获取右值的地址 [英] Getting the address of an rvalue

问题描述

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，以前的版本。

This doesn't work with GCC 4.6, while it used to work in a previous version.

现在说：取地址xvalue（rvalue reference）。

Now it says: taking address of xvalue (rvalue reference).

安全地将右值的地址传递给另一个函数？

Is there any way to securely pass the address of an rvalue to another function?

推荐答案

安全地将一个右值引用（临时的地址）传递给另一个函数，而不必将其存储在一个变量中。

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，如下例所示：

#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) );
}

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