user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<count_4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x84(%rbp)
  mov    -0x84(%rbp),%eax
  mov    %eax,-0x88(%rbp)
  mov    -0x88(%rbp),%eax
  mov    %eax,-0x8c(%rbp)
  mov    -0x8c(%rbp),%eax
  mov    %eax,-0x94(%rbp)
  lea    -0x80(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0xd,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  mov    %rdi,%rdx
  mov    %eax,(%rdx)
  add    $0x4,%rdx
  movl   $0x0,-0x90(%rbp)
  jmp    L244
L70:
  cmpl   $0x40,-0x98(%rbp)
  jle    L224
  cmpl   $0x7a,-0x98(%rbp)
  jg     L224
  mov    -0x98(%rbp),%eax
  mov    %eax,%edi
  call   <toupper@plt>
  mov    %eax,-0x98(%rbp)
  mov    -0x98(%rbp),%eax
  sub    $0x41,%eax
  movslq %eax,%rdx
  mov    -0x80(%rbp,%rdx,4),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0x80(%rbp,%rax,4)
  addl   $0x1,-0x8c(%rbp)
  movl   $0x1,-0x90(%rbp)
  jmp    L193
L224:
  cmpl   $0x2f,-0x98(%rbp)
  jle    L6
  cmpl   $0x39,-0x98(%rbp)
  jg     L6
  addl   $0x1,-0x88(%rbp)
  movl   $0x1,-0x90(%rbp)
  jmp    L193
L6:
  cmpl   $0x20,-0x98(%rbp)
  je     L196
  cmpl   $0x9,-0x98(%rbp)
  je     L196
  cmpl   $0xa,-0x98(%rbp)
  jne    L193
L196:
  movl   $0x0,-0x90(%rbp)
L193:
  cmpl   $0x0,-0x90(%rbp)
  jne    L244
  addl   $0x1,-0x84(%rbp)
  movl   $0x1,-0x90(%rbp)
L244:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x98(%rbp)
  cmpl   $0x7e,-0x98(%rbp)
  jne    L70
  mov    -0x88(%rbp),%ecx
  mov    -0x8c(%rbp),%edx
  mov    -0x84(%rbp),%eax
  mov    %eax,%esi
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x9c(%rbp)
  jmp    L83
L1:
  mov    -0x9c(%rbp),%eax
  cltq
  mov    -0x80(%rbp,%rax,4),%eax
  cmp    %eax,-0x94(%rbp)
  jge    L142
  mov    -0x9c(%rbp),%eax
  cltq
  mov    -0x80(%rbp,%rax,4),%eax
  mov    %eax,-0x94(%rbp)
L142:
  addl   $0x1,-0x9c(%rbp)
L83:
  cmpl   $0x19,-0x9c(%rbp)
  jle    L1
  mov    -0x94(%rbp),%eax
  mov    %eax,-0x9c(%rbp)
  jmp    L199
L32:
  mov    -0x9c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L105
L76:
  nop
  mov    -0x9c(%rbp),%eax
  cmp    -0x94(%rbp),%eax
  jle    L32
L105:
  movl   $0x0,-0x98(%rbp)
  jmp    L211
L237:
  mov    -0x98(%rbp),%eax
  cltq
  mov    -0x80(%rbp,%rax,4),%eax
  cmp    %eax,-0x9c(%rbp)
  jg     L256
  lea    D68(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L58
L256:
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L58:
  addl   $0x1,-0x98(%rbp)
L211:
  cmpl   $0x19,-0x98(%rbp)
  jle    L237
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  subl   $0x1,-0x9c(%rbp)
L199:
  cmpl   $0x0,-0x9c(%rbp)
  jns    L76
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x41,-0x9c(%rbp)
  jmp    L134
L121:
  mov    -0x9c(%rbp),%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x9c(%rbp)
L134:
  cmpl   $0x5a,-0x9c(%rbp)
  jle    L121
  nop
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L165
  call   <__stack_chk_fail@plt>
L165:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D57
data_type: string
name: unpack_data
arguments: label: D14
data_type: string
name: unpack_data
arguments: label: D68
data_type: string
name: unpack_data
arguments: label: D40
data_type: string
name: unpack_data
arguments: label: D9
data_type: string
name: unpack_data
arguments: label: D20
data_type: string
name: unpack_data
arguments: label: D23
data_type: string

tool: D57: So you had %d words with %d letters and %d numbers




tool: D14: Times x %d|

tool: D68:  * 

tool: D40:  - 

tool: D9: ----------------------------------------------------------------------------------------

tool: D20:   Letter |

tool: D23:  %c 

assistant: ```c
count_4(void) {
  int x, c, nr_maxim, state, lit, nre, words;
  nr_maxim = lit = nre = words = 0;
  int litere[27] = {0};
  state = 0;
  while ((c = getchar()) != '~') {
    if (c >= 'A' && c <= 'z') {
      c = toupper(c);
      litere[c - 'A']++;
      lit++;
      state = 1;
    } else if (c >= '0' && c <= '9') {
      nre++;
      state = 1;
    } else if (c == ' ' || c == '\t' || c == '\n')
      state = 0;
    if (state == 0) {
      words++;
      state = 1;
    }
  }
  printf("So you had %d words with %d letters and %d numbers\n\n\n", words, lit, nre);
  for (x = 0; x < 26; x++) {
    if (nr_maxim < litere[x]) {
      nr_maxim = litere[x];
    }
  }
  for (x = nr_maxim; x >= 0; x--) {
    while (x <= nr_maxim) {
      printf("Times x %d|", x + 1);
      break;
    }
    for (c = 0; c < 26; c++) {
      if (litere[c] >= x)
        printf(" * ");
      else
        printf(" - ");
    }
    putchar('\n');
    puts("----------------------------------------------------------------------------------------");
  }
  printf("  Letter |");
  for (x = 65; x < 91; x++)
    printf(" %c ", x);
}
```