作为std :: initializer_list对象的抽象类 [英] Abstract class as std::initializer_list object

问题描述

为了拥有更简洁的语法，我想使用 std :: initializer_list 将对象列表发送给构造函数。但是，对象是抽象的，这会引起问题：在VS 2013中，该对象失去了vfptr引用，并给出了 R6025：纯虚函数调用运行时错误，而在g ++中，它抱怨编译期间 无法分配抽象类型'base'的对象。我猜想编译器正在尝试复制对象（这是不希望的-它们可能很大），但是仅能成功复制基类，因此会出现错误。我的问题是：是否有一种解决方案（1）避免复制对象，而（2）不会过于冗长，从而消除了更干净的语法优势？下面的代码说明了我的问题：

In order to have a cleaner syntax, I would like to use an std::initializer_list to send a list of objects to a constructor. The objects, however, are abstract, which causes a problem: in VS 2013, it looses the vfptr reference, giving a "R6025: pure virtual function call" runtime error, and in g++ it complains that it "cannot allocate an object of abstract type ‘base’" during compilation. I surmise the compiler is trying to copy the objects (which is undesirable -- they may be big), but succeeds only in copying the base class, hence the error. My question is: Is there a solution which (1) avoids copying the objects and (2) isn't massively verbose, negating the "cleaner syntax" advantage? The code below illustrates my issue:

#include <cstdio>
#include <initializer_list>

struct base{
    virtual void foo() const = 0;
};

struct derived : public base{
    int i;
    derived(int i) : i(i) {}
    void foo() const{
        printf("bar %i", i);
    }
};

void foo_everything(const std::initializer_list<base> &list){
    for (auto i = list.begin(), iend = list.end(); i != iend; i++) i->foo();
}

int main(void){

    // Works fine
    derived d(0);
    base * base_ptr = &d;
    base_ptr->foo();    

    // Does not work fine
    foo_everything({ derived(1), derived(2), derived(3) });
}

请注意，使用base&自std :: initializer_list尝试 [形成a指向引用类型base& 的指针以来，并在使用base *时，然后获取每个派生类的地址实际上在模板错误中工作，它是通过获取临时地址来实现的，因此并不安全（g ++抱怨）。如果我在方法调用（我的临时解决方案）之外声明派生类，则后者确实可以工作，但是它比我希望的还要冗长。

Note that using base& in the template errors since std::initializer_list tries to "[form a] pointer to reference type base&", and while using base*, and then taking the address of each derived class does in-fact work, it does so by taking the address of temporaries, and thus isn't safe (g++ complains). The latter does work if I declare the derived classes outside of the method call (my provisional solution), but it still is more verbose than I hoped for.

推荐答案

使用 initializer_list< base *> 的某种骇人听闻的方法：

Somewhat hackish approach using a initializer_list<base *>:

template<class... Ts>
void foo_everything(Ts&&... args){
    std::initializer_list<base *> list = {&args...};
    for(auto i : list) i->foo();
}

然后从通话中删除括号：

Then remove the braces from the call:

foo_everything(derived(1), derived(2), derived(3));

---

如果您真的不需要转换为 base * 并执行虚拟调用，只想在传入的每个对象上调用 foo（），那么我们可以使用通常的pack-expansion-inside-initializer-list技巧：

---

If you don't really need to convert to base * and perform a virtual call, and just want to call foo() on each object passed in, then we can use the usual pack-expansion-inside-an-initializer-list trick:

template<class... Ts>
void foo_everything(Ts&&... args){
    using expander = int[];
    (void) expander { 0, ((void) std::forward<Ts>(args).foo(), 0)...};
}

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