user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D3(%rip),%rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%r14
  movq   $0x0,0x10(%rsp)
L8:
  movq   $0x0,0x8(%rsp)
  mov    %r14,%r13
L1:
  mov    0x8(%rsp),%r15
  mov    $0x61,%esi
  mov    %r14,%rdi
  xor    %r12d,%r12d
  lea    0x0(,%r15,4),%rbx
  mov    %rbx,%rdx
  call   <memset@plt>
  lea    0x1(%r15),%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  jmp    L69
L61:
  mov    %r12,%rax
  and    $0x1,%eax
  neg    %eax
  and    $0x62,%eax
L69:
  mov    %al,0x0(%r13)
  mov    %r12,%rax
  mov    %r14,%rdi
  and    $0x2,%eax
  movb   $0x0,0x4(%r13)
  neg    %rax
  sbb    %al,%al
  and    $0x63,%eax
  mov    %al,0x1(%r13)
  mov    %r12,%rax
  and    $0x4,%eax
  neg    %rax
  sbb    %al,%al
  and    $0x64,%eax
  mov    %al,0x2(%r13)
  mov    %r12,%rax
  and    $0x8,%eax
  neg    %rax
  sbb    %al,%al
  and    $0x65,%eax
  mov    %al,0x3(%r13)
  call   <strlen@plt>
  mov    0x0(%rbp,%r12,8),%r9
  mov    %rax,%r15
  add    %rbx,%r9
  cmp    %r9,%rax
  jne    L37
  mov    $0xffffffff,%esi
  mov    %r14,%rdi
  call   <strnlen@plt>
  cmp    %rax,%r15
  jne    L3
  add    $0x1,%r12
  cmp    $0x10,%r12
  jne    L61
  add    $0x4,%r13
  cmpq   $0x4,0x18(%rsp)
  jne    L2
  addq   $0x1,0x10(%rsp)
  mov    0x10(%rsp),%rax
  add    $0x1,%r14
  cmp    $0x20,%rax
  jne    L8
  xor    %eax,%eax
  jmp    L41
L2:
  mov    0x18(%rsp),%rax
  mov    %rax,0x8(%rsp)
  jmp    L1
L37:
  mov    0x8(%rsp),%rdx
  mov    0x10(%rsp),%rsi
  mov    %rax,%r8
  mov    %r12,%rcx
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
L41:
  mov    0x58(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L40
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L3:
  or     $0xffffffff,%esi
  mov    %r14,%rdi
  call   <strnlen@plt>
  mov    0x8(%rsp),%rdx
  mov    %r15,%r9
  mov    %r12,%rcx
  mov    0x10(%rsp),%rsi
  mov    %rax,%r8
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  jmp    L41
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D3
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D7
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D8
type_guessed: string

tool: 

tool: D7: strlen failed for base=%Zu, words=%Zu, and last=%Zu (is %zd, expected %zd)


tool: D8: strnlen failed for base=%Zu, words=%Zu, and last=%Zu (is %zd, expected %zd)


assistant: ```c
int main(int argc, char *argv[]) {
  static const size_t lens[] = {0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4};
  char basebuf[24 + 32];
  size_t base;
  for (base = 0; base < 32; ++base) {
    char *buf = basebuf + base;
    size_t words;
    for (words = 0; words < 4; ++words) {
      size_t last;
      memset(buf, 'a', words * 4);
      for (last = 0; last < 16; ++last) {
        buf[words * 4 + 0] = (last & 1) != 0 ? 'b' : '\0';
        buf[words * 4 + 1] = (last & 2) != 0 ? 'c' : '\0';
        buf[words * 4 + 2] = (last & 4) != 0 ? 'd' : '\0';
        buf[words * 4 + 3] = (last & 8) != 0 ? 'e' : '\0';
        buf[words * 4 + 4] = '\0';
        if (strlen(buf) != words * 4 + lens[last]) {
          printf("strlen failed for base=%Zu, words=%Zu, and last=%Zu (is %zd, expected %zd)\n", base, words, last, strlen(buf), words * 4 + lens[last]);
          return 1;
        }
        if (strnlen(buf, -1) != words * 4 + lens[last]) {
          printf("strnlen failed for base=%Zu, words=%Zu, and last=%Zu (is %zd, expected %zd)\n", base, words, last, strnlen(buf, -1), words * 4 + lens[last]);
          return 1;
        }
      }
    }
  }
  return 0;
}
```