c++ - 如何阅读这些表达 : *&pointer VS &*pointer

如果我有:

int j = 8;
int *pointer = &j;

那么如果我这样做:

&*pointer == *&pointer

返回 1 ( true )。

但我对第二个表达有疑问:

&*pointer 返回指针指向的地址(首先求值 *
然后 &)

*&pointer 返回指针地址，然后它指向什么......但这是变量而不是地址。所以这是我的疑问...

最佳答案

pointer “指向”内存中的某个地址；它驻留在内存中的某个其他地址。

&*pointer // (*pointer) - dereference `pointer`, now you have `j`
          // &(*pointer) - the address of `j`(that's the data that `pointer` has)

然而:

*&pointer //(&pointer) - the address of pointer(where pointer resides in memory)
          // *(&pointer) - deference that address and you get `pointer`

我总是发现用图片更容易追踪指针，所以也许这个插图将有助于理解为什么它们是相同的:

//In case of &*pointer, we start with the pointer, the dereference it giving us j
//Then taking the address of that brings us back to pointer:

                                           +--&(*pointer)-----------+
                                           |                        |
memory address            0x7FFF3210       |            0x7FFF0123  |
                        +------------+     |             +-----+    |
data present            | pointer =  | <---+        +->  | j=8 |----+
                        | 0x7FFF0123 | ->(*pointer)-+    +-----+
                        +------------+

//in the *&pointer case, we start with the pointer, take the address of it, then
//dereference that address bring it back to pointer


memory address           +------------>  0x7FFF3210 ----*(&pointer)--+
                         |                                           |
                         |              +------------+               |
data present             |              | pointer =  | <----------- -+
                         +--&pointer ---| 0x7FFF0123 |
                                        +------------+

关于c++ - 如何阅读这些表达 : *&pointer VS &*pointer，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/15882272/