What is the usage of lambda trailing return type a

Question

What is the usage of adding -> auto in []() -> auto { return 4; }?

For me - it is not different than []() { return 4; }

Answer 1

It is auto by default. The Standard, [expr.prim.lambda]/4, reads:

If a lambda-expression does not include a lambda-declarator, it is as if the lambda-declarator were (). The lambda return type is auto, which is replaced by the trailing-return-type if provided and/or deduced from return statements as described in [dcl.spec.auto].

My addition.

So, -> auto itself is not useful. However, we can form other return types with auto, namely: -> auto&, -> const auto&, -> auto&&, -> decltype(auto). Standard rules of return type deduction apply. One should bear in mind that auto is never deduced to be a reference type, so by default a lambda returns a non-reference type.

A few (trivial) examples:

// 1.
int foo(int);
int& foo(char);

int x;

auto lambda1 = [](auto& x) { return x; };
static_assert(std::is_same_v<decltype(lambda1(x)), int>);

auto lambda2 = [](auto& x) -> auto& { return x; };
static_assert(std::is_same_v<decltype(lambda2(x)), int&>);

// 2.
auto lambda3 = [](auto x) { return foo(x); };
static_assert(std::is_same_v<decltype(lambda3(1)), int>);
static_assert(std::is_same_v<decltype(lambda3('a')), int>);

auto lambda4 = [](auto x) -> decltype(auto) { return foo(x); };
static_assert(std::is_same_v<decltype(lambda4(1)), int>);
static_assert(std::is_same_v<decltype(lambda4('a')), int&>);

// 3.
auto lambda5 = [](auto&& x) -> auto&& { return std::forward<decltype(x)>(x); };
static_assert(std::is_same_v<decltype(lambda5(x)), int&>);
static_assert(std::is_same_v<decltype(lambda5(foo(1))), int&&>);
static_assert(std::is_same_v<decltype(lambda5(foo('a'))), int&>);

---

PiotrNycz's addition. As pointed out in comments (credit to @StoryTeller) - the real usage is version with auto& and const auto& and "The degenerate case is just not something worth bending backwards to disallow."

See:

int p = 7;
auto p_cr = [&]()  -> const auto& { return p; };
auto p_r = [&]()  -> auto& { return p; };
auto p_v = [&]()  { return p; }; 

const auto& p_cr1 = p_v(); // const ref to copy of p
const auto& p_cr2 = p_cr(); // const ref to p
p_r() = 9; // we change p here

std::cout << p_cr1 << "!=" << p_cr2 << "!\n";
// print 7 != 9 !