问题
What is the usage of adding -> auto
in []() -> auto { return 4; }
?
For me - it is not different than []() { return 4; }
回答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 isauto
, which is replaced by the trailing-return-type if provided and/or deduced fromreturn
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 !
来源:https://stackoverflow.com/questions/52077143/what-is-the-usage-of-lambda-trailing-return-type-auto