On Wednesday, June 26, 2019, 11:53:30 AM PDT, Chris Angelico
<ros...@gmail.com> wrote:
> I don't think you can define what "f->spam" means
Well, you can, but only the first half of what it means, so I don't think this
changes your point.
If f is a pointer, then f->spam means to dereference the pointer f, then access
the spam attribute of the result. If f is an instance of a user-defined class,
rather than a raw pointer, it can redefine ->. But only to change the
"dereference" part, not the "access the spam attribute" part. In other words,
f->spam means something like (*(f.operator->())).spam and you can overload the
operator->.
Of course, under very restricted conditions, you can do something like this:
#include <iostream>
struct pstatus;
struct status { int eggs, spam; }
struct pstatus { status *p; status* operator->() {
return reinterpret_cast<status*>(reinterpret_cast<char*>(p) - sizeof(int));
} };
pstatus frob() { return pstatus{new status{1, 2}}; }
int main() { auto f = frob(); // eggs = 1, spam = 2 std::cout
<< f->spam; // accesses eggs, and prints 1 rather than 2 return 0; }
Compile that with g++ or clang++ with -std=c++11 (we don't actually need any
C++11 feature here, the code's just a bit shorter and simpler with auto, etc.),
and it should compile without warnings, and print out 1 rather than 2 when you
run it.
The trick is that "access the spam attribute" actually means "access bytes 4-8
of the struct as an int", so if we have a pointer to 4 bytes before the start
of the real struct, you get the int at bytes 0-4 of the real struct, which is
the real eggs.
(Actually, we should be using the difference between offsetof(spam) and
offsetof(eggs), not sizeof(int), so it would be legal even with non-default
alignment. But so many tiny things could turn this code into undefined
behavior, even with that change, like just adding a virtual method to status,
or referencing f->eggs in code that doesn't even run… so let's not worry about
bulletproofing it.) _______________________________________________
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