我想结合两个向量,但是当我尝试在屏幕上写入结果时,我得到没有int数的结果,这是两个。我想得到结果:一两三四五十 你能帮帮我,怎么解决?谢谢
#include <iostream>
#include <string>
#include <vector>
using namespace std;
template<typename T>
class One
{
protected:
T word;
T word2;
public:
One() {word = "0"; word2 = "0";}
One(T w, T w2) {word = w; word2 = w2;}
virtual const void Show() {cout << word << endl; cout << word2 << endl;}
};
template<typename T>
class Two : public One<T>
{
protected:
int number;
public:
Two() {number = 0;}
Two(T w, T w2, int n) : One(w,w2) {number = n;}
virtual const void Show () {cout << word << endl; cout << word2 << endl; cout << number << endl; }
};
int main ()
{
vector<One<string>> x;
vector<Two<string>> x2;
One<string> css("one","two");
Two<string> csss("three","four",50);
x.push_back(css);
x2.push_back(csss);
x.insert(x.end(),x2.begin(),x2.end());
for (int i = 0; i < x.size(); i++)
{
x.at(i).Show();
}
cin.get();
cin.get();
return 0;
}
答案 0 :(得分:0)
请参阅“切片”的评论。如果你使用指针,你将会遇到这个问题。
#include <iostream>
#include <string>
#include <vector>
using namespace std;
template<typename T>
class One
{
protected:
T word;
T word2;
public:
One() {word = "0"; word2 = "0";}
One(T w, T w2) {word = w; word2 = w2;}
virtual const void Show() {cout << word << endl; cout << word2 << endl;}
};
template<typename T>
class Two : public One<T>
{
protected:
int number;
public:
Two() {number = 0;}
Two(T w, T w2, int n) : One(w,w2) {number = n;}
virtual const void Show () {cout << word << endl; cout << word2 << endl; cout << number << endl; }
};
int main ()
{
std::vector< One<string> * > x;
std::vector< Two<string> * > x2;
One<string> css("one","two");
Two<string> csss("three","four",50);
x.push_back(&css);
x2.push_back(&csss);
x.insert(x.end(),x2.begin(),x2.end());
for (size_t i = 0; i < x.size(); i++)
{
x.at(i)->Show();
}
cin.get();
cin.get();
return 0;
}
答案 1 :(得分:0)
您遇到了一个名为切片的问题。
问题是向量x
只能存储One<string>
类型的对象
当您插入类型为Two<string>
的对象时,对象将在复制上切片(因为当您将内容放入向量时,它们将被复制)。所以基本上你将Two<string>
类型的对象复制到一个只能容纳One<String>
的位置,这样你就会丢失额外的信息(它被切掉)。
// Example:
Two<string> two("plop","plop1",34);
two.show;
One<string> one("stop","stop1");
one.show;
one = two; // copy a two into a one.
one.show; // Notice no number this time.
答案 2 :(得分:0)
这不是你期待的多态性
x.at(i).Show();
只需拨打Show
的{{1}}即可。您没有调用类One
的方法Show
,