指针是一种在C/ c++中本地可用的变量类型，包含一个内存地址。像任何其他变量一样，它有自己的地址并占用内存(数量是特定于平台的)。

由于混淆，您将看到的一个问题是试图通过简单地按值传递指针来更改函数中的引用。这将复制函数作用域内的指针，对这个新指针“指向”的地方的任何更改都不会改变调用该函数的作用域内指针的引用。为了修改函数中的实际指针，通常会将一个指针传递给另一个指针。

A pointer value is an address. A pointer variable is an object that can store an address. This is true because that's what the standard defines a pointer to be. It's important to tell it to C novices because C novices are often unclear on the difference between a pointer and the thing it points to (that is to say, they don't know the difference between an envelope and a building). The notion of an address (every object has an address and that's what a pointer stores) is important because it sorts that out.

然而，标准在特定的抽象层次上进行讨论。作者所说的那些“知道地址是关于什么的”，但对C不熟悉的人，必须在不同的抽象级别上学习地址——也许是通过编写汇编语言。不能保证C实现使用与cpu操作码相同的地址表示(在本文中称为“存储地址”)，这些人已经知道。

He goes on to talk about "perfectly reasonable address manipulation". As far as the C standard is concerned there's basically no such thing as "perfectly reasonable address manipulation". Addition is defined on pointers and that is basically it. Sure, you can convert a pointer to integer, do some bitwise or arithmetic ops, and then convert it back. This is not guaranteed to work by the standard, so before writing that code you'd better know how your particular C implementation represents pointers and performs that conversion. It probably uses the address representation you expect, but it it doesn't that's your fault because you didn't read the manual. That's not confusion, it's incorrect programming procedure ;-)

简而言之，C使用了比作者更抽象的地址概念。

The author's concept of an address of course is also not the lowest-level word on the matter. What with virtual memory maps and physical RAM addressing across multiple chips, the number that you tell the CPU is "the store address" you want to access has basically nothing to do with where the data you want is actually located in hardware. It's all layers of indirection and representation, but the author has chosen one to privilege. If you're going to do that when talking about C, choose the C level to privilege!

Personally I don't think the author's remarks are all that helpful, except in the context of introducing C to assembly programmers. It's certainly not helpful to those coming from higher level languages to say that pointer values aren't addresses. It would be far better to acknowledge the complexity than it is to say that the CPU has the monopoly on saying what an address is and thus that C pointer values "are not" addresses. They are addresses, but they may be written in a different language from the addresses he means. Distinguishing the two things in the context of C as "address" and "store address" would be adequate, I think.

指针是一种在C/ c++中本地可用的变量类型，包含一个内存地址。像任何其他变量一样，它有自己的地址并占用内存(数量是特定于平台的)。

由于混淆，您将看到的一个问题是试图通过简单地按值传递指针来更改函数中的引用。这将复制函数作用域内的指针，对这个新指针“指向”的地方的任何更改都不会改变调用该函数的作用域内指针的引用。为了修改函数中的实际指针，通常会将一个指针传递给另一个指针。

指针只是另一个变量，用来保存内存位置的地址(通常是另一个变量的内存地址)。

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)。

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。