Why can I not assign a memory address number to a pointer?

Question

We know a pointer has a name and a value, the essence is variable, the value is hexadecimal number which is a memory address.

int a = 10;
int * p = &a;

cout << p << endl;  // 0x7ffee04fe2b8

int * q = 0x7ffee04fe2b8;  // why can not assign address number to the pointer `q`?

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If I assign 0x0, it works:

int * q = 0x0;

The 0x0 refer to NULL, whether because of the interpreter take special handling.

I use CLion editor.

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EDIT

I tried to declare as long, got issue too:

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EDIT

I use 64-bit macOS.

The pointer value stores a memory address, so I want to try to assign the address directly. Why can I not use this method?

int a = 10;
int * p = &a;

cout << p << endl;  // 0x7ffee04fe2b8

long q = 0x7ffee04fe2b8;
cout << * (int *)q << endl;  // prints nothing

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EDIT

I tried this post: Pointer to a specific fixed address:

int a = 10;
int * p = &a;

cout << p << endl;  // 0x7ffee04fe2b8

volatile unsigned int *myPointer = (volatile unsigned int *)0x7ffee04fe2b8;

cout << *myPointer << endl;  // print nothing

But why does it not print anything?

It builds successful, but when I run it, it prints nothing with *myPointer.

====================[ Build | untitled | Debug ]================================
/Applications/CLion.app/Contents/bin/cmake/mac/bin/cmake --build /Users/luowensheng/CLionProjects/untitled/cmake-build-debug --target untitled -- -j 4
[100%] Built target untitled

Build finished

Answer 1

In the C language you CAN assign addresses from constants. It is not a good idea. But you can do it.

Example compiler error:

$ cat c-ptr-test.c
int main() {
  long *p = 0x0011223344556677;
  return 0;
}
$ make c-ptr-test
cc -Wall -W -pedantic -g -O3 -flto -fno-fat-lto-objects -pthread -MMD  -std=c11   -pthread -flto -fno-fat-lto-objects  c-ptr-test.c   -o c-ptr-test
c-ptr-test.c: In function ‘main’:
c-ptr-test.c:2:13: warning: initialization of ‘long int *’ from ‘long int’ makes pointer from integer without a cast [-Wint-conversion]
    2 |   long *p = 0x0011223344556677;
      |             ^~~~~~~~~~~~~~~~~~
c-ptr-test.c:2:9: warning: unused variable ‘p’ [-Wunused-variable]
    2 |   long *p = 0x0011223344556677;
      |         ^

Two warnings but GCC compiled it successfully.

However, in the C++ language it requires an explicit cast. You're supposed to use reinterpret_cast but an old C style cast will also work.

Answer 2

It seems you are trying to assign pointer while not telling compiler you "want" to assign it.

Your compiler error should looks like this.

invalid conversion from 'long int' to 'int*'

To assign the addres, the syntax should be int *q = (int*)0x7ffee04fe2b8;,while (int*) means "this value is an address of pointer int*.

Or a more modern way : int * q = reinterpret_cast<int*>( 0x7ffebfc7019c );

However you must to know what are you doing. Unless you're coding low level code with hardware registers, never hardcoded address.

Also everytime you execute the code, its address might not be the same as last time.

Try following code: or on GCC 10.1 https://godbolt.org/z/8zn7j5

int main()
{
   int a = 10;
    int * p = &a;

    cout << p << endl;

    long addr = (long)p;
    int * q = reinterpret_cast<int*>( addr );
    cout << *q << endl;
    return 0;
}