2013-01-17 125 views
5

SimpleClass.h为什么多态在这种情况下不起作用?

class SimpleClass 
{ 
    int i; 

    public: 
    SimpleClass() : i(0) {} 
    SimpleClass(int j) : i(j) {} 
    friend std::ostream& operator<<(std::ostream&, const SimpleClass&); 
}; 

SimpleClass.cpp

#include <ostream> 
#include "SimpleClass.h" 

std::ostream& operator<<(std::ostream& out, const SimpleClass& obj) 
{ 
    out << "SimpleClass : " << obj.i << '\n'; 
    return out; 
} 

基类和派生Classes.h

class BaseClass 
{ 
    protected: 
    int i; 

    public: 
    BaseClass() : i(0) {} 
    BaseClass(int j) : i(j) {} 
    virtual void print(std::ostream& out) const { out << "BaseClass : " << i << '\n'; } 
}; 

class DerivedClass : public BaseClass 
{ 
    int j; 

    public: 
    DerivedClass() : BaseClass(), j(0) {} 
    DerivedClass(int m, int n) : BaseClass(m), j(n) {} 
    void print(std::ostream& out) { out << "DerivedClass : " << i << ' ' << j << '\n'; } 
}; 

std::ostream& operator<<(std::ostream&, const BaseClass&); 

基类和派生Classes.cpp

#include <ostream> 
#include "Base and Derived Classes.h" 

std::ostream& operator<<(std::ostream& out, const BaseClass& obj) 
{ 
    obj.print(out); 
    return out; 
} 

的main.cpp

#include <iostream> 
#include "SimpleClass.h" 
#include "Base and Derived Classes.h" 

int main() 
{ 
    SimpleClass simple(10); 
    std::cout << simple; 
    BaseClass base(100); 
    std::cout << base; 
    DerivedClass derived(100, 200); 
    std::cout << derived;     // Doesn't call derived.print(), but base.print() instead. Why ? 
} 
+0

什么(特别是)不工作?你在期待什么,这与结果有什么不同? – iamnotmaynard

+0

print()函数在基类中是常量,但派生类中是非常量。签名需要保持一致。 – tp1

+0

@ tp1这是正确的。我会接受你的回答。谢谢。 – Belloc

回答

20
virtual void print(std::ostream& out) const 

不受

void print(std::ostream& out) 

(因为const)覆盖。

+4

+1。成员函数的'const'是**签名的一部分**! – Nawaz

8

您忘记了派生类中print的定义const

5

在你DerivedClassprint()方法必须是const,就像它在BaseClass

void print(std::ostream& out) const 
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