q have different bit patterns.

甚至在某些环境中，指针携带除地址以外的其他信息，例如类型或权限信息。作为一名程序员，你很容易在生活中不会遇到这些问题。

在某些环境中，不同类型的指针具有不同的表示形式。你可以把它想象成不同类型的地址有不同的表示。例如，一些体系结构有字节指针和字指针，或者对象指针和函数指针。

总而言之，只要记住这一点，将指针视为地址并不太糟糕

它只有有效的，非空的地址指针; 同一个位置可以有多个地址; 你不能对地址进行算术运算，地址上也没有顺序; 指针还携带类型信息。

反过来就麻烦多了。并不是所有看起来像地址的东西都可以是指针。在深层的某个地方，任何指针都表示为可以作为整数读取的位模式，并且您可以说这个整数是一个地址。但反过来说，不是每个整数都是指针。

首先有一些众所周知的限制;例如，在程序地址空间之外指定位置的整数不能是有效指针。未对齐的地址不能为需要对齐的数据类型创建有效指针;例如，在int需要4字节对齐的平台上，0x7654321不能是有效的int*值。

然而，它远远不止于此，因为当您将指针设置为整数时，您就会遇到很多麻烦。这个问题的很大一部分是优化编译器在微优化方面比大多数程序员预期的要好得多，因此他们对程序如何工作的思维模型是严重错误的。仅仅因为指针具有相同的地址并不意味着它们是等价的。例如，考虑下面的代码片段:

unsigned int x = 0;
unsigned short *p = (unsigned short*)&x;
p[0] = 1;
printf("%u = %u\n", x, *p);

您可能会期望，在sizeof(int)==4和sizeof(short)==2的普通机器上，这要么打印1 = 1?(little-endian)还是65536 = 1?(大端)。但在我的64位Linux PC上，GCC 4.4:

$ c99 -O2 -Wall a.c && ./a.out 
a.c: In function ‘main’:
a.c:6: warning: dereferencing pointer ‘p’ does break strict-aliasing rules
a.c:5: note: initialized from here
0 = 1?

在这个简单的例子中，GCC会提醒我们哪里出了问题——在更复杂的例子中，编译器可能不会注意到。由于p与&x的类型不同，改变p指向的对象不会影响&x指向的对象(除了一些定义良好的异常)。因此，编译器可以自由地将x的值保存在寄存器中，而不会在*p更改时更新该寄存器。程序解引用两个指向相同地址的指针，得到两个不同的值!

The moral of this example is that thinking of a (non-null valid) pointer as an address is fine, as long as you stay within the precise rules of the C language. The flip side of the coin is that the rules of the C language are intricate, and difficult to get an intuitive feeling for unless you know what happens under the hood. And what happens under the hood is that the tie between pointers and addresses is somewhat loose, both to support “exotic” processor architectures and to support optimizing compilers.

因此，可以将指针作为地址作为理解的第一步，但不要过于遵循这种直觉。