调用 gsl_integration 时出现分段错误

Question

我正在尝试从 Numerical Recipes 中实现多维积分例程（代码取自 here, pg. 164). The use of wrappers to pass the integrand to void* params is from here（那里有进一步的参考）。我的 class 声明在 NRquad3d_Const_qags_2.C 中，如下所示

#include <gsl/gsl_integration.h>

gsl_integration_workspace * w1 = gsl_integration_workspace_alloc (100000);
gsl_integration_workspace * w2 = gsl_integration_workspace_alloc (100000);

double func3d(double x, double y, double z);

class NRf3 {
public:
  double xsav, ysav;

  double ret_int(double z)
  {  
    return func3d(xsav, ysav, z);
  }

  static double ret_int_wrapper(double z, void *params)
  {
   return static_cast<NRf3*>(params)->ret_int(z); 
  }

};

//integrates over the z-variable
class NRf2 {
private:
  double z1, z2;
  gsl_function F2; 

public:
  NRf3 f3;
  double resultf2, errorf2;

  NRf2 (double zz1, double zz2): z1(zz1), z2(zz2) {}

  double ret_int_2(double y) 
  {
    f3.ysav = y;        

    F2.function = &(f3.ret_int_wrapper);
    F2.params = &f3;

    gsl_integration_qags(&F2, z1, z2, 0, 1e-7, 10, w1, &resultf2, &errorf2);
    gsl_integration_workspace_free (w1);

    return resultf2; 
    }

  static double ret_int_2_wrapper(double y, void *params)
  {
   return static_cast<NRf2*>(params)->ret_int_2(y); 
  }

};
//integrates over the y-variable ==> gives function of x

class output {
private:  
  gsl_function F;
public:
  double result, error;

  double integrate(double x, double y1, double y2, double z1, double z2){

    NRf2 f2(z1, z2);
    f2.f3.xsav = x;

    F.function = &(f2.ret_int_2_wrapper);
    F.params = &f2;

    gsl_integration_qags(&F, y1, y2, 0, 1e-7, 10, w2, &result, &error);
    gsl_integration_workspace_free (w2);

    return result;
  }

};

我这样调用积分器：

#include <stdio.h>
#include </users/.../NRquad3d_Const_qags_2.C>

double func3d(double x, double y, double z)
  {
    double f = x*y*z;
    return f;
  }


int main (void)
{
  double result; 
  double error;
  double expected = -4.0;

  output out ; 
  result = out.integrate(1.0, -1.0, 1.0, -1.0, 1.0);

  printf ("result          = % .18f\n", result);
  printf ("exact result    = % .18f\n", expected); 
  printf ("actual error    = % .18f\n", result - expected);

  return 0;
}

代码编译正常，但是运行 a.out 给我一个段错误。来自 gdb 的回溯给出

#0  0x00007ffff7a4b430 in ?? () from /usr/lib64/libgsl.so.0
#1  0x0000000000400bed in NRf2::ret_int_2 (this=0x7fffffffdbd0, y=-0.97390652851717174) at /users/.../NRquad3d_Const_qags_2.C:45
#2  0x0000000000400c39 in NRf2::ret_int_2_wrapper (y=-0.97390652851717174, params=0x7fffffffdbd0) at /users/.../NRquad3d_Const_qags_2.C:53
#3  0x00007ffff7a49c6e in gsl_integration_qk () from /usr/lib64/libgsl.so.0
#4  0x00007ffff7a4997b in gsl_integration_qk21 () from /usr/lib64/libgsl.so.0
#5  0x00007ffff7a4b4c7 in ?? () from /usr/lib64/libgsl.so.0
#6  0x0000000000400d14 in output::integrate (this=0x7fffffffdc30, x=1, y1=-1, y2=1, z1=-1, z2=1) at /users/.../NRquad3d_Const_qags_2.C:73
#7  0x00000000004009ac in main () at crap_test.cpp:22
(gdb) frame 6
#6  0x0000000000400d14 in output::integrate (this=0x7fffffffdc30, x=1, y1=-1, y2=1, z1=-1, z2=1) at /users/.../NRquad3d_Const_qags_2.C:73
73          gsl_integration_qags(&F, y1, y2, 0, 1e-7, 10, w2, &result, &error);
(gdb) frame 1
#1  0x0000000000400bed in NRf2::ret_int_2 (this=0x7fffffffdbd0, y=-0.97390652851717174) at /users/.../NRquad3d_Const_qags_2.C:45
45          gsl_integration_qags(&F2, z1, z2, 0, 1e-7, 10, w1, &resultf2, &errorf2);

运行 gdb 已经帮助我消除了 w1 和 w2 地址中的冲突，但段错误仍然存在。由于错误似乎发生在 NRf2 中对 qags 的第二次调用中，我认为必须有一个我试图引用的已释放指针，但我似乎找不到它。有人可以帮忙吗？

Answer 1

运行 valgrind 给出以下形式的错误：

==26286== Invalid read of size 8
==26286==    at 0x4ED0239: gsl_integration_workspace_free (in /usr/lib64/libgsl.so.0.16.0)
==26286==    by 0x400A54: main (crap_test.cpp:32)
==26286==  Address 0x6053090 is 80 bytes inside a block of size 88 free'd
==26286==    at 0x4C29D4E: free (in /usr/lib64/valgrind/vgpreload_memcheck-amd64-linux.so)
==26286==    by 0x400C39: NRf2::ret_int_2(double) (NRquad3d_Const_qags_2.C:46)
==26286==    by 0x400C76: NRf2::ret_int_2_wrapper(double, void*) (NRquad3d_Const_qags_2.C:53)
==26286==    by 0x4ECBBA3: gsl_integration_qk (in /usr/lib64/libgsl.so.0.16.0)
==26286==    by 0x4ECB97A: gsl_integration_qk21 (in /usr/lib64/libgsl.so.0.16.0)
==26286==    by 0x4ECD4C6: ??? (in /usr/lib64/libgsl.so.0.16.0)
==26286==    by 0x400D51: output::integrate(double, double, double, double, double) (NRquad3d_Const_qags_2.C:73)
==26286==    by 0x4009CB: main (crap_test.cpp:25)

这暗示了我 trying to access freed memory。在我的脚本中，我尝试这样做的唯一地方是行

gsl_integration_workspace_free (w1);
.
.
.
.
gsl_integration_workspace_free (w2);

在 .C 文件中。然后我（重新）将这些行移动到我的 .cpp 文件的末尾并且它有效。

愚蠢的错误，但无论如何我都会回答这个问题，以防有人遇到这个确切的（或类似的！）错误并需要解决方案。

调用 gsl_integration 时出现分段错误

Segmentation fault while calling gsl_integration

c++

segmentation-fault

gsl