This is a self triggered question based on a self-answer I gave here.
This seems a pretty convincing explanation of why short-circuiting of logical operators is available in fold expressions, and of the fact that wrapping a fold expression in a function with a variadic argument seems to be non short-circuting (in fact, the answer explains, it's the function call which triggers the evaluation of all arguments, before the short-circuit can take place inside the function body).
However, the following code seems to me, proves that (at least when the arguments in a fold expression are 2) the short-circuiting doesn't happen:
#include <assert.h>
#include <optional>
constexpr auto all_r = [](auto const& ... ps){
return [&ps...](auto const& x){
return (ps(x) && ...);
};
};
constexpr auto all_l = [](auto const& ... ps){
return [&ps...](auto const& x){
return (... && ps(x));
};
};
constexpr auto has_value = [](std::optional<int> o){
return o.has_value();
};
constexpr auto has_positive = [](std::optional<int> o){
assert(o.has_value());
return o.value() > 0;
};
int main() {
assert(!(has_value(std::optional<int>{}) && has_positive(std::optional<int>{})));
//assert(!(has_positive(std::optional<int>{}) && has_value(std::optional<int>{}))); // expectedly fails at run-time
assert(!all_r(has_value, has_positive)(std::optional<int>{}));
assert(!all_l(has_value, has_positive)(std::optional<int>{})); // I expected this to fail at run-time
//assert(!all_r(has_positive, has_value)(std::optional<int>{}));
//assert(!all_l(has_positive, has_value)(std::optional<int>{})); // I expected this to succeed at run-time
}
