在关于C的一个有信誉的来源中,在讨论&操作符后给出了以下信息:

... 有点不幸的是,术语[地址的]仍然存在,因为它混淆了那些不知道地址是关于什么的人,并误导了那些知道地址的人:将指针视为地址通常会导致悲伤……

我读过的其他材料(来自同样有名望的来源,我想说)总是毫不掩饰地将指针和&操作符作为内存地址。我很愿意继续寻找事情的真相,但当有信誉的消息来源不同意时,这有点困难。

现在我有点困惑了——如果指针不是内存地址,那么它到底是什么?

P.S.

作者后来说:……不过,我将继续使用“地址”这个术语,因为发明一个不同的(术语)会更糟糕。


当前回答

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。

其他回答

A pointer, like any other variable in C, is fundamentally a collection of bits which may be represented by one or more concatenated unsigned char values (as with any other type of cariable, sizeof(some_variable) will indicate the number of unsigned char values). What makes a pointer different from other variables is that a C compiler will interpret the bits in a pointer as identifying, somehow, a place where a variable may be stored. In C, unlike some other languages, it is possible to request space for multiple variables, and then convert a pointer to any value in that set into a pointer to any other variable within that set.

Many compilers implement pointers by using their bits store actual machine addresses, but that is not the only possible implementation. An implementation could keep one array--not accessible to user code--listing the hardware address and allocated size of all of the memory objects (sets of variables) which a program was using, and have each pointer contain an index into an array along with an offset from that index. Such a design would allow a system to not only restrict code to only operating upon memory that it owned, but also ensure that a pointer to one memory item could not be accidentally converted into a pointer to another memory item (in a system that uses hardware addresses, if foo and bar are arrays of 10 items that are stored consecutively in memory, a pointer to the "eleventh" item of foo might instead point to the first item of bar, but in a system where each "pointer" is an object ID and an offset, the system could trap if code tried to index a pointer to foo beyond its allocated range). It would also be possible for such a system to eliminate memory-fragmentation problems, since the physical addresses associated with any pointers could be moved around.

Note that while pointers are somewhat abstract, they're not quite abstract enough to allow a fully-standards-compliant C compiler to implement a garbage collector. The C compiler specifies that every variable, including pointers, is represented as a sequence of unsigned char values. Given any variable, one can decompose it into a sequence of numbers and later convert that sequence of numbers back into a variable of the original type. Consequently, it would be possible for a program to calloc some storage (receiving a pointer to it), store something there, decompose the pointer into a series of bytes, display those on the screen, and then erase all reference to them. If the program then accepted some numbers from the keyboard, reconstituted those to a pointer, and then tried to read data from that pointer, and if user entered the same numbers that the program had earlier displayed, the program would be required to output the data that had been stored in the calloc'ed memory. Since there is no conceivable way the computer could know whether the user had made a copy of the numbers that were displayed, there would be no conceivable may the computer could know whether the aforementioned memory might ever be accessed in future.

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;
}

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.

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.

指针只是另一个变量,它通常包含另一个变量的内存地址。指针是一个变量,它也有一个内存地址。