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;
}

以下是我过去是如何向一些困惑的人解释的: 指针有两个影响其行为的属性。它有一个值(在典型环境中)是一个内存地址，还有一个类型(告诉您它所指向的对象的类型和大小)。

例如，给定:

union {
    int i;
    char c;
} u;

你可以有三个不同的指针都指向同一个对象:

void *v = &u;
int *i = &u.i;
char *c = &u.c;

如果你比较这些指针的值，它们都是相等的:

v==i && i==c

但是，如果对每个指针加1，就会发现它们所指向的类型变得相关了。

i++;
c++;
// You can't perform arithmetic on a void pointer, so no v++
i != c

此时，变量i和c将具有不同的值，因为i++使i包含下一个可访问的整数的地址，而c++使c指向下一个可寻址的字符。通常，整数比字符占用更多的内存，所以在它们都加一之后，i的值将比c的值更大。

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.

在理解指针之前，我们需要先理解对象。对象是存在的实体，具有一个称为地址的位置说明符。指针与C语言中的其他变量一样，是一个类型为指针的变量，其内容被解释为支持以下操作的对象的地址。

+ : A variable of type integer (usually called offset) can be added to yield a new pointer
- : A variable of type integer (usually called offset) can be subtracted to yield a new pointer
  : A variable of type pointer can be subtracted to yield an integer (usually called offset)
* : De-referencing. Retrieve the value of the variable (called address) and map to the object the address refers to.
++: It's just `+= 1`
--: It's just `-= 1`

指针是根据它当前引用的对象类型进行分类的。唯一重要的信息是物体的大小。

任何对象都支持& (address of)操作，该操作将对象的位置说明符(地址)作为指针对象类型检索。这将减少围绕命名的混乱，因为调用&作为对象的操作而不是作为结果类型为对象类型的指针的指针是有意义的。

注意:在整个解释中，我省略了内存的概念。

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。