Question

我不明白为什么这会给我一个段错误。有什么想法吗？

这是返回信号以停止程序的函数（加上在其中调用的其他函数）：

double bisect(double A0,double A1,double Sol[N],double tol,double c)
{
  double Amid,shot;

  while (A1-A0 > tol) {
    Amid = 0.5*(A0+A1);

    shot = shoot(Sol, Amid, c);

    if (shot==2.*Pi) {
      return Amid;
    }

    if (shot > 2.*Pi){
      A1 = Amid;
    }
    else if (shot < 2.*Pi){
      A0 = Amid;
    }
  }

  return 0.5*(A1+A0);
}

double shoot(double Sol[N],double A,double c)
{
  int i,j;

  /*Initial Conditions*/
  for (i=0;i<buff;i++)
    {
      Sol[i] = 0.;
    }
  for (i=buff+l;i<N;i++)
    {
      Sol[i] = 2.*Pi;
    }
  Sol[buff]= 0;
  Sol[buff+1]= A*exp(sqrt(1+3*c)*dx);


  for (i=buff+2;i<buff+l;i++)
    {
      Sol[i] = (dx*dx)*( sin(Sol[i-1]) + c*sin(3.*(Sol[i-1])) )
 - Sol[i-2] + 2.*Sol[i-1];
    }

  return Sol[i-1];
}

价值观 buff，l，N使用#define语句定义。 l = 401，buff = 50，N = 2000

以下是完整代码：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define w 10    /*characteristic width of a soliton*/
#define dx 0.05 /*distance between lattice sites*/
#define s (2*w)/dx  /*size of soliton shape*/
#define l (int)(s+1)   /*array length for soliton*/
#define N (int)2000    /*length of field array--lattice sites*/
#define Pi (double)4*atan(1)
#define buff (int)50


double shoot(double Sol[N],double A,double c);
double bisect(double A0,double A1,double Sol[N],double tol,double c);
void super_pos(double antiSol[N],double Sol[N],double phi[][N]);
void vel_ver(double phi[][N],double v,double c,int tsteps,double dt);



int main(int argc, char **argv)
{
  double c,Sol[N],antiSol[N],A,A0,A1,tol,v,dt;
  int  tsteps,i;
  FILE *fp1,*fp2,*fp3;

  fp1 = fopen("soliton.dat","w");
  fp2 = fopen("final-phi.dat","w");
  fp3 = fopen("energy.dat","w");
  printf("Please input the number of time steps:");
  scanf("%d",&tsteps);
  printf("Also, enter the time step size:");
  scanf("%lf",&dt);
  do{
    printf("Please input the parameter c in the interval [-1/3,1]:");
    scanf("%lf",&c);}
  while(c < (-1./3.) || c > 1.);
  printf("Please input the inital speed of eiter soliton:");
  scanf("%lf",&v);

  double phi[tsteps+1][N];

  tol = 0.0000001;
  A0 = 0.;
  A1 = 2.*Pi;
  A = bisect(A0,A1,Sol,tol,c);
  shoot(Sol,A,c);
  for (i=0;i<N;i++)
    {
      fprintf(fp1,"%d\t",i);
      fprintf(fp1,"%lf\n",Sol[i]);
    }
  fclose(fp1);
  super_pos(antiSol,Sol,phi);
  /*vel_ver(phi,v,c,tsteps,dt);
  for (i=0;i<N;i++){
    fprintf(fp2,"%d\t",i);
    fprintf(fp2,"%lf\n",phi[tsteps][i]);
    }*/


}


double shoot(double Sol[N],double A,double c)
{
  int i,j;

  /*Initial Conditions*/
  for (i=0;i<buff;i++)
    {
      Sol[i] = 0.;
    }
  for (i=buff+l;i<N;i++)
    {
      Sol[i] = 2.*Pi;
    }
  Sol[buff]= 0;
  Sol[buff+1]= A*exp(sqrt(1+3*c)*dx);


  for (i=buff+2;i<buff+l;i++)
    {
      Sol[i] = (dx*dx)*( sin(Sol[i-1]) + c*sin(3.*(Sol[i-1])) )
    - Sol[i-2] + 2.*Sol[i-1];
    }

  return Sol[i-1];
}


double bisect(double A0,double A1,double Sol[N],double tol,double c)
{
  double Amid,shot;

  while (A1-A0 > tol) {
    Amid = 0.5*(A0+A1);

    shot = shoot(Sol, Amid, c);

    if (shot==2.*Pi) {
      return Amid;
    }

    if (shot > 2.*Pi){
      A1 = Amid;
    }
    else if (shot < 2.*Pi){
      A0 = Amid;
    }
  }

  return 0.5*(A1+A0);
}


void super_pos(double antiSol[N],double Sol[N],double phi[][N])
{
  int i;

  /*for (i=0;i<N;i++)
    {
    phi[i]=0;
    }

  for (i=buffer+s;i<1950-s;i++)
    {
      phi[i]=2*Pi;
      }*/

  for (i=0;i<N;i++)
    {
      antiSol[i] = Sol[N-i];
    }

  /*for (i=0;i<s+1;i++)
    {
      phi[buffer+j] = Sol[j];
      phi[1549+j] = antiSol[j];
      }*/

  for (i=0;i<N;i++)
    {
      phi[0][i] = antiSol[i] + Sol[i] - 2.*Pi;
    }

}

/* This funciton will set the 2nd input array to the derivative at the time t, for all points x in the lattice */
void deriv2(double phi[][N],double DphiDx2[][N],int t)
{
  //double SolDer2[s+1];
  int x;
  for (x=0;x<N;x++)
    {
      DphiDx2[t][x] = (phi[buff+x+1][t] + phi[buff+x-1][t] - 2.*phi[x][t])/(dx*dx);
    }

  /*for (i=0;i<N;i++)
    {
      ptr[i] = &SolDer2[i];
      }*/

  //return DphiDx2[x];
  }




void vel_ver(double phi[][N],double v,double c,int tsteps,double dt)
{
  int t,x;
  double d1,d2,dp,DphiDx1[tsteps+1][N],DphiDx2[tsteps+1][N],dpdt[tsteps+1][N],p[tsteps+1][N];

  for (t=0;t<tsteps;t++){
    if (t==0){
      for (x=0;x<N;x++){//inital conditions
    deriv2(phi,DphiDx2,t);
    dpdt[t][x] = DphiDx2[t][x] - sin(phi[t][x]) - sin(3.*phi[t][x]);
    DphiDx1[t][x] = (phi[t][x+1] - phi[t][x])/dx;
    p[t][x] = -v*DphiDx1[t][x];
      }
    }
    for (x=0;x<N;x++){//velocity-verlet
      phi[t+1][x] = phi[t][x] + dt*p[t][x] + (dt*dt/2)*dpdt[t][x];
      p[t+1][x] = p[t][x] + (dt/2)*dpdt[t][x];
      deriv2(phi,DphiDx2,t+1);
      dpdt[t][x] = DphiDx2[t][x] - sin(phi[t+1][x]) - sin(3.*phi[t+1][x]);
      p[t+1][x] += (dt/2)*dpdt[t+1][x];
    }
  }  
}

所以，这真的不是因为我覆盖了Sol数组的末尾。我已经注释掉了我怀疑导致问题的两个函数（平分或拍摄）并插入了打印功能。发生了两件事。当我有如下代码时：

double A,Pi,B,c;

c=0;
Pi = 4.*atan(1.);
A = Pi;
B = 1./4.;
printf("%lf",B);
B = shoot(Sol,A,c);
printf("%lf",B);

我从该功能获得了一个段错误，拍摄。但是，如果我拿走了射击功能，那么我就有了：

double A,Pi,B,c;

c=0;
Pi = 4.*atan(1.);
A = Pi;
B = 1./4.;
printf("%lf",B);

它让我在printf上出现了段错...为什么！？

Answer 1

也许你正在写Sol数组的结尾？

我建议您首先使用调试程序（如gdb）来确定导致分段错误的行。

Answer 2

就我个人而言，我认为对这些常量使用#define会使您的函数不再可重用。为什么不通过这些？至少通过这种方式，您的函数用户可以通过查看函数来了解所需的内容。

我在很多地方看到了神奇的数字。我根本不关心这种方法。

Answer 3

所以，这已经解决了。问题在于记忆。我正在使用几个大尺寸phi [tsteps + 1] [N]（N = 2000）等阵列。当tsteps从用户（我自己的）输入进来时它大约是20,000。因此，内存过载导致程序崩溃运行时。感谢所有建议，大家好。

我的C程序中的分段错误

3 个答案: