Question

我正在尝试使用Numerical Recipes实现多维集成例程（代码取自here，第164页）。使用包装器将被积函数传递给1 3 3 1 2 2 3 1 3来自here（在那里进一步引用）。我的void* params声明位于class，如下所示

NRquad3d_Const_qags_2.C

我给集成商打电话如下：

#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;
  }

};

代码编译正常，但运行#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; }会给我一个段错误。来自a.out的Backtrace给出了

gdb

正在运行#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);已帮助我消除了gdb和w1地址中的冲突，但仍然存在段错误。由于错误似乎发生在w2中对qags的第二次调用中的调用中，我认为必须有一个我想要引用的解除分配的指针，但我似乎无法找到它。有人可以帮忙吗？

Answer 1

正在运行valgrind会出现以下格式错误：

==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)

暗示了我trying to access freed memory。我尝试这样做的脚本中唯一的地方是

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

<{1>}文件中的

。然后我（重新）将这些行移动到我的.C文件的末尾，它可以工作。

愚蠢的错误，但无论如何，如果有人有这个确切（或类似的！）错误并需要解决方案，我会回答这个问题。

调用gsl_integration时出现分段错误

1 个答案: