C ++ 11中“return {}”语句的含义是什么？

Question

What does the statement

return {};

in C++11 indicate, and when to use it instead of (say)

return NULL;

or

return nullptr;

Answer 1

return {}; indicates "return an object of the function's return type initialized with an empty list-initializer ". The exact behaviour depends on the returned object's type.

From cppreference.com (because the OP is tagged C++11, I excluded the rules in C++14 and C++17; refer to the link for further details):

• If the braced-init-list is empty and T is a class type with a default constructor, value-initialization is performed.
• Otherwise, if T is an aggregate type, aggregate initialization is performed.
• Otherwise, if T is a specialization of std::initializer_list, the T object is direct-initialized or copy-initialized, depending on context, from the braced-init-list.

• Otherwise, the constructors of T are considered, in two phases:

  • All constructors that take std::initializer_list as the only argument, or as the first argument if the remaining arguments have default values, are examined, and matched by overload resolution against a single argument of type std::initializer_list
  • If the previous stage does not produce a match, all constructors of T participate in overload resolution against the set of arguments that consists of the elements of the braced-init-list, with the restriction that only non-narrowing conversions are allowed. If this stage produces an explicit constructor as the best match for a copy-list-initialization, compilation fails (note, in simple copy-initialization, explicit constructors are not considered at all).

• Otherwise (if T is not a class type), if the braced-init-list has only one element and either T isn't a reference type or is a reference type that is compatible with the type of the element, T is direct-initialized (in direct-list-initialization) or copy-initialized (in copy-list-initialization), except that narrowing conversions are not allowed.

• Otherwise, if T is a reference type that isn't compatible with the type of the element. (this fails if the reference is a non-const lvalue reference)
• Otherwise, if the braced-init-list has no elements, T is value-initialized.

Before C++11, for a function returning a std::string , you would have written:

std::string get_string() {
    return std::string();
}

Using the brace syntax in C++11, you don't need to repeat the type:

std::string get_string() {
    return {}; // an empty string is returned
}

return NULL and return nullptr should be used when the function returns a pointer type:

any_type* get_pointer() {
    return nullptr;
}

However, NULL is deprecated since C++11 because it is just an alias to an integer value (0), while nullptr is a real pointer type:

int get_int() {
    return NULL; // will compile, NULL is an integer
}

int get_int() {
    return nullptr; // error: nullptr is not an integer
}

Answer 2

This is probably confusing:

int foo()
{
  return {};   // honestly, just return 0 - it's clearer
}

This is probably not:

SomeObjectWithADefaultConstructor foo()
{
  return {};
  // equivalent to return SomeObjectWithADefaultConstructor {};
}

Answer 3

return {}; means that {} is the initializer for the return value . The return value is list-initialized with an empty list.

---

Here is some background on the return value , based on [stmt.return] in the C++ Standard:

For a function that returns by value (ie the return type is not a reference and not void ), there is a temporary object called the return value . This object is created by the return statement, and its initializers depend on what was in the return statement.

The return value survives until the end of the full-expression in the code that called the function; if it has class type, then its destructor will run unless it has lifetime extended by the caller binding a reference directly to it.

The return value can be initialized in two different ways:

• return some_expression; - the return value is copy-initialized from some_expression
• return { possibly_empty_list }; - the return value is list-initialized from the list.

---

Assuming T is the function's return type, then note that return T{}; is different to return {} : in the former, a temporary T{} is created, and then the return value is copy-initialized from that temporary.

This will fail to compile if T has no accessible copy/move-constructor, but return {}; will succeed even if those constructors are not present. Accordingly, return T{}; may show side-effects of the copy-constructor etc., although this is a copy elision context so it may not.

---

Here's a brief recap of list-initialization in C++14 (N4140 [dcl.init.list]/3), where the initializer is an empty list:

• If T is an aggregate, then each member is initialized from its brace-or-equal-initializer if it had one, otherwise as if by {} (so apply these steps recursively).
• If T is a class type with a user-provided default constructor, that constructor is called.
• If T is a class type with an implicitly-defined, or = default ed default constructor, the object is zero-initialized and then the default constructor is called.
• If T is a std::initializer_list , the return value is an empty such list.
• Otherwise (ie T is a non-class type -- return types cannot be arrays), the return value is zero-initialized.

Answer 4

这是方法返回类型的新实例的一种简写。