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
如果预先知道模板方法的类型集,则'虚拟模板方法'有一个变通方法。
为了说明这个想法,在下面的例子中只使用了两种类型(int和double)。
在那里,一个“虚拟”模板方法(Base:: method)调用相应的虚拟方法(Base:: VMethod之一),后者反过来调用模板方法实现(Impl::TMethod)。
只需要在派生实现(AImpl, BImpl)中实现模板方法TMethod,并使用derived <*Impl>。
class Base
{
public:
virtual ~Base()
{
}
template <typename T>
T Method(T t)
{
return VMethod(t);
}
private:
virtual int VMethod(int t) = 0;
virtual double VMethod(double t) = 0;
};
template <class Impl>
class Derived : public Impl
{
public:
template <class... TArgs>
Derived(TArgs&&... args)
: Impl(std::forward<TArgs>(args)...)
{
}
private:
int VMethod(int t) final
{
return Impl::TMethod(t);
}
double VMethod(double t) final
{
return Impl::TMethod(t);
}
};
class AImpl : public Base
{
protected:
AImpl(int p)
: i(p)
{
}
template <typename T>
T TMethod(T t)
{
return t - i;
}
private:
int i;
};
using A = Derived<AImpl>;
class BImpl : public Base
{
protected:
BImpl(int p)
: i(p)
{
}
template <typename T>
T TMethod(T t)
{
return t + i;
}
private:
int i;
};
using B = Derived<BImpl>;
int main(int argc, const char* argv[])
{
A a(1);
B b(1);
Base* base = nullptr;
base = &a;
std::cout << base->Method(1) << std::endl;
std::cout << base->Method(2.0) << std::endl;
base = &b;
std::cout << base->Method(1) << std::endl;
std::cout << base->Method(2.0) << std::endl;
}
输出:
0
1
2
3
注: Base::Method对于实际代码来说实际上是多余的(VMethod可以被设为public并直接使用)。 我添加它,使它看起来像一个实际的“虚拟”模板方法。
其他回答
至少在gcc 5.4中,虚函数可以是模板成员,但必须是模板本身。
#include <iostream>
#include <string>
class first {
protected:
virtual std::string a1() { return "a1"; }
virtual std::string mixt() { return a1(); }
};
class last {
protected:
virtual std::string a2() { return "a2"; }
};
template<class T> class mix: first , T {
public:
virtual std::string mixt() override;
};
template<class T> std::string mix<T>::mixt() {
return a1()+" before "+T::a2();
}
class mix2: public mix<last> {
virtual std::string a1() override { return "mix"; }
};
int main() {
std::cout << mix2().mixt();
return 0;
}
输出
mix before a2
Process finished with exit code 0
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。
虽然很多人已经回答了一个老问题,但我相信一个简洁的方法,与其他发布的方法没有太大不同,就是使用一个小宏来帮助减轻类声明的重复。
// abstract.h
// Simply define the types that each concrete class will use
#define IMPL_RENDER() \
void render(int a, char *b) override { render_internal<char>(a, b); } \
void render(int a, short *b) override { render_internal<short>(a, b); } \
// ...
class Renderable
{
public:
// Then, once for each on the abstract
virtual void render(int a, char *a) = 0;
virtual void render(int a, short *b) = 0;
// ...
};
现在,要实现我们的子类:
class Box : public Renderable
{
public:
IMPL_RENDER() // Builds the functions we want
private:
template<typename T>
void render_internal(int a, T *b); // One spot for our logic
};
这样做的好处是,当添加一个新支持的类型时,它可以从抽象头文件中完成,而不必在多个源文件/头文件中进行修改。
在其他答案中,建议的模板函数是一个门面,并不能提供任何实际的好处。
模板函数对于只编写一次代码很有用 不同的类型。 虚函数对于为不同的类提供公共接口非常有用。
该语言不允许虚拟模板函数,但通过一个变通方法,可以同时拥有两者,例如,每个类都有一个模板实现和一个虚拟公共接口。
但是,有必要为每个模板类型组合定义一个虚拟包装器函数:
#include <memory>
#include <iostream>
#include <iomanip>
//---------------------------------------------
// Abstract class with virtual functions
class Geometry {
public:
virtual void getArea(float &area) = 0;
virtual void getArea(long double &area) = 0;
};
//---------------------------------------------
// Square
class Square : public Geometry {
public:
float size {1};
// virtual wrapper functions call template function for square
virtual void getArea(float &area) { getAreaT(area); }
virtual void getArea(long double &area) { getAreaT(area); }
private:
// Template function for squares
template <typename T>
void getAreaT(T &area) {
area = static_cast<T>(size * size);
}
};
//---------------------------------------------
// Circle
class Circle : public Geometry {
public:
float radius {1};
// virtual wrapper functions call template function for circle
virtual void getArea(float &area) { getAreaT(area); }
virtual void getArea(long double &area) { getAreaT(area); }
private:
// Template function for Circles
template <typename T>
void getAreaT(T &area) {
area = static_cast<T>(radius * radius * 3.1415926535897932385L);
}
};
//---------------------------------------------
// Main
int main()
{
// get area of square using template based function T=float
std::unique_ptr<Geometry> geometry = std::make_unique<Square>();
float areaSquare;
geometry->getArea(areaSquare);
// get area of circle using template based function T=long double
geometry = std::make_unique<Circle>();
long double areaCircle;
geometry->getArea(areaCircle);
std::cout << std::setprecision(20) << "Square area is " << areaSquare << ", Circle area is " << areaCircle << std::endl;
return 0;
}
输出:
方形面积为1,圆形面积为3.1415926535897932385
在这里试试
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。