user: 请从给定的汇编指令推导出C源代码：

```asm
<count_4>:
  endbr64
  push   %r15
  mov    $0xd,%ecx
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  mov    %rdi,0x8(%rsp)
  rep stos %rax,%es:(%rdi)
  movl   $0x0,(%rdi)
  jmp    L51
  nopl   0x0(%rax)
L79:
  call   <toupper@plt>
  add    $0x1,%r12d
  sub    $0x41,%eax
  cltq
  addl   $0x1,0x10(%rsp,%rax,4)
L80:
  mov    $0x1,%ebx
L51:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    $0x7e,%eax
  je     L38
  lea    -0x41(%rdi),%eax
  cmp    $0x39,%eax
  jbe    L79
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%eax
  ja     L11
  add    $0x1,%r13d
  jmp    L80
L11:
  cmp    $0x20,%edi
  sete   %al
  xor    $0x1,%ebx
  or     %bl,%al
  jne    L44
  sub    $0x9,%edi
  cmp    $0x1,%edi
  ja     L80
L44:
  add    $0x1,%ebp
  jmp    L80
L38:
  mov    %r12d,%edx
  mov    %r13d,%ecx
  mov    %ebp,%esi
  xor    %eax,%eax
  lea    D10(%rip),%rdi
  lea    D4(%rip),%r15
  call   <printf@plt>
  pxor   %xmm3,%xmm3
  mov    0x70(%rsp),%ebx
  movdqa 0x10(%rsp),%xmm1
  mov    0x74(%rsp),%edx
  lea    0x78(%rsp),%rbp
  lea    D5(%rip),%r13
  movdqa %xmm1,%xmm0
  lea    D23(%rip),%r12
  pcmpgtd 0x20(%rsp),%xmm0
  cmp    %edx,%ebx
  cmovl  %edx,%ebx
  pand   %xmm0,%xmm1
  pandn  0x20(%rsp),%xmm0
  por    %xmm1,%xmm0
  movdqa 0x30(%rsp),%xmm1
  movdqa %xmm1,%xmm2
  pcmpgtd 0x40(%rsp),%xmm2
  pand   %xmm2,%xmm1
  pandn  0x40(%rsp),%xmm2
  por    %xmm1,%xmm2
  movdqa %xmm0,%xmm1
  pcmpgtd %xmm2,%xmm1
  pand   %xmm1,%xmm0
  pandn  %xmm2,%xmm1
  movdqa 0x60(%rsp),%xmm2
  por    %xmm0,%xmm1
  movdqa %xmm2,%xmm0
  pcmpgtd 0x50(%rsp),%xmm0
  pand   %xmm0,%xmm2
  pandn  0x50(%rsp),%xmm0
  por    %xmm2,%xmm0
  movdqa %xmm0,%xmm2
  pcmpgtd %xmm3,%xmm2
  pand   %xmm2,%xmm0
  movdqa %xmm1,%xmm2
  pcmpgtd %xmm0,%xmm2
  pand   %xmm2,%xmm1
  pandn  %xmm0,%xmm2
  movdqa %xmm2,%xmm0
  por    %xmm1,%xmm0
  movdqa %xmm0,%xmm2
  psrldq $0x8,%xmm2
  movdqa %xmm2,%xmm1
  pcmpgtd %xmm0,%xmm1
  pand   %xmm1,%xmm2
  pandn  %xmm0,%xmm1
  por    %xmm2,%xmm1
  movdqa %xmm1,%xmm2
  psrldq $0x4,%xmm2
  movdqa %xmm2,%xmm0
  pcmpgtd %xmm1,%xmm0
  pand   %xmm0,%xmm2
  pandn  %xmm1,%xmm0
  por    %xmm2,%xmm0
  movd   %xmm0,%eax
  cmp    %eax,%ebx
  cmovl  %eax,%ebx
  nopl   0x0(%rax)
L74:
  mov    %ebx,%esi
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%r14
  jmp    L17
  nopl   0x0(%rax,%rax,1)
L3:
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%r14
  call   <printf@plt>
  cmp    %rbp,%r14
  je     L32
L17:
  cmp    %ebx,(%r14)
  jge    L3
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%r14
  call   <printf@plt>
  cmp    %rbp,%r14
  jne    L17
L32:
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D59(%rip),%rdi
  call   <puts@plt>
  sub    $0x1,%ebx
  jae    L74
  lea    D61(%rip),%rdi
  xor    %eax,%eax
  mov    $0x41,%ebx
  call   <printf@plt>
  lea    D30(%rip),%rbp
  nopl   0x0(%rax,%rax,1)
L142:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x5b,%ebx
  jne    L142
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x98,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D10
type_infered: string
name: retrieve_data
arguments: data_tag: D4
type_infered: string
name: retrieve_data
arguments: data_tag: D5
type_infered: string
name: retrieve_data
arguments: data_tag: D23
type_infered: string
name: retrieve_data
arguments: data_tag: D59
type_infered: string
name: retrieve_data
arguments: data_tag: D61
type_infered: string
name: retrieve_data
arguments: data_tag: D30
type_infered: string

tool: D10: So you had %d words with %d letters and %d numbers




tool: D4: Times x %d|

tool: D5:    

tool: D23:  * 

tool: D59: ----------------------------------------------------------------------------------------

tool: D61:   Letter |

tool: D30:  %c 

assistant: ```c
void 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);
      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);
}
```