为什么std :: transform使用构造函数？

Question

我已经使用std :: transform为列表中的现有值添加了一些值.下面的代码工作正常,我只是想知道是否有可能在执行转换时避免对复制构造函数的所有调用(参见程序的输出).如果我只是破解代码,并使for循环显式调用Base的+ =运算符,则不执行复制构造,并且更改位置更有效的值.

我可以使变换调用运算符+ = Base而不是复制构造吗？我应该专注于increment<Type>？

该程序:

#include <iostream>
#include<list>
#include <algorithm>
#include <iterator>

template<class T>
class Base;

template<class T>
std::ostream& operator << (std::ostream& os, const Base<T>& b);

template<class T>
class Base
{
    private: 
        T b_;
    public: 
        typedef T value_type;

        Base()
            :
                b_()
        { std::cout << "Base::def ctor" << std::endl; }

        Base (const T& b)
            : 
                b_(b)
        { std::cout << "Base::implicit conversion ctor: " << b_ << std::endl; }

        const T& value()
        {
            return b_;
        }

        const Base operator+(const Base& b) const
        {
            std::cout << "Base operator+ " << std::endl;
            return Base(b_ + b.b_);
        }

        const Base&  operator+=(const T& t) 
        {
            b_ += t;
            return *this;
        }

        friend std::ostream& operator<< <T> (std::ostream& os, const Base<T>& b);
};

template<class T>
std::ostream& operator<< (std::ostream& os, const Base<T>& b)
{
    os << b.b_; 
    return os;
}

template<class Type> 
class increment
{
    typedef typename Type::value_type T; 

    T initial_; 

    public: 

        increment()
            :
                initial_()
        {};

        increment(const T& t)
            :
                initial_(t)
        {}

        T operator()()
        {
            return initial_++;
        }
};

template<class Container>
void write(const Container& c)
{
    std::cout << "WRITE: " << std::endl;
    copy(c.begin(), c.end(), 
         std::ostream_iterator<typename Container::value_type > (std::cout, " "));
    std::cout << std::endl;
    std::cout << "END WRITE" << std::endl;
}

using namespace std;

int main(int argc, const char *argv[])
{
    typedef list<Base<int> > bList; 

    bList baseList(10); 

    cout << "GENERATE" << endl;
    generate_n (baseList.begin(), 10, increment<Base<int> >(10));
    cout << "END GENERATE" << endl;

    write(baseList); 

    // Let's add some integers to Base<int>

    cout << "TRANSFORM: " << endl;

    std::transform(baseList.begin(), baseList.end(), 
                   baseList.begin(), 
                   bind2nd(std::plus<Base<int> >(), 4)); 
    cout << "END TRANSFORM " << endl;

    write(baseList);

    // Hacking the code: 
    cout << "CODE HACKING: " << endl;
    int counter = 4;
    for (bList::iterator it = baseList.begin(); 
         it != baseList.end(); 
         ++it)
    {
        *it += counter; // Force the call of the operator+=
        ++counter;
    }
    write (baseList);
    cout << "END CODE HACKING" << endl;

    return 0;
}

Answer 1

Base (const T& b)它不是一个复制构造函数,它是一个Base<T>接受一个的构造函数const T&.复制构造函数通常具有签名Base(const Base& )

也就是说,每次Base<int>从an 创建一个new时,都会调用你的构造函数int,这是你在add运算符中所做的.

最后,std :: transform()使用输出迭代器赋值运算符将函数的结果赋给输出.如果你想完全避免复制,你应该使用std::for_each,以及std::bind2nd( std::mem_fun_ref(&Base<int>::operator +=), 4 )).这将避免复制,因为它将在引用上运行.