The following code fails to compile if you uncomment the commented line. (Compiler Explorer)
Why is that?
I imagine the reason is that the iota_view of Foos doesn't know how to advance a Foo. But how do I make it know?
#include <iostream>
#include <ranges>
struct Foo {
int x;
Foo() : x{-1} {}
Foo(int x) : x{x} {}
using difference_type = int;
friend auto operator<=>(Foo const& a, Foo const& b) {
return a.x <=> b.x;
}
friend bool operator==(Foo const& a, Foo const& b) {
return a.x == b.x;
}
auto& operator++() {
++x;
return *this;
}
auto operator++(int) {
auto r = *this;
++x;
return r;
}
};
int main () {
auto ints = std::views::iota(1, 30);
for (auto i : ints) {}
std::cout << ints[15] << std::endl;
auto foos = std::views::iota(Foo{1}, Foo{30});
for (auto i : foos) {}
//std::cout << foos[15] << std::endl;
}
The example uses std-ranges, but I'm interested in an answer for range-v3 too, should they differ.
Take a look at this page: https://en.cppreference.com/w/cpp/ranges/iota_view/iterator
To be a random access range, the type needs to be sufficiently "advanceable". This requires operator+, operator+=, operator- and operator-= (as well as operator-- to be "decrementable", and to be totally ordered which you already have with operator<=>)
Advancing an iterator by n will just add n to the current value with operator+= or operator+.
Example implementation:
struct Foo {
// ...
Foo& operator--() { --x; return *this; }
Foo operator--(int) { auto r = *this; --*this; return r; }
friend Foo operator+(Foo l, difference_type r) { l += r; return l; }
friend Foo operator+(difference_type l, Foo r) { r += l; return r; }
friend Foo operator-(Foo l, difference_type r) { l -= r; return l; }
friend difference_type operator-(Foo l, Foo r) { return l.x - r.x; }
Foo& operator+=(difference_type r) { x += r; return *this; }
Foo& operator-=(difference_type r) { x -= r; return *this; }
};
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