从子类的STL向量到基类的向量的转换 [英] Conversion from STL vector of subclass to vector of base class

问题描述

我想知道是否可以将派生类值的向量转换为基类值的向量。具体来说，我想能够传递一个基类对象的向量到一个函数，其形式参数接受基类的向量。它似乎不可能直接作为下面的代码示例产生错误（使用g ++）：

  #include< vector> ; 
 
 class A {
}; 
 
 B类：public A {
}; 
 
 
 void function（std :: vector< A> objs）{
} 
 
 int main（int argc，char ** argv）{ 
 std :: vector< B> objs_b; 
 objs_b.push_back（B（））; 
 function（objs_b）; 
} 
  

test.cc:16：error：conversion从'std :: vector< B，std :: allocator< B>'到非标量类型'std :: vector >

我想能够调用函数，而不必定义一个新的向量元素类型A，插入我的元素类型B或更改为向量

解决方案

不，不是。 不是从向量< A> 派生的向量 B 派生自 A 。你将不得不以某种方式改变你的函数。

一个更惯用的C ++方法可能是模板它和它需要一对迭代器 - 这就是为什么各种标准库函数（< algorithm> 等）工作方式，因为它将算法的实现与其操作的事物分开。

I am wondering if it is possible to convert a vector of derived class values to a vector of base class values. Specifically I want to be able to pass a vector of base class objects to a function whose formal parameters takes a vector of base class. It does not appear to be possible directly as the following code example produces an error (using g++):

#include <vector>

class A {
};

class B : public A {
};


void function(std::vector<A> objs) {
}

int main(int argc, char **argv) {
    std::vector<B> objs_b;
    objs_b.push_back(B());
    function(objs_b);
}

test.cc:16: error: conversion from ‘std::vector<B, std::allocator<B> >’ to non-scalar type ‘std::vector<A, std::allocator<A> >’ requested

I would like to be able to be able to call function without having to define a new vector with elements of type A, inserting my elements of type B or changing to a vector of pointers.

解决方案

No, it is not. vector<B> is not derived from vector<A>, regardless of the fact that B is derived from A. You will have to change your function somehow.

A more idiomatically C++ approach might be to template it and have it take a pair of iterators - this is why the various standard library functions (<algorithm> etc) work that way, because it separates the implementation of the algorithm from the kind of thing it's operating on.

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