user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r13
  lea    D19(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%r12
  call   <puts@plt>
  mov    %r12,%rdi
  call   <gets@plt>
  movzbl 0x30(%rsp),%eax
  test   %al,%al
  je     L51
  mov    %rsp,%rbp
  mov    $0x1,%r8d
  mov    %r12,%rsi
  xor    %ebx,%ebx
  mov    %rbp,%rdi
  xor    %edx,%edx
  sub    %r12,%r8
  jmp    L47
  nopl   0x0(%rax)
L110:
  add    $0x1,%edx
L128:
  lea    (%r8,%rsi,1),%r9
  movzbl 0x1(%rsi),%eax
  add    $0x1,%rsi
  test   %al,%al
  je     L113
L47:
  sub    $0x30,%eax
  cmp    $0x9,%al
  jbe    L110
  test   %edx,%edx
  je     L128
  movsbl -0x1(%rsi),%r9d
  sub    $0x30,%r9d
  cmp    $0x1,%edx
  je     L174
  lea    0x1(%rdx),%r10d
  mov    $0x2,%ecx
  nopl   (%rax)
L69:
  mov    $0x1,%eax
  mov    $0x1,%edx
  nopw   0x0(%rax,%rax,1)
L136:
  lea    (%rax,%rax,4),%eax
  add    $0x1,%edx
  add    %eax,%eax
  cmp    %ecx,%edx
  jne    L136
  mov    %rcx,%rdx
  add    $0x1,%rcx
  neg    %rdx
  movsbl (%rsi,%rdx,1),%edx
  sub    $0x30,%edx
  imul   %edx,%eax
  add    %eax,%r9d
  cmp    %rcx,%r10
  jne    L69
L174:
  mov    %r9d,(%rdi)
  lea    (%r8,%rsi,1),%r9
  movzbl 0x1(%rsi),%eax
  add    $0x1,%rsi
  add    $0x4,%rdi
  add    $0x1,%ebx
  xor    %edx,%edx
  test   %al,%al
  jne    L47
L113:
  test   %edx,%edx
  je     L53
  movsbl 0x2f(%rsp,%r9,1),%esi
  sub    $0x30,%esi
  cmp    $0x1,%edx
  je     L132
  lea    0x1(%rdx),%r8d
  mov    $0x2,%ecx
  add    %r9,%r12
  xchg   %ax,%ax
L143:
  mov    $0x1,%eax
  mov    $0x1,%edx
  nopw   0x0(%rax,%rax,1)
L70:
  lea    (%rax,%rax,4),%eax
  add    $0x1,%edx
  add    %eax,%eax
  cmp    %ecx,%edx
  jne    L70
  mov    %rcx,%rdx
  add    $0x1,%rcx
  neg    %rdx
  movsbl (%r12,%rdx,1),%edx
  sub    $0x30,%edx
  imul   %edx,%eax
  add    %eax,%esi
  cmp    %rcx,%r8
  jne    L143
L132:
  mov    %esi,(%rdi)
  add    $0x1,%ebx
  lea    D43(%rip),%rdi
  xor    %eax,%eax
  mov    %ebx,%esi
  call   <printf@plt>
L20:
  xor    %r12d,%r12d
  lea    D47(%rip),%r13
  nopw   0x0(%rax,%rax,1)
L8:
  mov    0x0(%rbp,%r12,4),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%r12
  call   <printf@plt>
  cmp    %r12d,%ebx
  jg     L8
L61:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L140
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L53:
  xor    %eax,%eax
  mov    %ebx,%esi
  lea    D43(%rip),%rdi
  call   <printf@plt>
  test   %ebx,%ebx
  jne    L20
  jmp    L61
L51:
  xor    %esi,%esi
  lea    D43(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L61
L140:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D19
type_infered: string
name: unpack_data
arguments: label: D43
type_infered: string
name: unpack_data
arguments: label: D47
type_infered: string

tool: D19: input a string:

tool: D43: There are %d numbers in this line.They are :


tool: D47: %d 

assistant: ```c
int main() {
  char str[50], *pstr;
  int i, j, k, m, e10, digit, ndigit, a[10], *pa;
  printf("input a string:\n");
  gets(str);
  pstr = &str[0];
  pa = &a[0];
  ndigit = 0;
  i = 0;
  j = 0;
  while (*(pstr + i) != '\0') {
    if (*(pstr + i) >= '0' && *(pstr + i) <= '9')
      j++;
    else {
      if (j > 0) {
        digit = *(pstr + i - 1) - 48;
        k = 1;
        while (k < j) {
          e10 = 1;
          for (m = 1; m <= k; m++)
            e10 = e10 * 10;
          digit = digit + (*(pstr + i - 1 - k) - 48) * e10;
          k++;
        }
        *pa = digit;
        pa++;
        ndigit++;
        j = 0;
      }
    }
    i++;
  }
  if (j > 0) {
    digit = *(pstr + i - 1) - 48;
    k = 1;
    while (k < j) {
      e10 = 1;
      for (m = 1; m <= k; m++)
        e10 = e10 * 10;
      digit = digit + (*(pstr + i - 1 - k) - 48) * e10;
      k++;
    }
    *pa = digit;
    ndigit++;
    j = 0;
  }
  printf("There are %d numbers in this line.They are :\n", ndigit);
  j = 0;
  pa = &a[0];
  for (j = 0; j < ndigit; j++)
    printf("%d ", *(pa + j));
  printf("\n");
  return 0;
}
```