我有这段C代码：
class Args {};
class MyClass {
public:
MyClass(Args& a) {}
MyClass(MyClass &&) = delete;
};
int main() {
Args a;
MyClass c1 = MyClass(a);
MyClass c2 = a;
MyClass c3(a);
return 0;
}
这不会编译,因为对象c1和c2的构造似乎涉及类的移动构造函数：
错误：使用已删除的函数’MyClass :: MyClass(MyClass&&)’
似乎编译器想要创建临时对象,然后将它们移动到c1和c2.为什么会这样？不应该所有三个语句都只调用MyClass(Args& a)构造函数吗？
解决方法
见
copy elision：
Under the following circumstances,the compilers are permitted,but not required to omit the copy- and move- (since C++11) construction of class objects even if the copy/move (since C++11) constructor and the destructor have observable side-effects. This is an optimization: even when it takes place and the copy-/move-constructor is not called,it still must be present and accessible (as if no optimization happened at all),otherwise the program is ill-formed.
从C 17开始：
They need not be present or accessible,as the language rules ensure that no copy/move operation takes place,even conceptually.
