[SOLVED] If-then-else vs ternary operator when returning full or empty std::optional

If-then-else vs ternary operator when returning full or empty std::optional

(I've not found much by searching for return statement, return deduce, and similar, with tags c++optional.)

Why does this work

#include <optional>

auto const f = [](bool b) {
    return b ? std::optional<int>(3) : std::nullopt;
};

while this doesn't?

#include <optional>

auto const f = [](bool b) {
    if (b) {
        return std::optional<int>(3);
    } else {
        return std::nullopt;
    }
};

Why can't the compiler deduce the type from the first return and see it's compatiple with the second return?

Solution

Lambda return type deduction require the type of all return expressions match basically exactly.

? does a relatively complex system to find a common type of the two cases. There is only one return statement, so so long as ? can figure it out, lambda return type deduction doesn't care.

Just different rules.

auto const f = [](bool b)->std::optional<int> {
  if (b) {
    return 3;
  } else {
    return std::nullopt;
  }
};

this is probably the cleanest.