Actually, no. These tag bits are usually stored in the lowest bits which are zero for all pointers (you would be mad not to align your data structurs to the four or eight byte boundaries your hardware uses for memory access). So you get the full width for pointers, but reduced width for your fixnums, because you have to set one of the least significant bits of the machine word to one to distinguish it from a pointer. That you still can't use the full 4GB of a 32 Bit address space is due to the fact that the OS needs some address space for itself, the details of this vary from OS to OS and what the runtime of your language does with the addresses the OS allows it to use. So beeing able to use more than 2GB on a 32 bit architecture should not be taken for granted.

Not to mention that many languages written in C use unions to describe their primitive -type- [ed: object]. The result is that the minimum number of bytes for storing an integer for instance, is the minimum number of bytes that can store a value of the largest type.

For example, if you have an string type, which keeps track of it's length, then you might need 8 bytes. 4 for the pointer to the string of chars, 4 for the integer to keep count.

Here's a better example from tinyscheme:

    struct cell {
      unsigned int _flag;
      union {
        struct {
          char   *_svalue;
          int   _length;
        } _string;
        num _number;
        port *_port;
        foreign_func _ff;
        struct {
          struct cell *_car;
          struct cell *_cdr;
        } _cons;
      } _object;
    };

As a minimum, each object takes up max(sizeof(_string), sizeof(num), sizeof(port), sizeof(_cons), sizeof(foreign_func)); And num is defined as follows:

    typedef struct num {
       char is_fixnum;
       union {
          long ivalue;
          double rvalue;
       } value;
    } num;