user: 这段汇编代码用C语言表示是什么样的：

```asm
<test_dobldobl_multiplicity_structure>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,%r12d
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  lea    0x1(%rdi),%eax
  mov    %rsp,%rcx
  cltq
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L62
L17:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L17
L62:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L0
L38:
  lea    D25(%rip),%rdi
  lea    -0x34(%rbp),%r13
  mov    %rsp,%r14
  call   <puts@plt>
  xor    %eax,%eax
  call   <syscon_read_dobldobl_system@plt>
  mov    %r13,%rdi
  call   <syscon_number_of_dobldobl_polynomials@plt>
  mov    -0x34(%rbp),%esi
  lea    D62(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  call   <solcon_read_dobldobl_solutions@plt>
  lea    -0x30(%rbp),%rdi
  call   <solcon_number_of_dobldobl_solutions@plt>
  mov    -0x30(%rbp),%esi
  lea    D56(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rdi
  call   <solcon_dimension_of_dobldobl_solutions@plt>
  mov    -0x34(%rbp),%esi
  lea    D35(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    -0x2c(%rbp),%rdx
  mov    %r14,%rcx
  mov    %r12d,%edi
  movsd  D14(%rip),%xmm0
  mov    $0x1,%esi
  call   <dobldobl_multiplicity_structure@plt>
  mov    -0x2c(%rbp),%esi
  lea    D33(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  lea    D11(%rip),%rdi
  call   <printf@plt>
  test   %r12d,%r12d
  js     L12
  xor    %ebx,%ebx
  lea    D0(%rip),%r13
  nopw   0x0(%rax,%rax,1)
L63:
  mov    (%r14,%rbx,4),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,%r12d
  jge    L63
L12:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nopl   (%rax)
L0:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L38
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D25
type: string
name: parse_data
arguments: data_tag: D62
type: string
name: parse_data
arguments: data_tag: D56
type: string
name: parse_data
arguments: data_tag: D35
type: string
name: parse_data
arguments: data_tag: D14
type: byte[8]
name: parse_data
arguments: data_tag: D33
type: string
name: parse_data
arguments: data_tag: D11
type: string
name: parse_data
arguments: data_tag: D0
type: string

tool: D25: 
Testing with dobldobl arithmetic ...

tool: D62: The system container has %d polynomials.


tool: D56: The solution container has size %d.


tool: D35: The solutions in the container have dimension %d.


tool: {"D14": [0x3a,0x8c,0x30,0xe2,0x8e,0x79,0x45,0x3e]}

tool: D33: The multiplicity : %d.


tool: D11: Values of the Hilbert function :

tool: D0:  %d

assistant: ```c
void test_dobldobl_multiplicity_structure(int order) {
  const double tol = 1.0e-8;
  int fail, dim, len, mult;
  int hilb[order + 1];
  printf("\nTesting with dobldobl arithmetic ...\n");
  fail = syscon_read_dobldobl_system();
  fail = syscon_number_of_dobldobl_polynomials(&dim);
  printf("The system container has %d polynomials.\n", dim);
  fail = solcon_read_dobldobl_solutions();
  fail = solcon_number_of_dobldobl_solutions(&len);
  printf("The solution container has size %d.\n", len);
  fail = solcon_dimension_of_dobldobl_solutions(&dim);
  printf("The solutions in the container have dimension %d.\n", dim);
  fail = dobldobl_multiplicity_structure(order, 1, tol, &mult, hilb);
  printf("The multiplicity : %d.\n", mult);
  printf("Values of the Hilbert function :");
  int k;
  for (k = 0; k < order + 1; k++)
    printf(" %d", hilb[k]);
  printf("\n");
}
```