user: 这段汇编代码用C语言表示是什么样的：

```asm
<loadfile>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x4,%edi
  push   %rbx
  sub    $0x18,%rsp
  call   <malloc@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  mov    %rax,%rsi
  call   <readfile@plt>
  mov    (%r12),%esi
  mov    %rax,%rdi
  mov    %rax,%r15
  call   <analyze@plt>
  mov    (%r12),%r8d
  test   %r8d,%r8d
  je     L49
  mov    %r8d,%ecx
  xor    %edx,%edx
  jmp    L71
L50:
  add    $0x1,%rdx
  cmp    %rdx,%rcx
  je     L49
L71:
  mov    (%r15,%rdx,8),%rax
  mov    %edx,%esi
  movzbl (%rax),%edi
  cmp    $0x40,%dil
  jne    L50
  cmp    %r8d,%edx
  jae    L49
  mov    %esi,%edx
  xor    %ebx,%ebx
  lea    (%r15,%rdx,8),%rcx
  cmp    $0x28,%dil
  je     L85
  nopw   0x0(%rax,%rax,1)
L64:
  lea    0x1(%rbx),%r13d
  mov    %ebx,%edx
  cmpb   $0x28,(%rax,%r13,1)
  mov    %r13,%rbx
  jne    L64
  lea    0x2(%rdx),%ebp
  mov    %rcx,(%rsp)
  mov    %ebp,%edi
  mov    %esi,0x8(%rsp)
  mov    %edx,0xc(%rsp)
  call   <malloc@plt>
  test   %r13d,%r13d
  mov    0x8(%rsp),%esi
  mov    (%rsp),%rcx
  mov    %rax,%r14
  mov    %ebp,%edi
  je     L144
  mov    0xc(%rsp),%edx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L75:
  mov    (%rcx),%r8
  movzbl (%r8,%rax,1),%r8d
  mov    %r8b,(%r14,%rax,1)
  mov    %rax,%r8
  add    $0x1,%rax
  cmp    %rdx,%r8
  jne    L75
  add    %r14,%r13
L45:
  movb   $0x0,0x0(%r13)
  mov    (%r12),%r8d
  lea    0x1(%rsi),%ebx
  cmp    %r8d,%ebx
  jae    L179
  mov    %ebx,%eax
  lea    (%r15,%rax,8),%rax
  jmp    L180
  nopl   0x0(%rax)
L38:
  add    $0x1,%ebx
  add    $0x8,%rax
  cmp    %r8d,%ebx
  jae    L179
L180:
  mov    (%rax),%rdx
  cmpb   $0x40,(%rdx)
  jne    L38
L179:
  mov    (%rcx),%rdx
  xor    %r13d,%r13d
  movzbl (%rdx,%rdi,1),%eax
  cmp    $0x29,%al
  setne  %r13b
  test   %al,%al
  je     L133
  nopl   0x0(%rax,%rax,1)
L92:
  cmp    $0x2c,%al
  sete   %al
  movzbl %al,%eax
  add    %eax,%r13d
  lea    0x1(%rbp),%eax
  mov    %rax,%rbp
  movzbl (%rdx,%rax,1),%eax
  test   %al,%al
  jne    L92
L133:
  mov    %ebx,%edx
  mov    %r15,%rdi
  mov    %r13d,%ecx
  call   <loadframe@plt>
  mov    %r13d,%edx
  mov    %r14,%rsi
  mov    %rax,%rdi
  call   <addframe@plt>
  cmp    %ebx,(%r12)
  jbe    L49
  mov    %ebx,%eax
  mov    %ebx,%esi
  xor    %ebx,%ebx
  mov    (%r15,%rax,8),%rax
  mov    %esi,%edx
  lea    (%r15,%rdx,8),%rcx
  movzbl (%rax),%edi
  cmp    $0x28,%dil
  jne    L64
L85:
  mov    $0x1,%edi
  mov    %rcx,(%rsp)
  mov    $0x1,%ebp
  mov    %esi,0x8(%rsp)
  call   <malloc@plt>
  mov    0x8(%rsp),%esi
  mov    (%rsp),%rcx
  mov    $0x1,%edi
  mov    %rax,%r14
  mov    %rax,%r13
  jmp    L45
  cs nopw 0x0(%rax,%rax,1)
L49:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L144:
  mov    %rax,%r13
  mov    $0x1,%ebp
  jmp    L45
```

assistant: ```c
void loadfile(FILE *fin) {
  unsigned int *len = malloc(sizeof(unsigned int));
  char **lines = readfile(fin, len);
  analyze(lines, *len);
  unsigned int i, ii, iii;
  for (i = 0; i < *len && lines[i][0] != '@'; i++)
    ;
  while (i < *len) {
    for (ii = 0; lines[i][ii] != '('; ii++)
      ;
    char *fname = malloc((ii + 1) * sizeof(char));
    for (iii = 0; iii < ii; iii++) {
      fname[iii] = lines[i][iii];
    }
    fname[ii] = '\0';
    unsigned int start = i;
    for (i = start + 1; i < *len && lines[i][0] != '@'; i++)
      ;
    int argnum = lines[start][ii + 1] != ')';
    for (ii = ii + 1; lines[start][ii] != '\0'; ii++) {
      if (lines[start][ii] == ',') {
        argnum++;
      }
    }
    struct frame *f = loadframe(lines, start, i, argnum);
    addframe(f, fname, argnum);
  }
}
```