user: 这段汇编代码用C语言表示是什么样的？

```asm
<fgen_copy_partial_bitstring>:
  endbr64
  test   %ecx,%ecx
  je     L12
  test   %edx,%edx
  lea    0x7(%rdx),%r11d
  mov    %ecx,%eax
  mov    %edx,%ecx
  cmovns %edx,%r11d
  push   %rbp
  push   %rbx
  sar    $0x3,%r11d
  mov    %r11d,%r8d
  and    $0x7,%ecx
  jne    L87
L145:
  cmp    $0x7,%eax
  jle    L122
  movslq %r8d,%rdx
  lea    (%rsi,%rdx,1),%rcx
  and    $0x3,%ecx
  je     L114
  lea    0x1(%rsi),%r9d
  jmp    L126
  nopw   0x0(%rax,%rax,1)
L40:
  add    %r9d,%ecx
  add    $0x1,%rdx
  and    $0x3,%ecx
  je     L114
L126:
  movzbl (%rdi,%rdx,1),%ecx
  sub    $0x8,%eax
  lea    0x1(%rdx),%r8d
  mov    %cl,(%rsi,%rdx,1)
  mov    %edx,%ecx
  cmp    $0x7,%eax
  jg     L40
L122:
  test   %eax,%eax
  jle    L150
  movslq %r8d,%r8
  xor    %ecx,%ecx
  mov    $0x1,%r10d
  add    %r8,%rsi
  add    %rdi,%r8
  movzbl (%rsi),%edx
  nopl   (%rax)
L119:
  mov    %r10d,%edi
  shl    %cl,%edi
  add    $0x1,%ecx
  mov    %edi,%r9d
  not    %r9d
  and    %r9d,%edx
  mov    %dl,(%rsi)
  and    (%r8),%dil
  or     %edi,%edx
  mov    %dl,(%rsi)
  cmp    %eax,%ecx
  jne    L119
L150:
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L87:
  movslq %r11d,%rbx
  mov    $0x1,%ebp
  lea    (%rsi,%rbx,1),%r9
  add    %rdi,%rbx
  movzbl (%r9),%edx
  nopl   0x0(%rax,%rax,1)
L158:
  mov    %ebp,%r8d
  shl    %cl,%r8d
  mov    %r8d,%r10d
  not    %r10d
  and    %r10d,%edx
  mov    %dl,(%r9)
  and    (%rbx),%r8b
  or     %r8d,%edx
  mov    %dl,(%r9)
  sub    $0x1,%eax
  je     L150
  add    $0x1,%ecx
  cmp    $0x8,%ecx
  jne    L158
  lea    0x1(%r11),%r8d
  jmp    L145
  nopl   0x0(%rax)
L12:
  ret
  nopl   0x0(%rax)
L114:
  movslq %r8d,%rdx
  mov    %eax,%ecx
  cmp    $0x1f,%eax
  jle    L73
  xchg   %ax,%ax
L63:
  mov    (%rdi,%rdx,1),%r9d
  sub    $0x20,%ecx
  mov    %r9d,(%rsi,%rdx,1)
  add    $0x4,%rdx
  cmp    $0x1f,%ecx
  jg     L63
  lea    -0x20(%rax),%r10d
  xor    %r9d,%r9d
  mov    %r10d,%ecx
  shr    $0x5,%ecx
  mov    %ecx,%edx
  neg    %edx
  shl    $0x5,%edx
  cmp    $0x1f,%eax
  cmovle %r9d,%edx
  shl    $0x2,%ecx
  add    %r10d,%edx
  cmp    $0x1f,%eax
  cmovle %r9d,%ecx
  mov    %edx,%eax
  lea    0x4(%r8,%rcx,1),%r8d
  cmp    $0x7,%edx
  jle    L122
L143:
  movslq %r8d,%rax
  mov    %edx,%ecx
L10:
  movzbl (%rdi,%rax,1),%r9d
  sub    $0x8,%ecx
  mov    %r9b,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    $0x7,%ecx
  jg     L10
  lea    -0x8(%rdx),%r10d
  xor    %r9d,%r9d
  mov    %r10d,%ecx
  shr    $0x3,%ecx
  mov    %ecx,%eax
  neg    %eax
  shl    $0x3,%eax
  cmp    $0x7,%edx
  cmovle %r9d,%eax
  add    %r10d,%eax
  cmp    $0x7,%edx
  cmovle %r9d,%ecx
  lea    0x1(%r8,%rcx,1),%r8d
  jmp    L122
L73:
  mov    %eax,%edx
  jmp    L143
```

assistant: ```c
void fgen_copy_partial_bitstring(const unsigned char *src, unsigned char *dest, int bitoffset, int number_of_bits) {
  int i, offset;
  if (number_of_bits == 0)
    return;
  offset = bitoffset / 8;
  if ((bitoffset & 7) != 0) {
    for (i = bitoffset & 7; i < 8; i++) {
      dest[offset] &= ~(1 << i);
      dest[offset] |= src[offset] & (1 << i);
      number_of_bits--;
      if (number_of_bits == 0)
        return;
    }
    offset++;
  }
  while (number_of_bits >= 8 && (((unsigned int)(&dest[offset])) & 3) != 0) {
    dest[offset] = src[offset];
    number_of_bits -= 8;
    offset++;
  }
  while (number_of_bits >= 32) {
    *(unsigned int *)&dest[offset] = *(unsigned int *)&src[offset];
    number_of_bits -= 32;
    offset += 4;
  }
  while (number_of_bits >= 8) {
    dest[offset] = src[offset];
    number_of_bits -= 8;
    offset++;
  }
  for (i = 0; i < number_of_bits; i++) {
    dest[offset] &= ~(1 << i);
    dest[offset] |= src[offset] & (1 << i);
  }
}
```