功能不返回什麼suppossed返回-C++

Question

我正在處理一個問題，這與c + +中的動態分配有關。我有以下函數來計算一個信號

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; 

} 

首先的均方根，如果正確地明白double* <var>表示該參數有望被動態定義的，這意味着它的大小將由限制硬件。

那麼我所做的就是在我的代碼中調用這個函數。樣本代碼是下面

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，在執行程序時，我得到從與邏輯均方根的功能的COUT，而程序結束之前我得到的RMS是0

任何想法或建議爲什麼發生這種情況？事情是，我必須堅持這個功能，因爲它屬於我必須堅持使用的庫...

我想改變函數返回double*而不是void，但沒有真正改變。 ..下面是修改的功能

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FindRMS只設置了*rms這是一樣的rms[0]功能。數組中的所有其他值都是未初始化的，這意味着它們的值將是不確定的（並且似乎是隨機的）。讀取這些未初始化的值將導致未定義的行爲。