Question

我正在处理一个与c ++中的动态分配有关的问题。我有以下函数来计算信号的均方根

void FindRMS(int points, double* signal_x, double* signal_y, int presamples, double* mean, double* rms)
{       
    //double fraction_presamples_RMS = 0.9; // fraction of presamples to calculate RMS
  //Safety condition
  if (presamples>points) {
    printf("\nERROR: Too many presamples!\n\n");
    return;
  }

  //Main procedure
  (*rms)  =0.;  
  (*mean) =0.;

  for (int i=0; i<presamples; i++) {
    (*mean)+=signal_y[i];
    (*rms)+=pow(signal_y[i],2);
  }

  (*mean)/=presamples;
  (*rms)=sqrt((*rms)/presamples-pow((*mean),2));
    cout << "RMS was found to be : " << (*rms) << endl;

}

首先，如果我理解正确，double* <var>意味着该参数应该是动态定义的，这意味着它的大小将受到硬件的限制。

然后我做的是在我的代码中调用这个函数。示例代码如下

void Analyze(unsigned int first_run, unsigned int last_run, unsigned int last-segment){

    int points = 9e6;//hSignal->GetNbinsX();
//double x[points], y[points], derivative[points]; // SIZE limited by COMPILER to the size of the stack frame
    double* x          = new double[points];           // SIZE limited only by OS/Hardware
    double* y          = new double[points];
    double* derivative = new double[points];
    double* mean       = new double[points];
    double* rms        = new double[points];
    for (int i = 0; i < points; i++){
        x[i] = hSignal->GetBinLowEdge(i+1);
        y[i] = hSignal->GetBinContent(i+1);
        //cout << " Bin Center " << hSignal->GetBinLowEdge(2) << endl;
    }
    FindRMS(points, x, y, 0.9*points, mean, rms);
    delete[] x;
    delete[] y;
    delete[] mean;
    cout << "The value of rms[10] = " << rms[10] << endl;

}

奇怪的是，当程序执行时，我从函数中得到一个带有逻辑rms的cout，而在程序结束之前我得到的rms是0。

有关为何发生这种情况的任何想法或建议？问题是我必须坚持使用该功能，因为它属于我必须坚持使用的库...

我想改变函数以返回double*而不是void，但实际上没有任何改变......这是修改后的函数

double* FindRMS(int points, double* signal_x, double* signal_y, int presamples, double* mean, double* rms)
{       
    //double fraction_presamples_RMS = 0.9; // fraction of presamples to calculate RMS
  //Safety condition
  if (presamples>points) {
    printf("\nERROR: Too many presamples!\n\n");
    //return;
  }

  //Main procedure
  (*rms)  =0.;  
  (*mean) =0.;

  for (int i=0; i<presamples; i++) {
    (*mean)+=signal_y[i];
    (*rms)+=pow(signal_y[i],2);
  }

  (*mean)/=presamples;
  (*rms)=sqrt((*rms)/presamples-pow((*mean),2));
    cout << "RMS was found to be : " << (*rms) << endl;

    return rms;

}

Answer 1

在函数FindRMS中，您只设置*rms，它与rms[0]相同。数组中的所有其他值都是未初始化的，这意味着它们的值将是不确定（并且似乎是随机的）。读取那些未初始化的值将导致未定义的行为。

函数不会返回返回c ++的内容

1 个答案: