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

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.

其他回答

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。

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。

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)。

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!

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。