user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<_get_long_sub_str>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  mov    %rdi,%r15
  mov    %rdi,0x18(%rsp)
  mov    %esi,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%rdi
  mov    $0xd,%ecx
  rep stos %rax,%es:(%rdi)
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jle    L61
  mov    %r15,%rdx
  lea    -0x1(%rax),%eax
  lea    0x1(%r15,%rax,1),%rsi
  mov    $0x0,%ecx
L0:
  movsbl (%rdx),%eax
  sub    $0x61,%eax
  cltq
  cmpl   $0x1,0x20(%rsp,%rax,4)
  adc    $0x0,%ecx
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L0
L68:
  cmp    %ecx,%r12d
  jg     L5
L110:
  mov    0x18(%rsp),%rcx
  movsbl (%rcx),%eax
  sub    $0x61,%eax
  cltq
  addl   $0x1,0x20(%rsp,%rax,4)
  cmp    $0x1,%ebx
  jle    L67
  lea    0x1(%rcx),%rax
  lea    -0x2(%rbx),%edx
  lea    0x2(%rcx,%rdx,1),%rcx
  mov    %rcx,0x8(%rsp)
  mov    %rax,%r14
  movl   $0x1,0x4(%rsp)
  movl   $0x0,0x14(%rsp)
  mov    $0x0,%ebx
  mov    $0x1,%edx
  sub    %eax,%edx
  mov    %edx,0x10(%rsp)
  jmp    L15
L61:
  mov    $0x0,%ecx
  jmp    L68
L5:
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  jmp    L110
L86:
  movsbl 0x0(%r13),%eax
  sub    $0x61,%eax
  cltq
  subl   $0x1,0x20(%rsp,%rax,4)
  add    $0x1,%ebx
L13:
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <is_valid@plt>
  test   %eax,%eax
  je     L86
  sub    %ebx,%r15d
  cmp    0x4(%rsp),%r15d
  jl     L38
  lea    0x1(%r15),%eax
  mov    %eax,0x4(%rsp)
  mov    %ebx,0x14(%rsp)
L38:
  add    $0x1,%r14
  cmp    0x8(%rsp),%r14
  je     L75
L15:
  mov    %r14,%r13
  movsbl (%r14),%eax
  sub    $0x61,%eax
  cltq
  addl   $0x1,0x20(%rsp,%rax,4)
  mov    0x10(%rsp),%eax
  lea    (%rax,%r14,1),%r15d
  lea    0x20(%rsp),%rbp
  jmp    L13
L67:
  movl   $0x1,0x4(%rsp)
  movl   $0x0,0x14(%rsp)
L75:
  mov    0x4(%rsp),%ecx
  mov    0x14(%rsp),%edx
  add    %edx,%ecx
  mov    0x18(%rsp),%rsi
  lea    D12(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  add    $0x98,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L55:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D18
data_type: string
name: retrieve_data
arguments: data_label: D12
data_type: string

tool: D18: 
 Error there are not enough unique characters in the string 

tool: D12: 
Max length sub str of %s starts from %d to %d

assistant: ```c
void _get_long_sub_str(char *str, int k) {
  int curr_start = 0, curr_end = 0, max_wind_start = 0, max_wind_size = 0;
  int count[26] = {};
  int len = strlen(str);
  int i = 0, unique = 0;
  for (i = 0; i < len; i++) {
    if (count[str[i] - 'a'] == 0) {
      unique++;
    }
  }
  if (unique < k) {
    printf("\n Error there are not enough unique characters in the string ");
    exit(-1);
  }
  max_wind_start = 0;
  max_wind_size = 1;
  curr_start = 0;
  curr_end = 0;
  count[str[0] - 'a']++;
  for (i = 1; i < len; i++) {
    count[str[i] - 'a']++;
    curr_end++;
    while (!is_valid(count, k)) {
      count[str[i] - 'a']--;
      curr_start++;
    }
    if ((curr_end - curr_start + 1) > max_wind_size) {
      max_wind_size = curr_end - curr_start + 1;
      max_wind_start = curr_start;
    }
  }
  printf("\nMax length sub str of %s starts from %d to %d", str, max_wind_start, max_wind_size + max_wind_start);
}
```