user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x302f,-0xd0(%rbp)
  movq   $0x0,-0xc8(%rbp)
  movq   $0x0,-0xc0(%rbp)
  movq   $0x0,-0xb8(%rbp)
  movq   $0x0,-0xb0(%rbp)
  movq   $0x0,-0xa8(%rbp)
  movq   $0x0,-0xa0(%rbp)
  movq   $0x0,-0x98(%rbp)
  movq   $0x0,-0x90(%rbp)
  movq   $0x0,-0x88(%rbp)
  movq   $0x0,-0x80(%rbp)
  movq   $0x0,-0x78(%rbp)
  movq   $0x0,-0x70(%rbp)
  movq   $0x0,-0x68(%rbp)
  movq   $0x0,-0x60(%rbp)
  movq   $0x0,-0x58(%rbp)
  movq   $0x0,-0x50(%rbp)
  movq   $0x0,-0x48(%rbp)
  movq   $0x0,-0x40(%rbp)
  movq   $0x0,-0x38(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  movl   $0x0,-0xdc(%rbp)
  mov    -0xdc(%rbp),%eax
  mov    %eax,-0xd4(%rbp)
  mov    -0xd4(%rbp),%eax
  mov    %eax,-0xd8(%rbp)
  mov    -0xd8(%rbp),%eax
  mov    %eax,-0xe0(%rbp)
  mov    $0x2,%esi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1f,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xd0(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <gets@plt>
  movl   $0x0,-0xe0(%rbp)
  jmp    L35
L3:
  mov    -0xe0(%rbp),%eax
  cltq
  movzbl -0xd0(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isalpha@plt>
  test   %eax,%eax
  je     L57
  addl   $0x1,-0xd8(%rbp)
L57:
  mov    -0xe0(%rbp),%eax
  cltq
  movzbl -0xd0(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %eax,%eax
  je     L43
  addl   $0x1,-0xd4(%rbp)
L43:
  mov    -0xe0(%rbp),%eax
  cltq
  movzbl -0xd0(%rbp,%rax,1),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  cmp    $0x9,%eax
  ja     L16
  addl   $0x1,-0xdc(%rbp)
L16:
  addl   $0x1,-0xe0(%rbp)
L35:
  cmpl   $0xc7,-0xe0(%rbp)
  jle    L3
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0xd4(%rbp),%ecx
  mov    -0xdc(%rbp),%edx
  mov    -0xd8(%rbp),%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D31
type_infered: string
name: retrieve_data
arguments: data_label: D13
type_infered: string
name: retrieve_data
arguments: data_label: D44
type_infered: string
name: retrieve_data
arguments: data_label: D20
type_infered: string
name: retrieve_data
arguments: data_label: D29
type_infered: string

tool: D31: 		The great letter counting program %c



tool: D13: Type here the sentence of which you want the letters to be counted: %c


tool: D44: 

	Here comes the results....

tool: D20: So....

tool: D29: 
In the previous sentence you had %d letters, %d digits and %d spaces!




assistant: ```c
int main() {
  char s[200] = {"/0"};
  int x, d, l, sp;
  x = l = sp = d = 0;
  printf("\t\tThe great letter counting program %c\n\n", 2);
  printf("Type here the sentence of which you want the letters to be counted: %c\n", 31);
  gets(s);
  for (x = 0; x < 200; x++) {
    if (isalpha(s[x]))
      l++;
    if (isspace(s[x]))
      sp++;
    if (isdigit(s[x]))
      d++;
  }
  printf("\n\n\tHere comes the results....\n");
  printf("So....");
  printf("\nIn the previous sentence you had %d letters, %d digits and %d spaces!\n\n\n", l, d, sp);
}
```