我听说c++的类成员函数模板不能是虚的。这是真的吗?
如果它们可以是虚拟的,那么有什么场景可以使用这样的函数呢?
我听说c++的类成员函数模板不能是虚的。这是真的吗?
如果它们可以是虚拟的,那么有什么场景可以使用这样的函数呢?
当前回答
不,他们不能。但是:
template<typename T>
class Foo {
public:
template<typename P>
void f(const P& p) {
((T*)this)->f<P>(p);
}
};
class Bar : public Foo<Bar> {
public:
template<typename P>
void f(const P& p) {
std::cout << p << std::endl;
}
};
int main() {
Bar bar;
Bar *pbar = &bar;
pbar -> f(1);
Foo<Bar> *pfoo = &bar;
pfoo -> f(1);
};
如果您想要做的只是拥有一个公共接口并将实现推迟到子类,则效果大致相同。
其他回答
回答问题的第二部分:
如果它们可以是虚拟的,那么有什么场景可以使用这样的函数呢?
这并不是一件不合理的事情。例如,Java(每个方法都是虚的)使用泛型方法没有问题。
c++中需要虚函数模板的一个例子是接受泛型迭代器的成员函数。或接受泛型函数对象的成员函数。
这个问题的解决方案是使用boost::any_range和boost::function的类型擦除,这将允许您接受泛型迭代器或函子,而不需要使您的函数成为模板。
下面的代码可以在windows 7上使用mingwg++ 3.4.5编译并正常运行:
#include <iostream>
#include <string>
using namespace std;
template <typename T>
class A{
public:
virtual void func1(const T& p)
{
cout<<"A:"<<p<<endl;
}
};
template <typename T>
class B
: public A<T>
{
public:
virtual void func1(const T& p)
{
cout<<"A<--B:"<<p<<endl;
}
};
int main(int argc, char** argv)
{
A<string> a;
B<int> b;
B<string> c;
A<string>* p = &a;
p->func1("A<string> a");
p = dynamic_cast<A<string>*>(&c);
p->func1("B<string> c");
B<int>* q = &b;
q->func1(3);
}
输出为:
A:A<string> a
A<--B:B<string> c
A<--B:3
后来我又添加了一个新类X:
class X
{
public:
template <typename T>
virtual void func2(const T& p)
{
cout<<"C:"<<p<<endl;
}
};
当我试图在main()中像这样使用类X时:
X x;
x.func2<string>("X x");
g++报告以下错误:
vtempl.cpp:34: error: invalid use of `virtual' in template declaration of `virtu
al void X::func2(const T&)'
所以很明显:
虚成员函数可以在类模板中使用。编译器可以很容易地构造虚表 将类模板成员函数定义为虚函数是不可能的,如你所见,很难确定函数签名和分配虚表项。
不,他们不能。但是:
template<typename T>
class Foo {
public:
template<typename P>
void f(const P& p) {
((T*)this)->f<P>(p);
}
};
class Bar : public Foo<Bar> {
public:
template<typename P>
void f(const P& p) {
std::cout << p << std::endl;
}
};
int main() {
Bar bar;
Bar *pbar = &bar;
pbar -> f(1);
Foo<Bar> *pfoo = &bar;
pfoo -> f(1);
};
如果您想要做的只是拥有一个公共接口并将实现推迟到子类,则效果大致相同。
虚函数表
让我们从虚函数表及其工作原理的一些背景知识开始(来源):
[20.3] What's the difference between how virtual and non-virtual member functions are called? Non-virtual member functions are resolved statically. That is, the member function is selected statically (at compile-time) based on the type of the pointer (or reference) to the object. In contrast, virtual member functions are resolved dynamically (at run-time). That is, the member function is selected dynamically (at run-time) based on the type of the object, not the type of the pointer/reference to that object. This is called "dynamic binding." Most compilers use some variant of the following technique: if the object has one or more virtual functions, the compiler puts a hidden pointer in the object called a "virtual-pointer" or "v-pointer." This v-pointer points to a global table called the "virtual-table" or "v-table." The compiler creates a v-table for each class that has at least one virtual function. For example, if class Circle has virtual functions for draw() and move() and resize(), there would be exactly one v-table associated with class Circle, even if there were a gazillion Circle objects, and the v-pointer of each of those Circle objects would point to the Circle v-table. The v-table itself has pointers to each of the virtual functions in the class. For example, the Circle v-table would have three pointers: a pointer to Circle::draw(), a pointer to Circle::move(), and a pointer to Circle::resize(). During a dispatch of a virtual function, the run-time system follows the object's v-pointer to the class's v-table, then follows the appropriate slot in the v-table to the method code. The space-cost overhead of the above technique is nominal: an extra pointer per object (but only for objects that will need to do dynamic binding), plus an extra pointer per method (but only for virtual methods). The time-cost overhead is also fairly nominal: compared to a normal function call, a virtual function call requires two extra fetches (one to get the value of the v-pointer, a second to get the address of the method). None of this runtime activity happens with non-virtual functions, since the compiler resolves non-virtual functions exclusively at compile-time based on the type of the pointer.
我的问题,或者我是怎么来的
我尝试使用类似这样的东西,现在cubefile基类与模板优化加载函数,这将实现不同类型的立方体(一些存储像素,一些通过图像等)。
一些代码:
virtual void LoadCube(UtpBipCube<float> &Cube,long LowerLeftRow=0,long LowerLeftColumn=0,
long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1) = 0;
virtual void LoadCube(UtpBipCube<short> &Cube, long LowerLeftRow=0,long LowerLeftColumn=0,
long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1) = 0;
virtual void LoadCube(UtpBipCube<unsigned short> &Cube, long LowerLeftRow=0,long LowerLeftColumn=0,
long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1) = 0;
我想要它是什么,但它不会编译由于虚拟模板组合:
template<class T>
virtual void LoadCube(UtpBipCube<T> &Cube,long LowerLeftRow=0,long LowerLeftColumn=0,
long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1) = 0;
我最终将模板声明移到了类级别。这种解决方案将迫使程序在读取数据之前了解它们将要读取的特定类型的数据,这是不可接受的。
解决方案
警告,这不是很漂亮,但它允许我删除重复的执行代码
1)在基类中
virtual void LoadCube(UtpBipCube<float> &Cube,long LowerLeftRow=0,long LowerLeftColumn=0,
long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1) = 0;
virtual void LoadCube(UtpBipCube<short> &Cube, long LowerLeftRow=0,long LowerLeftColumn=0,
long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1) = 0;
virtual void LoadCube(UtpBipCube<unsigned short> &Cube, long LowerLeftRow=0,long LowerLeftColumn=0,
long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1) = 0;
2)和在儿童班
void LoadCube(UtpBipCube<float> &Cube, long LowerLeftRow=0,long LowerLeftColumn=0,
long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1)
{ LoadAnyCube(Cube,LowerLeftRow,LowerLeftColumn,UpperRightRow,UpperRightColumn,LowerBand,UpperBand); }
void LoadCube(UtpBipCube<short> &Cube, long LowerLeftRow=0,long LowerLeftColumn=0,
long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1)
{ LoadAnyCube(Cube,LowerLeftRow,LowerLeftColumn,UpperRightRow,UpperRightColumn,LowerBand,UpperBand); }
void LoadCube(UtpBipCube<unsigned short> &Cube, long LowerLeftRow=0,long LowerLeftColumn=0,
long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1)
{ LoadAnyCube(Cube,LowerLeftRow,LowerLeftColumn,UpperRightRow,UpperRightColumn,LowerBand,UpperBand); }
template<class T>
void LoadAnyCube(UtpBipCube<T> &Cube, long LowerLeftRow=0,long LowerLeftColumn=0,
long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1);
注意,LoadAnyCube没有在基类中声明。
下面是另一个堆栈溢出的答案: 需要一个虚拟模板成员解决方案。
从c++模板的完整指南:
Member function templates cannot be declared virtual. This constraint is imposed because the usual implementation of the virtual function call mechanism uses a fixed-size table with one entry per virtual function. However, the number of instantiations of a member function template is not fixed until the entire program has been translated. Hence, supporting virtual member function templates would require support for a whole new kind of mechanism in C++ compilers and linkers. In contrast, the ordinary members of class templates can be virtual because their number is fixed when a class is instantiated