c++ - 我可以在不查看每个元素的情况下将 std::vector 转换为 std::vector 吗？

Question

我有一个基类，有几个扩展它的类。我有一些通用库实用程序，它们创建一个包含指向基类的指针的 vector ，以便任何子类都可以工作。如何将 vector 的所有元素都转换为特定的子类？

// A method is called that assumes that a vector containing
// Dogs casted to Animal is passed.
void myDogCallback(vector<Animal*> &animals) {
    // I want to cast all of the elements of animals to
    // be dogs.
    vector<Dog*> dogs = castAsDogs(animals);
}

我天真的解决方案看起来像这样:

// A method is called that assumes that a vector containing
// Dogs casted to Animal is passed.
void myDogCallback(vector<Animal*> &animals) {
    // I want to cast all of the elements of animals to
    // be dogs.
    vector<Dog*> dogs;
    vector<Animal*>::iterator iter;
    for ( iter = animals.begin(); iter != animals.end(); ++iter ) {
        dogs.push_back(dynamic_cast<Dog*>(*iter));
    }
}

Answer 1

您可以使用 std::transform。它仍然在内部使用 for()，但是你会得到两个字符串的实现:

#include <vector>
#include <algorithm>
using namespace std;

struct Animal { virtual ~Animal() {} };
struct Dog : Animal { virtual ~Dog() {} };

template<typename Target>
struct Animal2Target { Target* operator ()( Animal* value ) const { return dynamic_cast<Target*>(value); } };

void myDogCallback(vector<Animal*> &animals) {
{
    vector<Dog*> dogs;
    transform( animals.begin(), animals.end(), dogs.begin(), Animal2Target<Dog>() );
}