user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<inverttable>:
  endbr64
  push   %r14
  mov    %ecx,%r14d
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  lea    (%rsi,%rsi,1),%edi
  push   %rbx
  movslq %edx,%rbx
  sub    $0x10,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %r13d,%r13d
  jle    L72
  lea    -0x1(%r13),%eax
  cmp    $0x6,%eax
  jbe    L13
  mov    %r13d,%edx
  movdqa D22(%rip),%xmm0
  mov    %r12,%rax
  shr    $0x3,%edx
  shl    $0x4,%rdx
  add    %r12,%rdx
  nopl   (%rax)
L108:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rax,%rdx
  jne    L108
  mov    %r13d,%eax
  and    $0xfffffff8,%eax
  mov    %eax,%edx
  cmp    %r13d,%eax
  je     L72
L29:
  mov    %r13d,%ecx
  sub    %eax,%ecx
  lea    -0x1(%rcx),%esi
  cmp    $0x2,%esi
  jbe    L87
  mov    D22(%rip),%rsi
  mov    %rsi,(%r12,%rax,2)
  mov    %ecx,%eax
  and    $0xfffffffc,%eax
  add    %eax,%edx
  cmp    %eax,%ecx
  je     L72
L87:
  movslq %edx,%rax
  mov    $0xffff8000,%edi
  mov    %di,(%r12,%rax,2)
  lea    (%rax,%rax,1),%rcx
  lea    0x1(%rdx),%eax
  cmp    %eax,%r13d
  jle    L72
  mov    $0xffff8000,%esi
  add    $0x2,%edx
  mov    %si,0x2(%r12,%rcx,1)
  cmp    %edx,%r13d
  jle    L72
  mov    $0xffff8000,%eax
  mov    %ax,0x4(%r12,%rcx,1)
L72:
  test   %ebx,%ebx
  jle    L77
  pxor   %xmm3,%xmm3
  pxor   %xmm4,%xmm4
  cvtsi2sd %r14d,%xmm3
  xor    %r14d,%r14d
  cvtsi2sd %r13d,%xmm4
  movsd  %xmm3,(%rsp)
  movsd  %xmm4,0x8(%rsp)
  nopl   0x0(%rax,%rax,1)
L82:
  pxor   %xmm0,%xmm0
  cvtsi2sdl 0x0(%rbp,%r14,4),%xmm0
  divsd  (%rsp),%xmm0
  mulsd  0x8(%rsp),%xmm0
  call   <rint@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %r14d,%xmm0
  divsd  D29(%rip),%xmm0
  mov    %rax,%r13
  subsd  D9(%rip),%xmm0
  mulsd  D27(%rip),%xmm0
  movslq %r13d,%r13
  call   <rint@plt>
  mov    $0x7fff,%edx
  cmp    %edx,%eax
  cmovg  %edx,%eax
  mov    $0xffff8001,%edx
  cmp    %edx,%eax
  cmovl  %edx,%eax
  add    $0x1,%r14
  mov    %ax,(%r12,%r13,2)
  cmp    %rbx,%r14
  jne    L82
L77:
  add    $0x10,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L13:
  xor    %eax,%eax
  xor    %edx,%edx
  jmp    L29
```

tool_calls: name: disassemble_data
arguments: data_tag: D22
type_guessed: byte[16]
name: disassemble_data
arguments: data_tag: D29
type_guessed: byte[8]
name: disassemble_data
arguments: data_tag: D9
type_guessed: byte[8]
name: disassemble_data
arguments: data_tag: D27
type_guessed: byte[8]

tool: {"D22": [0x00,0x80,0x00,0x80,0x00,0x80,0x00,0x80,0x00,0x80,0x00,0x80,0x00,0x80,0x00,0x80]}

tool: {"D29": [0x00,0x00,0x00,0x00,0x00,0x6a,0xe8,0x40]}

tool: {"D9": [0x00,0x00,0x00,0x00,0x00,0x00,0x10,0x40]}

tool: {"D27": [0x00,0x00,0x00,0x00,0x00,0x00,0xc0,0x40]}

assistant: ```c
short *inverttable(int *table, int inversesize, int tablesize, int cumulative) {
  int i, inverseindex, inversevalue;
  short *inverse;
  double findex, fvalue;
  inverse = (short *)malloc(inversesize * sizeof(short));
  for (i = 0; i < inversesize; ++i) {
    inverse[i] = -32768;
  }
  for (i = 0; i < tablesize; ++i) {
    findex = ((double)i / (double)50000) - ((32767 / 8192) + 1);
    fvalue = (double)table[i] / (double)cumulative;
    inverseindex = (int)rint(fvalue * inversesize);
    inversevalue = (int)rint(findex * 8192);
    if (inversevalue <= -32768)
      inversevalue = -32768 + 1;
    if (inversevalue > 32767)
      inversevalue = 32767;
    inverse[inverseindex] = inversevalue;
  }
  return inverse;
}
```