Chaining three optional stages

#include #include int main() { std::optional a = 5; std::optional b = std::nullopt; std::optional c = 10; auto result = a.and_then([](int x) { return std::optional(x * 2); }).and_then([](int x) { return std::optional(x + 3); }).and_then([](int x) { return std::optional(x * 2); }); if (result) { std::cout << result.value() << std::endl; } else { std::cout << "nullopt" << std::endl; } return 0; } ```

Understanding C++ optional chaining with and_then

This is a medium-difficulty C++ language question that tests your grasp of optional monadic operations — specifically the and_then combinator and how it propagates values through a chain of transformations. It appears frequently in quant interviews where candidates write code in modern C++17 or later.

The question requires you to trace through sequential optional operations, understand when a chain short-circuits (upon encountering nullopt), and predict the final state. Mastery here demonstrates familiarity with functional composition patterns in C++ and the semantics of optional lifting — critical for writing expressive, safe code in trading systems where absent values are common and silence-by-default can hide bugs.

std::optional value semantics and construction
The and_then monadic bind operation
Short-circuit evaluation and chain termination
Lambda capture and return type deduction

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Understanding C++ optional chaining with and_then

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