如何取消对临时指针的引用？

Question

Here is a piece of C++ code that confuses me.

For a variable var, (int*)(&var) can get var's address and *(int*)(&var) can get var's value. However, if there is a pointer ptr point to var, WHY *(int*)(&ptr) cannot get var's address???

int var = 2;
cout << "var' address and value: " << &var << " " << var << endl;
cout << (int*)(&var) << endl;
cout << *(int*)(&var) << endl;
cout << (int*)*(int*)(&var) << endl;


int *ptr = &var;
cout << "ptr's address： " << &ptr << endl;
cout << (int*)(&ptr) << endl;
cout << *(int*)(&ptr) << endl;//NOT var's address, WHY?
cout << (int*)*(int*)(&ptr) << endl; //var's address,
cout << *(int*)*(int*)(&ptr) << endl; //var's value,

Answer 1

WhozCraig is write, it's just because you are casting a pointer to a int . If you are running on a 32bits machine, this may work fine (because pointers are on 8 bytes and int too), even if not recommended at all. But on 64bits machine it won't, as pointers ( int* , void* ) use 8 bytes, while int uses 4 bytes.

To illustrate, I'm using a 64bits machine, When I run your program as posted (using int ), I get:

var' address and value: 0x7ffd12c3dbd4 2
0x7ffd12c3dbd4
2
0x2
ptr's address： 0x7ffd12c3dbd8
0x7ffd12c3dbd8
314825684
0x12c3dbd4
Segmentation fault

When I replace all int by size_t (which is 64bits on my machine, and will be 32bits on a 32bits machine):

size_t var = 2;
cout << "var' address and value: " << &var << " " << var << endl;
cout << (size_t*)(&var) << endl;
cout << *(size_t*)(&var) << endl;
cout << (size_t*)*(size_t*)(&var) << endl;

size_t *ptr = &var;
cout << "ptr's address： " << &ptr << endl;
cout << (size_t*)(&ptr) << endl;
cout << *(size_t*)(&ptr) << endl;// var's address, OK
cout << (size_t*)*(size_t*)(&ptr) << endl; //var's address,
cout << *(size_t*)*(size_t*)(&ptr) << endl; //var's value, OK

I get:

var' address and value: 0x7ffd5c258de0 2
0x7ffd5c258de0
2
0x2
ptr's address： 0x7ffd5c258de8
0x7ffd5c258de8
140726149418464
0x7ffd5c258de0
2

No more segmentation fault, and *(size_t*)(&ptr) reports 140726149418464 which is equal to 0x7ffd5c258de0 .

Edit: See Marian Spanik comment, there may be other appropriate types than size_t . But what's sure, is that int is not appropriate.