user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1c0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L30:
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x1b4(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1b4(%rbp),%eax
  test   %eax,%eax
  js     L30
  mov    -0x1b4(%rbp),%eax
  cmp    $0x63,%eax
  jg     L30
  movl   $0x0,-0x1ac(%rbp)
  jmp    L126
L46:
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x1a0(%rbp),%rdx
  mov    -0x1ac(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x1ac(%rbp)
L126:
  mov    -0x1b4(%rbp),%eax
  cmp    %eax,-0x1ac(%rbp)
  jl     L46
  lea    D63(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x1b0(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x1ac(%rbp)
  jmp    L106
L81:
  mov    -0x1ac(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x1b0(%rbp),%eax
  cmp    %eax,%edx
  jne    L44
  movl   $0x1,-0x1a8(%rbp)
  mov    -0x1ac(%rbp),%eax
  mov    %eax,-0x1a4(%rbp)
L44:
  addl   $0x1,-0x1ac(%rbp)
L106:
  mov    -0x1b4(%rbp),%eax
  cmp    %eax,-0x1ac(%rbp)
  jge    L92
  cmpl   $0x0,-0x1a8(%rbp)
  je     L81
L92:
  cmpl   $0x0,-0x1a8(%rbp)
  jne    L119
  lea    D68(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L116
L34:
  mov    -0x1a4(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x1a4(%rbp),%eax
  cltq
  mov    %edx,-0x1a0(%rbp,%rax,4)
  addl   $0x1,-0x1a4(%rbp)
L119:
  mov    -0x1b4(%rbp),%eax
  cmp    %eax,-0x1a4(%rbp)
  jl     L34
  movl   $0x0,-0x1ac(%rbp)
  jmp    L0
L51:
  mov    -0x1ac(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x1ac(%rbp)
L0:
  mov    -0x1b4(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x1ac(%rbp)
  jl     L51
L116:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L35
  call   <__stack_chk_fail@plt>
L35:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D16
type_guessed: string
name: unpack_data
arguments: data_label: D17
type_guessed: string
name: unpack_data
arguments: data_label: D45
type_guessed: string
name: unpack_data
arguments: data_label: D63
type_guessed: string
name: unpack_data
arguments: data_label: D68
type_guessed: string

tool: D16: inserire il numero di elementi 

tool: D17: %d

tool: D45: inserire un valore 

tool: D63: inserire il numero da cercare 

tool: D68: num non trovato

assistant: ```c
int main() {
  int k;
  int num;
  int i;
  int t;
  int pos;
  int v[99];
  do {
    printf("inserire il numero di elementi \n");
    scanf("%d", &k);
  } while (k < 0 || k > 99);
  for (i = 0; i < k; i++) {
    printf("inserire un valore \n");
    scanf("%d", &v[i]);
  }
  printf("inserire il numero da cercare \n");
  scanf("%d", &num);
  i = 0;
  while (i < k && t == 0) {
    if (v[i] == num) {
      t = 1;
      pos = i;
    }
    i++;
  }
  if (t == 0) {
    printf("num non trovato");
  } else {
    while (pos < k) {
      v[pos] = v[pos + 1];
      pos++;
    }
    for (i = 0; i < (k - 1); i++) {
      printf("%d", v[i]);
    }
  }
}
```