error: cast from 'void*' to 'int' loses precision

You can cast it to an intptr_t type. It's an int type guaranteed to be big enough to contain a pointer. Use #include <cstdint> to define it.

Again, all of the answers above missed the point badly. The OP wanted to convert a pointer value to a int value, instead, most the answers, one way or the other, tried to wrongly convert the content of arg points to to a int value. And, most of these will not even work on gcc4.

The correct answer is, if one does not mind losing data precision,

int x = *((int*)(&arg));

This works on GCC4.

The best way is, if one can, do not do such casting, instead, if the same memory address has to be shared for pointer and int (e.g. for saving RAM), use union, and make sure, if the mem address is treated as an int only if you know it was last set as an int.

Instead of:

int x = (int)arg;

use:

int x = (long)arg;

On most platforms pointers and longs are the same size, but ints and pointers often are not the same size on 64bit platforms. If you convert (void*) to (long) no precision is lost, then by assigning the (long) to an (int), it properly truncates the number to fit.

There's no proper way to cast this to int in general case. C99 standard library provides intptr_t and uintptr_t typedefs, which are supposed to be used whenever the need to perform such a cast comes about. If your standard library (even if it is not C99) happens to provide these types - use them. If not, check the pointer size on your platform, define these typedefs accordingly yourself and use them.