获得向量< Derived *>转换成期望向量< Base *>的函数。 [英] Getting a vector<Derived*> into a function that expects a vector<Base*>

问题描述

考虑这些类。

class Base
{
   ...
};

class Derived : public Base
{
   ...
};

此功能

void BaseFoo( std::vector<Base*>vec )
{
    ...
}

最后我的矢量

std::vector<Derived*>derived;

我想通过派生 code> BaseFoo ，但是编译器不让我。我如何解决这个问题，而不是将整个向量复制到 std :: vector< Base *> ？

I want to pass derived to function BaseFoo, but the compiler doesn't let me. How do I solve this, without copying the whole vector to a std::vector<Base*>?

推荐答案

向量< Base *> 和向量< Derived *> 是不相关的类型，所以你不能这样做。有关说明，请参阅C ++常见问题这里。

vector<Base*> and vector<Derived*> are unrelated types, so you can't do this. This is explained in the C++ FAQ here.

您可以将变量从向量< Derived *> 更改为向量< ; Base *> 并插入派生的对象。

You could change your variable from a vector<Derived*> to a vector<Base*> and insert Derived objects into it.

应该通过const引用而不是值传递向量：

Also, you should pass the vector by const-reference, not by value:

void BaseFoo( const std::vector<Base*>& vec )
{
    ...
}

最后，为了避免内存泄漏，并使代码异常安全，请考虑使用一个用于处理堆分配对象的容器，例如：

Finally, to avoid memory leaks, and make your code exception-safe, consider using a container designed to handle heap-allocated objects, e.g:

#include <boost/ptr_container/ptr_vector.hpp>
boost::ptr_vector<Base> vec;

或者，更改向量以保存智能指针，而不是使用原始指针：

Alternatively, change the vector to hold a smart pointer instead of using raw pointers:

#include <memory>
std::vector< std::shared_ptr<Base*> > vec;

或

#include <boost/shared_ptr.hpp>
std::vector< boost::shared_ptr<Base*> > vec;

在每种情况下，您都需要修改 BaseFoo 函数。

In each case, you would need to modify your BaseFoo function accordingly.

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