从向量中删除，通过引用传递

debugcn 发表于 Dev

基思·麦迪逊

我正在尝试按索引擦除向量中的元素，该向量已通过引用传递给某些函数。通常，我会这样：

void erase_element(vector<int> &my_vector, int index){
    my_vector.erase(my_vector.begin() + index);
}

void make_vector(){
    vector<int> my_vector = {1,2,3,4,5};
    erase_element(my_vector, 3);
}

而且，的确，为了检查我的理智，我运行了它并且它起作用了。

但是由于某种原因，此代码抛出segfault。有问题的线路似乎是，其中从矢量擦除元件a1，a2和a3。

我的代码：

double maximum(double a, double b, double c) 
{ 
   return max(max(a, b), c); 
} 
  
 
double minimum(double a, double b, double c) 
{ 
   return min(min(a, b), c); 
} 

void findClosestTriplet(vector<double> &a1, vector<double> &a2, vector<double> &a3, TH2 *histo){

    double res_min, res_max, res_mid;
    int i = 0, j = 0, k = 0;

    int a1Size = a1.size();
    int a2Size = a2.size();
    int a3Size = a3.size();

    double diff = DBL_MAX;
    while(i < a1Size && j < a2Size && k < a3Size){

        double sum = a1[i] + a2[j] + a3[k];

        double min = minimum(a1[i], a2[j], a3[k]);
        double max = maximum(a1[i], a2[j], a3[k]);

        if(min == a1[i]){
            ++i;
        } else if (min == a2[j]){
            ++j;
        } else {
            ++k;
        }

        if (diff > (max-min)){ 
                    diff = max - min; 
                    res_max = max; 
                    res_mid = sum - (max + min); 
                    res_min = min; 
            } 

    }


    a1.erase(a1.begin() + i); a2.erase(a2.begin() + j); a3.erase(a3.begin() + k);

}

void minimizer(){

    TH2 *histo = new TH2D("Histo","Histo",1000,-0.1,0.1,1000,-0.1,0.1);

    ROOT::RDataFrame f1("D","data1.root");
    ROOT::RDataFrame f2("D","data2.root");
    ROOT::RDataFrame f3("D","data3.root");

    vector<double> a1,a2,a3;
    a1.reserve(1E+6); a2.reserve(1E+6); a3.reserve(1E+6);

    f1.Foreach([&](double tstamp){a1.push_back(tstamp);},{"UNIX"});
    f2.Foreach([&](double tstamp){a2.push_back(tstamp);},{"UNIX"});
    f3.Foreach([&](double tstamp){a3.push_back(tstamp);},{"UNIX"});

    int maxiter = std::max(std::max(a1.size(), a2.size()), a3.size());

//  std::ios_base::sync_with_stdio(false);
//  std::cin.tie(NULL);

    for(int i = 0; i < maxiter; ++i){
        findClosestTriplet(a1,a2,a3,histo);

    }

}

我试图做一个最小的可复制示例，但是失败了。这有效：

void findClosestTriplet(vector<double> &a1, vector<double> &a2, vector<double> &a3, TH2 *histo){

        int i = 1; int j = 2; int k = 3;

        a1.erase(a1.begin() + i); a2.erase(a2.begin() + j); a3.erase(a3.begin() + k);

}

void minimizer(){

        TH2 *histo = new TH2D("Histo","Histo",1000,-0.1,0.1,1000,-0.1,0.1);

        ROOT::RDataFrame f1("D","data1.root");
        ROOT::RDataFrame f2("D","data2.root");
        ROOT::RDataFrame f3("D","data3.root");

        vector<double> a1,a2,a3;
        a1.reserve(1E+6); a2.reserve(1E+6); a3.reserve(1E+6);

        f1.Foreach([&](double tstamp){a1.push_back(tstamp);},{"UNIX"});
        f2.Foreach([&](double tstamp){a2.push_back(tstamp);},{"UNIX"});
        f3.Foreach([&](double tstamp){a3.push_back(tstamp);},{"UNIX"});

        for(int i = 0; i < a1.size(); ++i){
                findClosestTriplet(a1,a2,a3,histo);
                std::cout << i << "\n";
        }

        histo->Draw("colz");

}

向量a1，a2和a3在两种情况下都是相同的。整数i,j,k不在循环中，因此似乎无关。而且，我知道在发生错误时，i,j,k它们的大小不大于a1,a2,a3，也不小于0。