user: 这段汇编代码用C语言表示是什么样的?

```asm
<bpconv>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %r8d,%ebx
  mov    0x38(%rsp),%rax
  sub    $0x2,%ebx
  mov    %ebx,-0x1c(%rsp)
  js     L46
  movslq %esi,%r11
  mov    %r11,%r10
  neg    %r10
  lea    0x0(,%r10,8),%r13
  movslq %r8d,%r10
  lea    0x0(,%r10,8),%r14
  add    %r11,%r10
  mov    %r13,-0x10(%rsp)
  lea    0x0(,%r10,8),%r11
  lea    -0x18(%r14),%rbx
  lea    -0x18(%r11),%r10
  mov    %rbx,-0x18(%rsp)
  add    %r9,%rbx
  lea    -0x10(%rax,%r14,1),%r15
  lea    (%r9,%r10,1),%rbp
  add    %rax,%r10
  cmpl   $0x1,-0x1c(%rsp)
  seta   %r12b
  cmp    $0xfffffffffffffff8,%r13
  setne  %r13b
  and    %r12d,%r13d
  cmp    %r15,%rbx
  setne  %r12b
  and    %r13d,%r12d
  cmp    %r15,%rbp
  setne  %r13b
  test   %r13b,%r12b
  je     L28
  lea    -0x8(%rax,%r11,1),%r11
  lea    -0x8(%r9,%r14,1),%r12
  cmp    %r11,%rbx
  setae  %r11b
  cmp    %r12,%r10
  setae  %r12b
  or     %r11b,%r12b
  je     L28
  lea    -0x1(%r8),%r12d
  mov    -0x18(%rsp),%r14
  xor    %r11d,%r11d
  mov    %r12d,%r13d
  shr    %r13d
  add    %rax,%r14
  neg    %r13
  shl    $0x4,%r13
  cs nopw 0x0(%rax,%rax,1)
L32:
  movupd 0x0(%rbp,%r11,1),%xmm3
  movups %xmm3,(%rbx,%r11,1)
  movupd (%r10,%r11,1),%xmm4
  movups %xmm4,(%r14,%r11,1)
  sub    $0x10,%r11
  cmp    %r13,%r11
  jne    L32
  mov    -0x1c(%rsp),%r10d
  mov    %r12d,%r11d
  and    $0xfffffffe,%r11d
  sub    %r11d,%r10d
  cmp    %r11d,%r12d
  je     L46
  lea    (%rsi,%r10,1),%r11d
  movslq %r10d,%r10
  movslq %r11d,%r11
  movsd  (%r9,%r11,8),%xmm0
  movsd  %xmm0,(%r9,%r10,8)
  movsd  (%rax,%r11,8),%xmm0
  movsd  %xmm0,(%rax,%r10,8)
L46:
  test   %esi,%esi
  jle    L88
  lea    -0x1(%r8),%r11d
  movslq %r8d,%r10
  movslq %esi,%rsi
  xor    %ebp,%ebp
  movslq %r11d,%r12
  mov    %r11d,%r11d
  pxor   %xmm2,%xmm2
  sub    %r11,%r10
  shl    $0x3,%r12
  lea    -0x10(,%r10,8),%rbx
  nop
L65:
  lea    0x0(,%rbp,8),%r11
  mov    %r12,%r10
  movapd %xmm2,%xmm0
  test   %r8d,%r8d
  jle    L104
  nopl   0x0(%rax)
L22:
  movsd  (%r9,%r11,1),%xmm1
  mulsd  (%rdx,%r10,1),%xmm1
  addsd  %xmm1,%xmm0
  movsd  (%rax,%r11,1),%xmm1
  mulsd  (%rcx,%r10,1),%xmm1
  sub    $0x8,%r10
  add    $0x8,%r11
  addsd  %xmm1,%xmm0
  cmp    %r10,%rbx
  jne    L22
L104:
  movsd  %xmm0,(%rdi,%rbp,8)
  add    $0x1,%rbp
  cmp    %rbp,%rsi
  jne    L65
L88:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L28:
  mov    -0x10(%rsp),%r15
  mov    %r9,%rbx
  mov    %rax,%r11
  movslq -0x1c(%rsp),%r10
  sub    %r15,%rbx
  sub    %r15,%r11
  nopl   0x0(%rax)
L89:
  movsd  (%rbx,%r10,8),%xmm0
  movsd  %xmm0,(%r9,%r10,8)
  movsd  (%r11,%r10,8),%xmm0
  movsd  %xmm0,(%rax,%r10,8)
  sub    $0x1,%r10
  test   %r10d,%r10d
  jns    L89
  jmp    L46
```

assistant: ```c
bpconv(double *x_out, int lx, double *g0, double *g1, int lh, double *x_inl, double *x_inh) {
  int i, j;
  double x0;
  for (i = lh - 2; i > -1; i--) {
    x_inl[i] = x_inl[lx + i];
    x_inh[i] = x_inh[lx + i];
  }
  for (i = 0; i < lx; i++) {
    x0 = 0;
    for (j = 0; j < lh; j++)
      x0 = x0 + x_inl[j + i] * g0[lh - 1 - j] + x_inh[j + i] * g1[lh - 1 - j];
    x_out[i] = x0;
  }
}
```