hkB*_*sai 14 c++ copy-constructor copy-assignment deleted-functions implicit-methods
我有一个用例,我的对象不得以任何方式复制.我在下面写了一个夸张的复制构造函数和复制赋值运算符删除的完整列表.有这么多,我无法确定使用哪些,有时这让我变得偏执.我不必在我的代码中全部写出来,是吗?因此,为了防止任何类型的对象复制,我应该使用哪些对象?
MyClass ( MyClass &) = delete;
MyClass (const MyClass &) = delete;
MyClass ( MyClass &&) = delete;
MyClass (const MyClass &&) = delete;
MyClass operator=( MyClass &) = delete;
MyClass operator=(const MyClass &) = delete;
const MyClass operator=( MyClass &) = delete;
const MyClass operator=(const MyClass &) = delete;
MyClass & operator=( MyClass &) = delete;
MyClass & operator=(const MyClass &) = delete;
const MyClass & operator=( MyClass &) = delete;
const MyClass & operator=(const MyClass &) = delete;
MyClass && operator=( MyClass &) = delete;
MyClass && operator=(const MyClass &) = delete;
const MyClass && operator=( MyClass &) = delete;
const MyClass && operator=(const MyClass &) = delete;
MyClass operator=( MyClass &&) = delete;
MyClass operator=(const MyClass &&) = delete;
const MyClass operator=( MyClass &&) = delete;
const MyClass operator=(const MyClass &&) = delete;
MyClass & operator=( MyClass &&) = delete;
MyClass & operator=(const MyClass &&) = delete;
const MyClass & operator=( MyClass &&) = delete;
const MyClass & operator=(const MyClass &&) = delete;
MyClass && operator=( MyClass &&) = delete;
MyClass && operator=(const MyClass &&) = delete;
const MyClass && operator=( MyClass &&) = delete;
const MyClass && operator=(const MyClass &&) = delete;
Run Code Online (Sandbox Code Playgroud)
Tar*_*ama 23
您只需将单个复制构造函数和复制赋值运算符标记为delete.复制版本的存在将阻止移动构造函数和移动赋值运算符的隐式声明,并且声明一种形式的复制特殊成员函数会抑制其他形式的隐式声明.
MyClass (const MyClass&) = delete;
MyClass& operator= (const MyClass&) = delete;
Run Code Online (Sandbox Code Playgroud)
请注意,在C++之后的11中,赋值运算符的默认隐式定义已被弃用,而应将其定义为已删除.
MyClass (const MyClass &) = delete;
Run Code Online (Sandbox Code Playgroud)
MyClass & operator=(const MyClass &) = delete;
Run Code Online (Sandbox Code Playgroud)
这些是隐式定义的唯一复制构造函数和复制分配运算符.