Question

// ... snipped includes for iostream and fusion ...
namespace fusion = boost::fusion;

class Base
{
protected: int x;
public: Base() : x(0) {}
    void chug() { 
        x++;
        cout << "I'm a base.. x is now " << x << endl;
    }
};

class Alpha : public Base
{
public:
    void chug() { 
        x += 2;
        cout << "Hi, I'm an alpha, x is now " << x << endl;
    }
};

class Bravo : public Base
{
public:
    void chug() { 
        x += 3;
        cout << "Hello, I'm a bravo; x is now " << x << endl; 
    }
};

struct chug {
    template<typename T>
    void operator()(T& t) const
    {
        t->chug();
    }
};

int main()
{
    typedef fusion::vector<Base*, Alpha*, Bravo*, Base*> Stuff;
    Stuff stuff(new Base, new Alpha, new Bravo, new Base);

    fusion::for_each(stuff, chug());     // Mutates each element in stuff as expected

    /* Output:
       I'm a base.. x is now 1
       Hi, I'm an alpha, x is now 2
       Hello, I'm a bravo; x is now 3
       I'm a base.. x is now 1
    */

    cout << endl;

    // If I don't put 'const' in front of Stuff...
    typedef fusion::result_of::push_back<const Stuff, Alpha*>::type NewStuff;

    // ... then this complains because it wants stuff to be const:
    NewStuff newStuff = fusion::push_back(stuff, new Alpha);

    // ... But since stuff is now const, I can no longer mutate its elements :(
    fusion::for_each(newStuff, chug());

    return 0;
};

如何让for_each（newStuff，chug（））起作用？

（注意：我只是假设来自boost :: fusion的overly brief documentation我每次调用push_back时都应该创建一个新的向量。）

Answer 1

（注意：我只是从关于boost :: fusion的过于简短的文档中假设我每次调用push_back时都应该创建一个新的向量。）

您没有创建新的矢量。 push_back在扩展序列上返回一个延迟评估的view。如果你想创建一个新的矢量，那么例如typedef NewStuff为

typedef fusion::vector<Base*, Alpha*, Bravo*, Base*, Alpha*> NewStuff;

你的程序就可以了。

Btw，fusion是一个非常实用的设计。我认为如果你存储实际的对象而不是指针并使用transform，那将更像融合。然后，chug逻辑将从类中移出到struct chug，每个类型都有operator()个{{1}}。然后，不必创建新的向量，您可以使用延迟评估的视图。

boost :: fusion :: push_back的正确用法是什么？

1 个答案: