Question

我试着在一个函数内向导向量中添加一个向量：该函数获取向量的向量作为指针（称为r），并将其添加到另一个向量。

std::vector<double> b;

b.push_back(0);
b.push_back(0);
b.push_back(0);
b.push_back(v0*sin(theta));
b.push_back(0);
b.push_back(v0*cos(theta));
b.push_back(0);
cout<<"befor"<<endl;
(*r).push_back(b);
cout<<"after"<<endl;

但是当它尝试将b添加到r时，该函数崩溃。我相信这是因为b是一个内部对象，但是为什么程序在尝试添加b而不是之后（当它结束函数时）崩溃？为了检查这一点，我在“之前”和“之后”添加了一条打印行（它打印在“之前”但不是“之后”然后它崩溃了。）

任何人都可以解释一下将向量添加到函数内的向量向量的最佳方法吗？

完整功能如下所示：

 void solve_all(double v0,double theta,double omega,double phi,double 
 dt_in,double total_time,std::string solver, double eps_in, 
  std::vector<std::vector<double>>* r)
  {
   double delta,eta,miu,diff,time,eps_f,dt_f,eps_t;
   std::vector<double> r_1(7),r_2(7),r_temp(7);
   int i,good;
   mone_f=0;
std::vector<double> b;

b.push_back(0);
b.push_back(0);
b.push_back(0);
b.push_back(v0*sin(theta));
b.push_back(0);
b.push_back(v0*cos(theta));
b.push_back(0);
cout<<"befor"<<endl;
(*r).push_back(b);
cout<<"after"<<endl;
i=1;
time=b[6];
dt_f=dt_in;
eps_f=eps_in;
delta=eps_f*dt_f/(total_time-time);
eta=delta/4;
miu=delta/2;
eps_t=1e-10;

while(total_time-time>eps_t)
{
    cout<<time<<endl;
    good=0;
    while(good==0)
    {
        if(dt_f>total_time-time){dt_f=total_time-time;}
        solve(&b,&r_1,dt_f,omega,phi,solver);
        solve(&b,&r_temp,dt_f/2,omega,phi,solver);
        solve(&r_temp,&r_2,dt_f/2,omega,phi,solver);
        diff=0; 
        for(int j=0; j<=5; j++)
        {
            double subs;
            subs=r_2[j]-r_1[j];
            if(subs<0){subs=subs*(-1);}
            if(subs>diff){diff=subs;}
        }
        if(diff>delta)
        {
            dt_f=dt_f/2;
        }
        else
        {
            if(miu<diff&&diff<delta)
            {
                good=1;
                dt_f=0.9*dt_f;
            }
            else
            {
                if(eta<diff&&diff<miu)
                {
                    good=1;
                }
                else
                {
                    if(eta>diff)
                    {
                        good=1;
                        dt_f=1.1*dt_f;
                    }
                }
            }
        }
        i++;
        b.clear();
        if(solver.compare("RK2"))
        {
            for(int j=0; j<=6;i++)
            {
                b.push_back((4*r_2[j]-r_1[j])/3);
            }
        }

        if(solver.compare("RK3"))
        {
            for(int j=0; j<=6;i++)
            {
                b.push_back((8*r_2[j]-r_1[j])/7);
            }
        }
        (*r).push_back(b);
        eps_f=eps_f-diff;
        time=b[6];
        delta=eps_f*dt_f/(total_time-time);
        eta=delta/4;
        miu=delta/2;
    }
}



}

相关功能：

 double big_f(double v)
 {
     return (0.0039+0.0058/(1+exp((v-35)/5)));
 }



 double f1(double vx,double vy,double vz,double omega ,double phi)
 {
     double v;
     mone_f++;
     v=sqrt(pow(vx,2)+pow(vy,2)+pow(vz,2));
     return (-big_f(v)*v*vx+B*omega*(vz*sin(phi)-vy*cos(phi)));
 }



 double f2(double vx,double vy,double vz,double omega ,double phi)
 {
     double v;
     mone_f++;
     v=sqrt(pow(vx,2)+pow(vy,2)+pow(vz,2));
     return (-big_f(v)*v*vy+B*omega*(vx*cos(phi)));
 }


  double f3(double vx,double vy,double vz,double omega ,double phi)
  {
     double v;
     mone_f++;
     v=sqrt(pow(vx,2)+pow(vy,2)+pow(vz,2));
    return (-g-big_f(v)*v*vz-B*omega*(vx*sin(phi)));
  }

 void solve(std::vector<double>* r,std::vector<double>* r_new,double 
 l,double omega, double phi, std::string solver)
 {
    double vx,vy,vz,vx2,vy2,vz2,vx3,vy3,vz3;
   if (solver.compare("RK2")==0)
    {
       vx=(*r)[3];
       vy=(*r)[4];
      vz=(*r)[5];
       vx2=vx+0.5*l*f1(vx,vy,vz,omega,phi);
       vy2=vy+0.5*l*f2(vx,vy,vz,omega,phi);
       vz2=vz+0.5*l*f3(vx,vy,vz,omega,phi);
       (*r_new).push_back((*r)[0]+l*vx2);
       (*r_new).push_back((*r)[1]+l*vy2);
       (*r_new).push_back((*r)[2]+l*vz2);
       (*r_new).push_back((*r)[3]+l*f1(vx2,vy2,vz2,omega,phi));
       (*r_new).push_back((*r)[4]+l*f2(vx2,vy2,vz2,omega,phi));
       (*r_new).push_back((*r)[5]+l*f3(vx2,vy2,vz2,omega,phi));
       (*r_new).push_back((*r)[6]+l);
   }
   if(solver.compare("RK3")==0)
   {
       vx=(*r)[3];
       vy=(*r)[4];
       vz=(*r)[5];
       vx2=vx+0.5*l*f1(vx,vy,vz,omega,phi);
       vy2=vy+0.5*l*f2(vx,vy,vz,omega,phi);
       vz2=vz+0.5*l*f3(vx,vy,vz,omega,phi);
       vx3=vx+2*l*f1(vx2,vy2,vz2,omega,phi)-l*f1(vx,vy,vz,omega,phi);
       vy3=vy+2*l*f2(vx2,vy2,vz2,omega,phi)-l*f2(vx,vy,vz,omega,phi);
       vz3=vz+2*l*f3(vx2,vy2,vz2,omega,phi)-l*f3(vx,vy,vz,omega,phi);
       (*r_new).push_back((*r)[0]+l*(vx+4*vx2+vx3)/6);
       (*r_new).push_back((*r)[1]+l*(vx+4*vx2+vx3)/6);
       (*r_new).push_back((*r)[2]+l*(vx+4*vx2+vx3)/6);
       (*r_new).push_back((*r)[3]+l(f1(vx,vy,vz,omega,phi)+4*f1(vx2,vy2,vz2,omega,phi)+f1(vx3,vy3,vz3,omega,phi))/6);
    (*r_new).push_back((*r)[4]+l*(f2(vx,vy,vz,omega,phi)+4*f2(vx2,vy2,vz2,omega,phi)+f2(vx3,vy3,vz3,omega,phi))/6);
    (*r_new).push_back((*r)[5]+l*(f3(vx,vy,vz,omega,phi)+4*f3(vx2,vy2,vz2,omega,phi)+f3(vx3,vy3,vz3,omega,phi))/6);
    (*r_new).push_back((*r)[6]+l);
}
}

Answer 1

如果您使用引用std::vector<std::vector<double>>&或指针std::vector<std::vector<double>>*，您可以push_back尽可能多地使用std::vector<double> push_back。如果您要推回堆栈变量并不重要，r会创建副本。

您的问题可能是nullptr指针无效。尝试检查它不是drawCallback=JS(" function ( settings ) { var api = this.api(); var mydata = api.rows( {page:'current'} ).data(); var last=null; api.column(0,{page:'current'}).data().each( function ( value, index ) { if ( value == last) { mydata[index][0] = '' api.rows({ page: 'current' }).invalidate(); } last=value; }); }" )，或者它是否指向超出范围的堆栈变量 - 在这种情况下它确实很重要。或者，最好更改函数定义以获取引用，并查看编译器是否抱怨。

如何在c ++

1 个答案: