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

user_5825294 0
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(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?

return-type-deduction option-type c++17 return c++
1 Answers

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.

user_17746670
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