首页 > 解决方案 > ***检测到堆栈粉碎***:C中的终止错误

问题描述

我正在尝试运行包含此函数的代码,并且收到“检测到***堆栈粉碎***:已终止”。

void Lapenta_Markidis( long double v[3], long double E[3], long double B[3], long double c2,  long double upart[3] ){
   
   long double   upartk[3], vbar[3];
   long double  tmp[3], Fk[3], C1[3], C2[3] ;
   long double  dupartk[3];
   long double gL = gamma_v(v, c2);
   

    for(int i=0;i<3;i++){       
        upart[i] = v[i]*gL;       // momentum at start of time step
        upartk[i] = upart[i];
    }


    /* Start of the nonlinear cycle */

    long double     abserr = 1.0;
    long double     tol=1e-14;
    int             nkmax=10;
    int             nk = 0; //


    do{
        long double  J11, J12, J13,J21, J22,  J23, J31, J32, J33, Det;
        long double  iJ11, iJ12, iJ13,iJ21, iJ22,  iJ23, iJ31, iJ32, iJ33;
        long double  gL_new;

        nk     =  nk+1;
        gL_new = gamma_u(upartk, c2);

        for(int i=0;i<3;i++){  
            vbar[i] = (upart[i] + upartk[i])/(gL_new + gL);
        }
        
        crossP(vbar,B,tmp);

        // Compute residual vector
        for(int i=0;i<3;i++){ 
            Fk[i] = upartk[i] - upart[i] - q*dt/mp * (E[i] + tmp[i]);
        }

         // Compute auxiliary coefficients
        for(int i=0;i<3;i++){
            C1[i] = (gL_new + gL - (upartk[i]*(upartk[i] + upart[i])) / (gL_new*c2) )/ pow((gL + gL_new),2) ;
            C2[i] = -( upartk[i] / (gL_new*c2) )/ ((gL_new + gL),2) ;
        }

    
        // Compute Jacobian
          J11 = 1. - (q*dt/mp) * (C2[1] * (upartk[2] + upart[2]) * B[3] - C2[1] * (upartk[3] + upart[3]) * B[2]) ;
          J12 = - (q*dt/mp)*(C1[2] * B[3] - C2[2] * (upartk[3] + upart[3]) * B[2]) ;
          J13 = - (q*dt/mp) * (C2[3] * (upartk[2] + upart[2]) * B[3] - C1[3] * B[2]) ;
          J21 = - q*dt/mp * (- C1[1] * B[3] + C2[1] * (upartk[3] + upart[3]) * B[1]) ;
          J22 = 1. - q*dt/mp * (- C2[2] * (upartk[1] + upart[1]) * B[3] + C2[2] * (upartk[3] + upart[3]) * B[1]) ;
          J23 = - q*dt/mp * (- C2[3] * (upartk[1] + upart[1]) * B[3] + C1[3] * B[1]) ;
          J31 = - q*dt/mp * (C1[1] * B[2] - C2[1] * (upartk[2] + upart[2]) * B[1]) ;
          J32 = - q*dt/mp * (C2[2] * (upartk[1] + upart[1]) * B[2] - C1[2] * B[1]) ;
          J33 = 1. - q*dt/mp * (C2[3] * (upartk[1] + upart[1]) * B[2] - C2[3] * (upartk[2] + upart[2]) * B[1]) ;

          // Compute inverse Jacobian
           Det = J11*J22*J33 + J21*J32*J13 + J31*J12*J23 - J11*J32*J23 - J31*J22*J13 - J21*J12*J33;
           iJ11 = (J22*J33 - J23*J32) / Det ;
           iJ12 = (J13*J32 - J12*J33) / Det ;
           iJ13 = (J12*J23 - J13*J22) / Det ;
           iJ21 = (J23*J31 - J21*J33) / Det ;
           iJ22 = (J11*J33 - J13*J31) / Det ;
           iJ23 = (J13*J21 - J11*J23) / Det ;
           iJ31 = (J21*J32 - J22*J31) / Det ;
           iJ32 = (J12*J31 - J11*J32) / Det ;
           iJ33 = (J11*J22 - J12*J21) / Det ;

        // Compute new upartk = upartk - J^(-1) * F(upartk)
          dupartk[1] = - (iJ11 * Fk[1] + iJ12 * Fk[2] + iJ13 * Fk[3]);
          dupartk[2] = - (iJ21 * Fk[1] + iJ22 * Fk[2] + iJ23 * Fk[3]);
          dupartk[3] = - (iJ31 * Fk[1] + iJ32 * Fk[2] + iJ33 * Fk[3]);

        // Check convergence
        for(int i=0;i<3;i++){
          upartk[i] +=  dupartk[i] ;
        }
          abserr = sqrt(dotP(dupartk, dupartk));
    } while(abserr > tol && nk < nkmax); // End of while -> end of cycle

    // Update velocity 
    for(int i=0;i<3;i++){
        upart[i] = upartk[i];
    }
   
    
}

我正在尝试运行包含此函数的代码,并且收到“检测到***堆栈粉碎***:已终止”。

关于我做错了什么的任何建议?我对 C 语法不太熟悉,我是否以错误的方式声明了一个变量、矩阵?

标签: cloopsdo-while

解决方案


您混合了基于 1 的索引和基于 0 的索引。但是 C 数组使用从 0 开始的索引。

在您使用的几个位置variable[3],仅variable[2]允许:

      dupartk[3] = - (iJ31 * Fk[1] + iJ32 * Fk[2] + iJ33 * Fk[3]);
//            ^                                               ^

将所有这些索引移一,这些访问可以了:

      dupartk[0] = - (iJ11 * Fk[0] + iJ12 * Fk[1] + iJ13 * Fk[2]);
      dupartk[1] = - (iJ21 * Fk[0] + iJ22 * Fk[1] + iJ23 * Fk[2]);
      dupartk[2] = - (iJ31 * Fk[0] + iJ32 * Fk[1] + iJ33 * Fk[2]);

但请记住,还有其他几个错误的索引,例如B[3]C[3]。检查每个索引。


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