user: 这段汇编代码用C语言表示是什么样的：

```asm
<http_request_recv>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  mov    (%rcx),%r13
  mov    %edi,0xc(%rsp)
  mov    %edx,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0xa0d0a0d,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  test   %r13,%r13
  je     L144
L183:
  lea    0x13(%rsp),%r15
  jmp    L145
L25:
  test   %eax,%eax
  jle    L74
  movslq %eax,%r13
  mov    %rsi,%rax
  sub    %rbx,%rax
  cmp    %rax,%r13
  jge    L2
  mov    %rbx,%rdi
  call   <strcpy@plt>
  lea    0x1(%r14),%eax
  lea    (%rbx,%r13,1),%rsi
L143:
  mov    %rbx,0x0(%rbp)
L55:
  mov    0x8(%rsp),%edx
  mov    0xc(%rsp),%edi
  sub    %eax,%edx
  mov    D8(%rip),%rax
  mov    (%rax),%ecx
  call   <recv@plt>
  test   %eax,%eax
  js     L52
  je     L1
  add    %r14d,%eax
  cltq
  movb   $0x0,(%rbx,%rax,1)
  mov    0x0(%rbp),%r13
L145:
  mov    %r15,%rsi
  mov    %r13,%rdi
  call   <strstr@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L149
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    0x0(%rbp),%rsi
  mov    %eax,%r14d
  cmp    %rbx,%rsi
  jne    L25
  movslq %eax,%r13
  mov    0x8(%rsp),%eax
  lea    (%rbx,%r13,1),%rsi
  sub    $0x1,%eax
  cmp    %r14d,%eax
  je     L90
  lea    0x1(%r14),%eax
  jmp    L55
  nopl   (%rax)
L74:
  movslq %r14d,%r13
  lea    0x1(%rax),%eax
  lea    (%rbx,%r13,1),%rsi
  jmp    L143
  nop
L2:
  movzbl (%rsi),%edx
  mov    $0x1,%eax
  mov    %rbx,%rsi
  xor    %r13d,%r13d
  xor    %r14d,%r14d
  test   %dl,%dl
  je     L57
  nopl   (%rax)
L140:
  mov    %dl,-0x1(%rbx,%rax,1)
  mov    0x0(%rbp),%rdx
  mov    %rax,%r13
  mov    %eax,%r14d
  movzbl (%rdx,%rax,1),%edx
  lea    0x1(%r13),%rax
  test   %dl,%dl
  jne    L140
  lea    (%rbx,%r13,1),%rsi
  lea    0x1(%r14),%eax
L57:
  movb   $0x0,(%rsi)
  jmp    L143
  nopw   0x0(%rax,%rax,1)
L149:
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  xor    %esi,%esi
  mov    %eax,%edx
  call   <memset@plt>
  mov    %r15,%rdi
  call   <strlen@plt>
  cltq
  add    %r12,%rax
  mov    %r13,%r12
  mov    %rax,0x0(%rbp)
L52:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L161
  add    $0x28,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L144:
  mov    %rsi,0x0(%rbp)
  movb   $0x0,(%rsi)
  mov    (%rcx),%r13
  jmp    L183
  nopw   0x0(%rax,%rax,1)
L1:
  test   %r14d,%r14d
  je     L52
  mov    0x0(%rbp),%r12
L126:
  lea    (%r12,%r13,1),%r9
  mov    %r9,0x0(%rbp)
  jmp    L52
L90:
  mov    %rbx,%r12
  jmp    L126
L161:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *http_request_recv(int fd, char *buf, int buflen, char **data) {
  const char delim[] = "\r\n"
                       "\r\n";
  char *retval, *split;
  int len, recvlen;
  if (!*data) {
    *data = buf;
    *buf = '\0';
  }
  while ((split = strstr(*data, delim)) == NULL) {
    len = strlen(*data);
    if (*data == buf && len == (buflen - 1)) {
      break;
    }
    if (*data != buf) {
      if (len > 0) {
        if (len < *data - buf) {
          strcpy(buf, *data);
        } else {
          for (len = 0; (*data)[len] != '\0'; ++len)
            buf[len] = (*data)[len];
          buf[len] = '\0';
        }
      }
      *data = buf;
    }
    recvlen = recv(fd, buf + len, buflen - (len + 1), MSG_DONTWAIT);
    if (recvlen < 0) {
      return NULL;
    } else if (recvlen == 0) {
      if (len == 0) {
        return NULL;
      }
      break;
    }
    len += recvlen;
    buf[len] = '\0';
  }
  retval = *data;
  if (split) {
    memset(split, '\0', strlen(delim));
    *data = split + strlen(delim);
  } else {
    *data += len;
  }
  return retval;
}
```