user: 请从给定的汇编指令推导出C源代码:

```asm
<priklad_dva>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x38(%rbp)
  jmp    L38
L52:
  mov    -0x38(%rbp),%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x20(%rbp),%rdx
  mov    -0x38(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x38(%rbp)
L38:
  cmpl   $0x4,-0x38(%rbp)
  jle    L52
  lea    -0x20(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  movl   $0x0,-0x34(%rbp)
  jmp    L47
L51:
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%edx
  mov    -0x34(%rbp),%eax
  mov    %eax,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x30(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  jge    L11
  lea    -0x20(%rbp),%rdx
  mov    -0x34(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,-0x30(%rbp)
L11:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  jle    L19
  lea    -0x20(%rbp),%rdx
  mov    -0x34(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,-0x28(%rbp)
L19:
  addl   $0x1,-0x34(%rbp)
L47:
  cmpl   $0x4,-0x34(%rbp)
  jle    L51
  lea    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  sub    %rdx,%rax
  sar    $0x2,%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x20(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  sub    %rdx,%rax
  sar    $0x2,%rax
  mov    %rax,%rdx
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L56
  call   <__stack_chk_fail@plt>
L56:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D50
type_guessed: string
name: disassemble_data
arguments: data_tag: D16
type_guessed: string
name: disassemble_data
arguments: data_tag: D44
type_guessed: string
name: disassemble_data
arguments: data_tag: D19
type_guessed: string
name: disassemble_data
arguments: data_tag: D34
type_guessed: string
name: disassemble_data
arguments: data_tag: D29
type_guessed: string
name: disassemble_data
arguments: data_tag: D32
type_guessed: string

tool: D50: ========== PRIKLAD 2 ==========

tool: D16: pole[%d] = 

tool: D44: %d

tool: D19: pole[%d] = %d     

tool: D34: 

MIN = %d   je na pole[%d]


tool: D29: MAX = %d   je na pole[%d]


tool: D32: ========== KONEC PRIKLADU 2 ==========

assistant: ```c
void priklad_dva() {
  printf("========== PRIKLAD 2 ==========\n");
  int pole[5 - 1];
  for (int i = 0; i < 5; i++) {
    printf("pole[%d] = ", i);
    scanf("%d", &pole[i]);
  }
  int *max, *min;
  max = &pole[0];
  min = &pole[0];
  for (int i = 0; i < 5; i++) {
    printf("pole[%d] = %d     ", i, pole[i]);
    if (*max < pole[i]) {
      max = &pole[i];
    }
    if (*min > pole[i]) {
      min = &pole[i];
    }
  }
  printf("\n\nMIN = %d   je na pole[%d]\n", *min, (min - &pole[0]));
  printf("MAX = %d   je na pole[%d]\n", *max, (max - &pole[0]));
  printf("========== KONEC PRIKLADU 2 ==========\n");
}
```