如何缩短此代码以避免冗余？

问题描述

这是我的代码片段：

string trend()
{
    double emaTrend0 = iMA (NULL,200,MODE_EMA,PRICE_CLOSE,0);
    double emaTrend1 = iMA (NULL,1); 
    double emaTrend2 = iMA (NULL,2); 
    double emaTrend3 = iMA (NULL,3); 
    double emaTrend4 = iMA (NULL,4); 
    double emaTrend5 = iMA (NULL,5); 
    double emaTrend6 = iMA (NULL,6); 
    double emaTrend7 = iMA (NULL,7); 
    double emaTrend8 = iMA (NULL,8);  
    double emaTrend9 = iMA (NULL,9); 
    string signal;
    
    double HighCandle0 = High[0];
    double HighCandle1 = High[1];
    double HighCandle2 = High[2];
    double HighCandle3 = High[3];
    double HighCandle4 = High[4];
    double HighCandle5 = High[5];
    double HighCandle6 = High[6];
    double HighCandle7 = High[7];
    double HighCandle8 = High[8];
    double HighCandle9 = High[9];
    
    double LowCandle0 = Low[0];
    double LowCandle1 = Low[1];
    double LowCandle2 = Low[2];
    double LowCandle3 = Low[3];
    double LowCandle4 = Low[4];
    double LowCandle5 = Low[5];
    double LowCandle6 = Low[6];
    double LowCandle7 = Low[7];
    double LowCandle8 = Low[8];
    double LowCandle9 = Low[9];
    
    if (emaTrend0 > HighCandle0 && emaTrend1 > HighCandle1 && emaTrend2 > HighCandle2 &&emaTrend3 > HighCandle3 &&
    emaTrend4 > HighCandle4 &&emaTrend5 > HighCandle5 &&emaTrend6 > HighCandle6 &&emaTrend7 > HighCandle7 &&
    emaTrend8 > HighCandle8 && emaTrend9 > HighCandle9 )  
      signal = "downtrend";
                                     
    else if (emaTrend0 < LowCandle0 && emaTrend1 < LowCandle1 && emaTrend2 < LowCandle2 &&emaTrend3 < LowCandle3 &&
    emaTrend4 < LowCandle4 &&emaTrend5 < LowCandle5 &&emaTrend6 < LowCandle6 &&emaTrend7 < LowCandle7 &&
    emaTrend8 < LowCandle8 && emaTrend9 < LowCandle9 )  
      signal = "uptrend";
    return signal;
}

我想要发生的是循环变量，并且在产生诸如下降趋势或上升趋势之类的信号之前，每个条件都必须是true。

下面的代码是我的尝试，但还是不行。当条件之一为 true 时，我的尝试产生了下降趋势 或上升趋势 的信号，但我需要的是一切都在 true 之前产生信号。

string trend() {
    string signal = "";
    for (int i = 0; i<=9 ; i++){
        double emaTrend = iMA (NULL,i);
        double highCandle = High[i];
        double lowCandle = Low[i];
        
            if (emaTrend > highCandle){
            signal = "downtrend";
            }
            else if (emaTrend < lowCandle){
            signal = "uptrend";
            }
    }
    return signal;
}

我们将不胜感激所有回复。谢谢！

解决方法

这是一个可能的解决方案，通过计算下降和上升趋势的数量，然后比较这些计数（未测试）：

std::string trend() {
    const std::size_t max_count = 10;

    std::size_t count_downtrend = 0;
    std::size_t count_uptrend = 0;

    for (std::size_t i = 0; i != max_count; ++i) {
        const double emaTrend = iMA (NULL,200,MODE_EMA,PRICE_CLOSE,i);
        const double highCandle = High[i];
        const double lowCandle = Low[i];
        
        if (emaTrend > highCandle) {
            ++count_downtrend;
        } else if (emaTrend < lowCandle) {
            ++count_uptrend;
        }
    }

    std::string signal;
    if (count_downtrend == max_count) {
        signal = "downtrend";
    } else if (count_uptrend == max_count) {
        signal = "uptrend";
    }

    return signal;
}

,

这与常规的“线性搜索循环”非常相似，如果趋势被打破，则提前突破：

enum Direction
{
    Up,Down,None
};

Direction direction(double value,double low,double high)
{
    return value < low ? Down : (value > high ? Up : None);
}

double emaClose(int i)
{
    return iMA(NULL,i);
}

string trend()
{
    const Direction d = direction(emaClose(0),Low[0],High[0]);
    if (d == None)
    {
        return "";
    }
    for (int i = 1; i < 10 ; i++)
    {
        if (direction(emaClose(i),Low[i],High[i]) != d)
        {
            return "";
        }
    }
    return d == Up ? "uptrend" : "downtrend";
}