user: 这段汇编代码用C语言表示是什么样的?

```asm
<LPC_inverse_pred_gain_QA>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %edx,-0x44(%rbp)
  mov    -0x44(%rbp),%eax
  cltq
  and    $0x1,%eax
  shl    $0x6,%rax
  mov    %rax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x38(%rbp),%rax
  movl   $0x40000000,(%rax)
  mov    -0x44(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x2c(%rbp)
  jmp    L15
L20:
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0xfff0,%eax
  jg     L52
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0xffff0010,%eax
  jge    L61
L52:
  mov    $0x1,%eax
  jmp    L6
L61:
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  shl    $0xf,%eax
  neg    %eax
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x24(%rbp),%eax
  cltq
  imul   %rdx,%rax
  sar    $0x20,%rax
  mov    %eax,%edx
  mov    $0x3fffffff,%eax
  sub    %edx,%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  mov    $0x2e,%esi
  mov    %eax,%edi
  call   <SKP_INVERSE32_varQ@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%eax
  cltq
  imul   %rdx,%rax
  sar    $0x20,%rax
  lea    0x0(,%rax,4),%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x2c(%rbp),%eax
  cltq
  and    $0x1,%eax
  shl    $0x6,%rax
  mov    %rax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <SKP_Silk_CLZ32@plt>
  sub    $0x1,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,%ecx
  shll   %cl,-0x1c(%rbp)
  movl   $0x0,-0x28(%rbp)
  jmp    L23
L99:
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x2c(%rbp),%eax
  sub    -0x28(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  movslq %eax,%rcx
  mov    -0x24(%rbp),%eax
  cltq
  imul   %rcx,%rax
  sar    $0x20,%rax
  lea    (%rax,%rax,1),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x1c(%rbp),%eax
  cltq
  imul   %rdx,%rax
  sar    $0x20,%rax
  mov    %eax,%esi
  mov    $0x10,%eax
  sub    -0x18(%rbp),%eax
  mov    -0x28(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x10(%rbp),%rdx
  add    %rcx,%rdx
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x28(%rbp)
L23:
  mov    -0x28(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L99
  subl   $0x1,-0x2c(%rbp)
L15:
  cmpl   $0x0,-0x2c(%rbp)
  jg     L20
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0xfff0,%eax
  jg     L91
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0xffff0010,%eax
  jge    L29
L91:
  mov    $0x1,%eax
  jmp    L6
L29:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  shl    $0xf,%eax
  neg    %eax
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x24(%rbp),%eax
  cltq
  imul   %rdx,%rax
  sar    $0x20,%rax
  mov    %eax,%edx
  mov    $0x3fffffff,%eax
  sub    %edx,%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%eax
  cltq
  imul   %rdx,%rax
  sar    $0x20,%rax
  lea    0x0(,%rax,4),%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
L6:
  leave
  ret
```

assistant: ```c
int LPC_inverse_pred_gain_QA(int *invGain_Q30, int A_QA[2][16], const int order) {
  int k, n, headrm;
  int rc_Q31, rc_mult1_Q30, rc_mult2_Q16, tmp_QA;
  int *Aold_QA, *Anew_QA;
  Anew_QA = A_QA[order & 1];
  *invGain_Q30 = (1 << 30);
  for (k = order - 1; k > 0; k--) {
    if ((Anew_QA[k] > ((int)((0.99975) * ((long long)1 << (16)) + 0.5))) || (Anew_QA[k] < -((int)((0.99975) * ((long long)1 << (16)) + 0.5)))) {
      return 1;
    }
    rc_Q31 = -((Anew_QA[k]) << (31 - 16));
    rc_mult1_Q30 = (0x7FFFFFFF >> 1) - (int)((((long long)((rc_Q31)) * ((rc_Q31)))) >> (32));
    ;
    ;
    rc_mult2_Q16 = SKP_INVERSE32_varQ(rc_mult1_Q30, 46);
    *invGain_Q30 = (((int)((((long long)((*invGain_Q30)) * ((rc_mult1_Q30)))) >> (32))) << (2));
    ;
    ;
    Aold_QA = Anew_QA;
    Anew_QA = A_QA[k & 1];
    headrm = SKP_Silk_CLZ32(rc_mult2_Q16) - 1;
    rc_mult2_Q16 = ((rc_mult2_Q16) << (headrm));
    for (n = 0; n < k; n++) {
      tmp_QA = Aold_QA[n] - (((int)((((long long)((Aold_QA[k - n - 1])) * ((rc_Q31)))) >> (32))) << (1));
      Anew_QA[n] = (((int)((((long long)((tmp_QA)) * ((rc_mult2_Q16)))) >> (32))) << (16 - headrm));
    }
  }
  if ((Anew_QA[0] > ((int)((0.99975) * ((long long)1 << (16)) + 0.5))) || (Anew_QA[0] < -((int)((0.99975) * ((long long)1 << (16)) + 0.5)))) {
    return 1;
  }
  rc_Q31 = -((Anew_QA[0]) << (31 - 16));
  rc_mult1_Q30 = (0x7FFFFFFF >> 1) - (int)((((long long)((rc_Q31)) * ((rc_Q31)))) >> (32));
  *invGain_Q30 = (((int)((((long long)((*invGain_Q30)) * ((rc_mult1_Q30)))) >> (32))) << (2));
  ;
  ;
  return 0;
}
```