user: 这段汇编代码用C语言表示是什么样的?

```asm
<lengthOfLongestSubstring>:
  endbr64
  push   %r15
  mov    $0x40,%ecx
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x228,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x218(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r15
  cmpb   $0x0,(%r12)
  mov    %r15,%rdi
  rep stos %rax,%es:(%rdi)
  je     L75
  movl   $0x0,0xc(%rsp)
  xor    %ebp,%ebp
  xor    %r13d,%r13d
  lea    D29(%rip),%rbx
  movl   $0x0,0x8(%rsp)
L35:
  mov    0x8(%rsp),%ecx
  mov    %ebp,%edx
  mov    %r13d,%esi
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  xor    %r14d,%r14d
  call   <printf@plt>
  jmp    L111
  xchg   %ax,%ax
L65:
  add    $0x1,%r14
  cmp    $0x80,%r14
  je     L25
L111:
  mov    (%r15,%r14,4),%edx
  mov    %r14d,%esi
  test   %edx,%edx
  je     L65
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x1,%r14
  call   <printf@plt>
  cmp    $0x80,%r14
  jne    L111
L25:
  mov    $0xa,%edi
  call   <putchar@plt>
  movsbq (%r12),%rax
  mov    0x10(%rsp,%rax,4),%eax
  cmp    0x8(%rsp),%eax
  jle    L105
  cmp    %ebp,%r13d
  mov    %eax,0x8(%rsp)
  cmovl  %ebp,%r13d
  mov    0xc(%rsp),%ebp
  sub    %eax,%ebp
L105:
  mov    0x8(%rsp),%ecx
  mov    %ebp,%edx
  mov    %r13d,%esi
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  xor    %r14d,%r14d
  call   <printf@plt>
  jmp    L73
  nopl   (%rax)
L36:
  add    $0x1,%r14
  cmp    $0x80,%r14
  je     L55
L73:
  mov    (%r15,%r14,4),%edx
  mov    %r14d,%esi
  test   %edx,%edx
  je     L36
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x1,%r14
  call   <printf@plt>
  cmp    $0x80,%r14
  jne    L73
L55:
  mov    $0xa,%edi
  add    $0x1,%r12
  add    $0x1,%ebp
  call   <putchar@plt>
  addl   $0x1,0xc(%rsp)
  movsbq (%r12),%rdx
  mov    0xc(%rsp),%ecx
  mov    %ecx,0x10(%rsp,%rdx,4)
  test   %dl,%dl
  jne    L35
  cmp    %ebp,%r13d
  mov    %ebp,%eax
  cmovge %r13d,%eax
L101:
  mov    0x218(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L78
  add    $0x228,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L75:
  xor    %eax,%eax
  jmp    L101
L78:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D29
type_infered: string
name: parse_data
arguments: data_label: D7
type_infered: string
name: parse_data
arguments: data_label: D14
type_infered: string

tool: D29: pos[%c] = %d 

tool: D7: bef: longest = %d, currentLength = %d, start = %d


tool: D14: aft: longest = %d, currentLength = %d, start = %d


assistant: ```c
int lengthOfLongestSubstring(char *s) {
  int longest = 0, currentLength = 0, start = 0, position[128] = {0};
  int i, j;
  for (i = 0; s[i]; position[s[i]] = ++i, currentLength++) {
    printf("bef: longest = %d, currentLength = %d, start = %d\n", longest, currentLength, start);
    for (j = 0; j < 128; ++j) {
      if (position[j]) {
        printf("pos[%c] = %d ", j, position[j]);
      }
    }
    printf("\n");
    if (position[s[i]] > start) {
      longest = (longest > currentLength) ? longest : currentLength;
      start = position[s[i]];
      currentLength = i - position[s[i]];
    }
    printf("aft: longest = %d, currentLength = %d, start = %d\n", longest, currentLength, start);
    for (j = 0; j < 128; ++j) {
      if (position[j]) {
        printf("pos[%c] = %d ", j, position[j]);
      }
    }
    printf("\n");
  }
  return (longest > currentLength) ? longest : currentLength;
}
```