内容简介：代码日志版权声明：翻译自：http://stackoverflow.com/questions/40039379/why-is-move-constructor-not-picked-when-returning-a-local-object-of-type-derived
(中继版本)拒绝：
#include <memory>
struct Base
{
Base() = default;
Base(Base const&) = delete;
Base(Base&&) = default;
};
struct Derived : Base
{
Derived() = default;
Derived(Derived const&) = delete;
Derived(Derived&&) = default;
};
auto foo()
-> Base
{
Derived d;
return d; // ERROR HERE
}
根据[class.copy] / 32：
When the criteria for elision of a copy/move operation are met, but not for an exception-declaration, and the object to be copied is designated by an lvalue, or when the expression in a return statement is a (possibly parenthesized) id-expression that names an object with automatic storage duration declared in the body or parameter-declaration-clause of the innermost enclosing function or lambda-expression, overload resolution to select the constructor for the copy is first performed as if the object were designated by an rvalue
如果上面的句子意图被解析为(复制elision标准符合& lvalue)|| (id-expression指定一个自动对象),正如 this CWG defect 所示,为什么这不是最后一个条件呢？在Clang和GCC中是否有编译器错误？
另一方面,如果这个句子意图被解析为(copy elision criteria met&(lvalue || id-expression指定一个自动对象)),这不是一个非常误导性的字面意义吗？
[class.copy] / 32继续：
[…] if the type of the first parameter of the selected constructor is not an rvalue reference to the object’s type (possibly cv-qualified), overload resolution is performed again, considering the object as an lvalue.
第一个重载分辨率,以d作为值,选择Base :: Base(Base&&).然而,所选构造函数的第一个参数的类型不是派生和&但是Base&&&&&&&&&&&&&&&第二个重载分辨率选择删除的拷贝构造函数.
代码日志版权声明：
翻译自：http://stackoverflow.com/questions/40039379/why-is-move-constructor-not-picked-when-returning-a-local-object-of-type-derived
以上所述就是小编给大家介绍的《c++ 为什么在返回从函数的返回类型派生的类型的本地对象时,没有选择move构造函数？》，希望对大家有所帮助，如果大家有任何疑问请给我留言，小编会及时回复大家的。在此也非常感谢大家对 码农网 的支持！
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