If they really want explicit mention of out parameters at the call site, they should actually require that instead of hacking around it by trying to make pointers mean something they don't. Pointers don't imply modification any more than references do, and it's not uncommon to pass pointers for non-modified objects.
One potential way to express out parameters explicitly:
template
struct Out {
explicit Out(T& obj) : base(obj) {}
T& operator*() { return base; }
T* operator->() { return &base; }
private:
T& base;
};
template
Out out(T& obj) {
return Out(obj);
}
void f(Out n) {
++*n;
}
int main() {
int n = 3;
f(out(n));
cout << n << '\n';
}
And as a temporary measure until they change old code to this, you can make the Out convertible to a pointer and/or reference:
// in class definition
operator T*() { return &base; }
operator T&() { return base; }
// elsewhere
void old(int *p);
void g() {
int n;
old(out(n));
}
I went ahead and wrote the various classes required for this, and for in-out parameters, in a way that should degrade nicely. I doubt I'll be using that convention any time soon (in C++, at least), but it'll work for anyone that wants to make call sites explicit.
