c – 获取右值的地址

class MyClass {
    public: MyClass(int a) : a(a) { }
    int a;
};

#include <iostream>
void print(MyClass* a) { std::cout << a->a << std::endl; }

int main() {
    print(&static_cast<MyClass&&>(MyClass(1337)));
    return 0;
}

这不适用于GCC 4.6,而它曾用于以前的版本.

现在它说：获取xvalue(rvalue reference)的地址.

有没有办法安全地将右值的地址传递给另一个函数？

解决方法:

is: there is anyway to securely pass an rvalue reference (a.k.a. address of temporary) to another function without boringly storing it in a variable just to do that?

是的,就像下一个例子一样：

#include <iostream>

class MyClass {
       public: MyClass(int a) : a(a) { }
       int a;
   };


void print(MyClass&& a) { std::cout << a.a << std::endl; }

int main() {
    print( MyClass(1337) );
}

标签：rvalue,c,c11,gcc
来源： https://codeday.me/bug/20190826/1730194.html