Question

我有以下函数，它应该可以正常使用for循环来分析一系列双重结果是否已“稳定”。我所写的内容总共得到了3分，我想知道我如何能够接受一个名为“minNumberOfSamples”的变量并将其与那么多结果进行比较？

    bool MeasurementStabilized(double newResult, double percentageThreshold, double limitRange)
{
    static double meas1 = 1E100;
    static double meas2 = 1E100;

    if ((abs(newResult - meas2) / limitRange >= percentageThreshold) ||
        (abs(meas2 - meas1) / limitRange >= percentageThreshold))
        return TRUE;

    meas1 = meas2;
    meas2 = newResult;
    return FALSE;
}

Answer 1

我想我现在在@John Coleman的帮助下离开了。这似乎可以完成这项工作，但如果有人能想到任何改进，那就寻找一些改进：

static double* gArr;

void MeasurementStabilizedCleanup(void)
{
    free(gArr); 
}

static bool MeasurementStabilized(double newResult, double percentageThreshold, double limitRange, int minNumberOfSamples)
{
    static bool firstRun = TRUE;
    bool retVal = TRUE;
    int x;

    //init variables
    if (firstRun)
    {
        gArr = malloc((minNumberOfSamples + 1) * sizeof(double)); 
        for(x = 0; x <= minNumberOfSamples; x++)
        {
            gArr[x] = 1E100;
        }
        firstRun = FALSE;
    }

    //loop through each value comparing it to the latest result
    for(x = minNumberOfSamples - 1; x >= 0; x--)
    {
        gArr[x + 1] = gArr[x];  //4 -> 5 , 3 -> 4, ...
        if (fabs(newResult - gArr[x]) / limitRange >= percentageThreshold)
            retVal = FALSE;
    }
    gArr[0] = newResult;    //dont forget to update the latest sample   
    return retVal;  
}

void main(void)
{
    double doubles[] = {1, 2, 4, 4, 4, 4, 4};
    int x;

    for (x = 0; x <= 6; x++)
    {
        if(MeasurementStabilized(doubles[x], 0.01, 4, 4))
        {
            ;
        }
    }
    MeasurementStabilizedCleanup();
}

如何在C中使函数“可扩展”？

1 个答案: