Why I can not return initializer list from lambda

问题:

Why this code is not valid?

  auto foo=[](){
    return {1,2};     
  };

However, this is valid since the initializer list is used just to initialize a vector not to return itself:

auto foo=[]()->std::vector<int>{
  return {1,2};     
};

Why I can not return initializer list? It could be useful. For example, a lambda that can be used to initialize a vector or a list or ... with some default values for something.

回答1:

Lambda return type deduction uses the auto rules, which normally would have deduced std::initializer_list just fine. However, the language designers banned deduction from a braced initializer list in a return statement ([dcl.spec.auto]/7):

If the deduction is for a return statement and the initializer is a braced-init-list ([dcl.init.list]), the program is ill-formed.

The reason for this is that std::initializer_list has reference semantics ([dcl.init.list]/6).
[]() -> std::initializer_list<int> { return {1, 2}; } is every bit as bad as
[]() -> const int & { return 1; }. The lifetime of the backing array of the initializer_list object ends when the lambda returns, and you are left with a dangling pointer (or two).

Demo:

#include <vector>

struct Noisy {
    Noisy()  { __builtin_printf("%s\n", __PRETTY_FUNCTION__); }
    Noisy(const Noisy&) { __builtin_printf("%s\n", __PRETTY_FUNCTION__); }
    ~Noisy() { __builtin_printf("%s\n", __PRETTY_FUNCTION__); }
};

int main()
{
    auto foo = []() -> std::initializer_list<Noisy> { return {Noisy{}, Noisy{}}; };
    std::vector<Noisy> bar{foo()};
}

Output:

Noisy::Noisy()
Noisy::Noisy()
Noisy::~Noisy()
Noisy::~Noisy()
Noisy::Noisy(const Noisy&)
Noisy::Noisy(const Noisy&)
Noisy::~Noisy()
Noisy::~Noisy()

Note how the copy constructors are called after all the Noisy objects created so far have been destroyed already.

回答2:

std::initializer_list can't be deduced by a template argument, which means you'll have to tell the lambda what it is explicitly:

#include <initializer_list>
#include <iostream>
#include <vector>

int main()
{
    auto foo = []() -> std::initializer_list<int> { return {1, 2}; };
    std::vector<int> bar{foo()};
    for (int x : bar) { std::cout << x << "  "; };
}

Demo. Here's the rationale behind this from the initializer list proposal:

Can an initializer list be used as a template argument? Consider:

template<class T> void f(const T&);
f({ }); // error
f({1});
f({1,2,3,4,5,6});
f({1,2.0}); // error
f(X{1,2.0}); // ok: T is X

There is obviously no problem with the last call (provided X{1,2.0} itself is valid)
because the template argument is an X. Since we are not introducing arbitrary lists of
types (product types), we cannot deduce T to be {int,double} for f({1,2.0}), so that call is
an error. Plain {} does not have a type, so f({}) is also an error.

This leaves the homogeneous lists. Should f({1}) and f({1,2,3,4,5,6}) be accepted? If so,
with what meaning? If so, the answer must be that the deduced type, T, is
initializer_list. Unless someone comes up with at least one good use of this simple
feature (a homogeneous list of elements of type E is deduced to be an
initializer_list), we won’t propose it and all the examples will be errors: No template
argument can be deduced from an (unqualified) initializer list. One reason to be cautious
here is that we can imagine someone getting confused about the possible interpretations
of single-element lists. For example, could f({1}) invoke f<int>(1)? No, that would be
quite inconsistent.

回答3:

You can return an initializer_list from the function either like that:

return std::initializer_list<int>{1, 2};

or

auto ret = {1, 2};
return ret;

The reason is, that the auto variable declaration uses different rules than the auto return type deduction. The first one has a special rule for this case, and the second one uses plain template type deduction.

This is discussed at length in Scott Meyers Effective Modern C++, Item 2. There is also a video and slides from him about the topic.