表达式模板-C ++模板：完整指南

Question

I'm studying c++ templates and reading <<C++ Templates: The Complete Guide>>. I don't understand the flowing about expression template:

The code as following:

//exprarray.h
#include <stddef.h>
#include <cassert>
#include "sarray.h"

template<typename T>
class A_Scale
{
public:
    A_Scale(T const& t):value(t){}
    T operator[](size_t) const
    {
        return value;
    }
    size_t size() const
    {
        return 0;
    }
private:
    T const& value;
};

template<typename T>
class A_Traits
{
public:
    typedef T const& exprRef;
};
template<typename T>
class A_Traits<A_Scale<T> >
{
public:
    typedef A_Scale<T> exprRef;
};

template<typename T,typename L1,typename R2>
class A_Add
{
private:
    typename A_Traits<L1>::exprRef op1;
    typename A_Traits<R2>::exprRef op2;
public:
    A_Add(L1 const& a,R2 const& b):op1(a),op2(b)
    {
    }
    T operator[](size_t indx) const
    {
        return op1[indx] + op2[indx];
    }
    size_t size() const
    {
        assert(op1.size()==0 || op2.size()==0 || op1.size() == op2.size());
        return op1.size() != 0 ? op1.size() : op2.size();
    }
};

template<typename T,typename L1,typename R2>
class A_Mul
{
private:
    typename A_Traits<L1>::exprRef op1;
    typename A_Traits<R2>::exprRef op2;
public:
    A_Mul(L1 const& a,R2 const& b):op1(a),op2(b)
    {
    }
    T operator[](size_t indx) const
    {
        return op1[indx] * op2[indx];
    }
    size_t size() const
    {
        assert(op1.size()==0 || op2.size()==0 || op1.size() == op2.size());
        return op1.size() != 0 ? op1.size():op2.size();
    }
};

template<typename T,typename Rep = SArray<T> >
class Array
{
public:
    explicit Array(size_t N):expr_Rep(N){}
    Array(Rep const& rep):expr_Rep(rep){}
    Array& operator=(Array<T> const& orig)
    {
        assert(size() == orig.size());
        for (size_t indx=0;indx < orig.size();indx++)
        {
            expr_Rep[indx] = orig[indx];
        }
        return *this;
    }
    template<typename T2,typename Rep2>
    Array& operator=(Array<T2,Rep2> const& orig)
    {
        assert(size() == orig.size());
        for (size_t indx=0;indx<orig.size();indx++)
        {
            expr_Rep[indx] = orig[indx];
        }
        return *this;
    }
    size_t size() const
    {
        return expr_Rep.size();
    }
    T operator[](size_t indx) const
    {
        assert(indx < size());
        return expr_Rep[indx];
    }
    T& operator[](size_t indx)
    {
        assert(indx < size());
        return expr_Rep[indx];
    }
    Rep const& rep() const
    {
        return expr_Rep;
    }
    Rep& rep()
    {
        return expr_Rep;
    }
private:
    Rep expr_Rep;
};

template<typename T,typename L1,typename R2>
Array<T,A_Add<T,L1,R2> >
operator+(Array<T,L1> const& a,Array<T,R2> const& b)
{
    return Array<T,A_Add<T,L1,R2> >(A_Add<T,L1,R2>(a.rep(),b.rep()));
}

template<typename T,typename L1,typename R2>
Array<T,A_Mul<T,L1,R2> >
operator*(Array<T,L1> const& a,Array<T,R2> const& b)
{
    return Array<T,A_Mul<T,L1,R2> >(A_Mul<T,L1,R2>(a.rep(),b.rep()));
}

template<typename T,typename R2>
Array<T,A_Mul<T,A_Scale<T>,R2> >
operator*(T const& a,Array<T,R2> const& b)
{
    return Array<T,A_Mul<T,A_Scale<T>,R2> >(A_Mul<T,A_Scale<T>,R2>(A_Scale<T>(a),b.rep()));
}

The test code:

//test.cpp
#include "exprarray.h"
#include <iostream>
using namespace std;

template <typename T>
void print (T const& c)
{
    for (int i=0; i<8; ++i) {
        std::cout << c[i] << ' ';
    }
    std::cout << "..." << std::endl;
}

int main()
{
    Array<double> x(1000), y(1000);

    for (int i=0; i<1000; ++i) {
        x[i] = i;
        y[i] = x[i]+x[i];
    }

    std::cout << "x: ";
    print(x);

    std::cout << "y: ";
    print(y);

    x = 1.2 * x;
    std::cout << "x = 1.2 * x: ";
    print(x);

    x = 1.2*x + x*y;
    std::cout << "1.2*x + x*y: ";
    print(x);

    x = y;
    std::cout << "after x = y: ";
    print(x);

    return 0;
}

My questions is why A_Traits for A_Scale is by value not by reference.

template<typename T>
class A_Traits
{
public:
    typedef T const& exprRef;
};
template<typename T>
class A_Traits<A_Scale<T> >
{
public:
    typedef A_Scale<T> exprRef;
};

The reason from the book as following:

This is necessary because of the following: In general, we can declare them to be references because most temporary nodes are bound in the top-level expression and therefore live until the end of the evaluation of that complete expression. The one exception are the A_Scalar nodes. They are bound within the operator functions and might not live until the end of the evaluation of the complete expression. Thus, to avoid that the members refer to scalars that don't exist anymore, for scalars the operands have to get copied "by value."

More detail please refer to the chapter 18 of C++ Templates: The Complete Guide

Answer 1

Consider, for example, the right hand side of

x = 1.2*x + x*y;

What the quote says is that this is composed of two different categories.

The heavy array x and y objects are not defined within this expression, but rather before it:

Array<double> x(1000), y(1000);

So, as you build expressions using them, you don't have to worry whether they're still alive - they were defined beforehand. Since they're heavy, you want to capture them by reference, and, fortunately, their lifetime makes that possible.

Conversely, the lightweight A_Scale objects are generated within the expression (eg, implicitly by the 1.2 above). Since they're temporaries, you have to worry about their lifetime. Since they're lightweight, it's not a problem.

That's the rationale for the traits class differentiating between them: the former are by reference, and the latter are by value (they are copied).