c - Why no switch on pointers?

Question

For instance:

#include <stdio.h>

void why_cant_we_switch_him(void *ptr)
{
    switch (ptr) {
        case NULL:
            printf("NULL!
");
            break;
        default:
            printf("%p!
", ptr);
            break;
    }
}

int main(void)
{
    void *foo = "toast";
    why_cant_we_switch_him(foo);
    return 0;
}

gcc test.c -o test
test.c: In function 'why_cant_we_switch_him':
test.c:5: error: switch quantity not an integer
test.c:6: error: pointers are not permitted as case values

Just curious. Is this a technical limitation?

EDIT

People seem to think there is only one constant pointer expression. Is that is really true, though? For instance, here is a common paradigm in Objective-C (it is really only C aside from NSString, id and nil, which are merely a pointers, so it is still relevant —?I just wanted to point out that there is, in fact, a common use for it, despite this being only a technical question):

#include <stdio.h>
#include <Foundation/Foundation.h>

static NSString * const kMyConstantObject = @"Foo";

void why_cant_we_switch_him(id ptr)
{
    switch (ptr) {
        case kMyConstantObject: // (Note that we are comparing pointers, not string values.)
            printf("We found him!
");
            break;
        case nil:
            printf("He appears to be nil (or NULL, whichever you prefer).
");
            break;
        default:
            printf("%p!
", ptr);
            break;
    }
}

int main(void)
{
    NSString *foo = @"toast";
    why_cant_we_switch_him(foo);
    foo = kMyConstantObject;
    why_cant_we_switch_him(foo);

    return 0;
}

gcc test.c -o test -framework Foundation
test.c: In function 'why_cant_we_switch_him':
test.c:5: error: switch quantity not an integer
test.c:6: error: pointers are not permitted as case values

It appears that the reason is that switch only allows integral values (as the compiler warning said). So I suppose a better question would be to ask why this is the case? (though it is probably too late now.)

Answer 1

Because there is only one constant pointer expression

Given that only a single constant pointer expression exists, the switch statement has little to offer pointer expressions. You have cited essentially the only possible construction.