使用std :: accumulate [英] using of std::accumulate

问题描述

需要更漂亮的解决方案，但使用std :: accumulate。

Need prettier solution of below example but with std::accumulate.

#include <algorithm>
#include <vector>
#include <iostream>

class Object
{
public:
    Object( double a, double b ):
        a_( a ),
        b_( b )
    {}

    double GetA() const { return a_; }
    double GetB() const { return b_; }
    // other methods
private:
    double a_;
    double b_;
};

class Calculator
{
public:
    Calculator( double& result ):
        result_( result )
    {}

    void operator() ( const Object& object )
    {
        // some formula
        result_ += object.GetA() * object.GetB();
    }
private:
    double& result_;
};

int main()
{
    std::vector< Object > collection;
    collection.push_back( Object( 1, 2 ) );
    collection.push_back( Object( 3, 4 ) );

    double result = 0.0;
    std::for_each( collection.begin(), collection.end(),
                   Calculator( result ) );

    std::cout << "result = " << result << std::endl;

    return 0;
}

推荐答案

函数。

struct Calculator
{
    double operator() ( double result, const Object& obj )
    {
    	return result + ( obj.GetA() * obj.GetB());
    }

};

int main()
{
    std::vector< Object > collection;
    collection.push_back( Object( 1, 2 ) );
    collection.push_back( Object( 3, 4 ) );

    double result = std::accumulate( collection.begin(), collection.end(), 0, Calculator() );
    std::cout << "result = " << result << std::endl;

    return 0;
}

也可以更好：

double sumABProduct( double result, const Object& obj )
{
    return result + ( obj.GetA() * obj.GetB());
}

double result = std::accumulate( collection.begin(), collection.end(), 0, sumABProduct );

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