C++ 如果我覆盖它，我可以调用基类的虚函数吗？
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Can I call a base class's virtual function if I'm overriding it?
提问by Alex
Say I have classes Fooand Barset up like this:
假设我有课程Foo并Bar设置如下：
class Foo
{
public:
int x;
virtual void printStuff()
{
std::cout << x << std::endl;
}
};
class Bar : public Foo
{
public:
int y;
void printStuff()
{
// I would like to call Foo.printStuff() here...
std::cout << y << std::endl;
}
};
As annotated in the code, I'd like to be able to call the base class's function that I'm overriding. In Java there's the super.funcname()syntax. Is this possible in C++?
正如代码中所注释，我希望能够调用我覆盖的基类函数。在 Java 中有super.funcname()语法。这在 C++ 中可能吗？
回答by sth
The C++ syntax is like this:
C++ 的语法是这样的：
class Bar : public Foo {
// ...
void printStuff() {
Foo::printStuff(); // calls base class' function
}
};
回答by Alex B
Yes,
是的，
class Bar : public Foo
{
...
void printStuff()
{
Foo::printStuff();
}
};
It is the same as superin Java, except it allows calling implementations from different bases when you have multiple inheritance.
它与superJava 中的相同，只是当您具有多重继承时，它允许从不同的基础调用实现。
class Foo {
public:
virtual void foo() {
...
}
};
class Baz {
public:
virtual void foo() {
...
}
};
class Bar : public Foo, public Baz {
public:
virtual void foo() {
// Choose one, or even call both if you need to.
Foo::foo();
Baz::foo();
}
};
回答by AlwaysTraining
Sometimes you need to call the base class' implementation, when you aren't in the derived function...It still works:
有时你需要调用基类的实现，当你不在派生函数中时......它仍然有效：
struct Base
{
virtual int Foo()
{
return -1;
}
};
struct Derived : public Base
{
virtual int Foo()
{
return -2;
}
};
int main(int argc, char* argv[])
{
Base *x = new Derived;
ASSERT(-2 == x->Foo());
//syntax is trippy but it works
ASSERT(-1 == x->Base::Foo());
return 0;
}
回答by MartinStettner
Just in case you do this for a lot of functions in your class:
以防万一您对班级中的许多功能执行此操作：
class Foo {
public:
virtual void f1() {
// ...
}
virtual void f2() {
// ...
}
//...
};
class Bar : public Foo {
private:
typedef Foo super;
public:
void f1() {
super::f1();
}
};
This might save a bit of writing if you want to rename Foo.
如果您想重命名 Foo，这可能会节省一些写作时间。
回答by Tunvir Rahman Tusher
If you want to call a function of base class from its derived class you can simply call inside the overridden function with mentioning base class name(like Foo::printStuff()).
如果您想从其派生类调用基类的函数，您可以简单地在重写的函数内部调用并提及基类名称（如Foo::printStuff()）。
code goes here
代码在这里
#include <iostream>
using namespace std;
class Foo
{
public:
int x;
virtual void printStuff()
{
cout<<"Base Foo printStuff called"<<endl;
}
};
class Bar : public Foo
{
public:
int y;
void printStuff()
{
cout<<"derived Bar printStuff called"<<endl;
Foo::printStuff();/////also called the base class method
}
};
int main()
{
Bar *b=new Bar;
b->printStuff();
}
Again you can determine at runtime which function to call using the object of that class(derived or base).But this requires your function at base class must be marked as virtual.
同样，您可以在运行时使用该类的对象（派生或基类）确定要调用哪个函数。但这要求基类中的函数必须标记为虚拟。
code below
下面的代码
#include <iostream>
using namespace std;
class Foo
{
public:
int x;
virtual void printStuff()
{
cout<<"Base Foo printStuff called"<<endl;
}
};
class Bar : public Foo
{
public:
int y;
void printStuff()
{
cout<<"derived Bar printStuff called"<<endl;
}
};
int main()
{
Foo *foo=new Foo;
foo->printStuff();/////this call the base function
foo=new Bar;
foo->printStuff();
}
回答by joseAndresGomezTovar
check this...
检查这个...
#include <stdio.h>
class Base {
public:
virtual void gogo(int a) { printf(" Base :: gogo (int) \n"); };
virtual void gogo1(int a) { printf(" Base :: gogo1 (int) \n"); };
void gogo2(int a) { printf(" Base :: gogo2 (int) \n"); };
void gogo3(int a) { printf(" Base :: gogo3 (int) \n"); };
};
class Derived : protected Base {
public:
virtual void gogo(int a) { printf(" Derived :: gogo (int) \n"); };
void gogo1(int a) { printf(" Derived :: gogo1 (int) \n"); };
virtual void gogo2(int a) { printf(" Derived :: gogo2 (int) \n"); };
void gogo3(int a) { printf(" Derived :: gogo3 (int) \n"); };
};
int main() {
std::cout << "Derived" << std::endl;
auto obj = new Derived ;
obj->gogo(7);
obj->gogo1(7);
obj->gogo2(7);
obj->gogo3(7);
std::cout << "Base" << std::endl;
auto base = (Base*)obj;
base->gogo(7);
base->gogo1(7);
base->gogo2(7);
base->gogo3(7);
std::string s;
std::cout << "press any key to exit" << std::endl;
std::cin >> s;
return 0;
}
output
输出
Derived
Derived :: gogo (int)
Derived :: gogo1 (int)
Derived :: gogo2 (int)
Derived :: gogo3 (int)
Base
Derived :: gogo (int)
Derived :: gogo1 (int)
Base :: gogo2 (int)
Base :: gogo3 (int)
press any key to exit
the best way is using the base::functionas say @sth
最好的方法是像@sth 一样使用base::function
回答by Atif
Yes you can call it. C++ syntax for calling parent class function in child class is
是的，你可以调用它。在子类中调用父类函数的 C++ 语法是
class child: public parent {
// ...
void methodName() {
parent::methodName(); // calls Parent class' function
}
};
Read more about function overriding.
阅读有关函数覆盖的更多信息。
