ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x5f5e100,-0x1c(%rbp)
  movq   $0x0,-0x10(%rbp)
  movq   $0x0,-0x18(%rbp)
  movl   $0x1,-0x28(%rbp)
  jmp    L1
L39:
  mov    -0x28(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x28(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x20(%rbp)
L64:
  mov    -0x20(%rbp),%eax
  imul   %eax,%eax
  add    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jg     L22
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <reverse@plt>
  cmp    %eax,-0x24(%rbp)
  jne    L2
  mov    -0x18(%rbp),%rax
  mov    -0x24(%rbp),%edx
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <seen@plt>
  test   %eax,%eax
  jne    L2
  mov    -0x24(%rbp),%eax
  cltq
  add    %rax,-0x10(%rbp)
  mov    -0x24(%rbp),%edx
  lea    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <append@plt>
L2:
  addl   $0x1,-0x20(%rbp)
  jmp    L64
L22:
  nop
  addl   $0x1,-0x28(%rbp)
L1:
  mov    -0x28(%rbp),%eax
  imul   %eax,%eax
  cmp    %eax,-0x1c(%rbp)
  jg     L39
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D0
type_guessed: string

tool: D0: %ld


assistant: ```c
int main(int argc, char **argv) {
  const int limit = 100000000;
  long sum = 0;
  struct node *results = 0;
  for (int i = 1; i * i < limit; i++) {
    int candidate = i * i;
    for (int j = i + 1;; j++) {
      candidate += j * j;
      if (candidate > limit) {
        break;
      }
      if (candidate == reverse(candidate) && !seen(results, candidate)) {
        sum += candidate;
        append(&results, candidate);
      }
    }
  }
  printf("%ld\n", sum);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    $0x1,%r12d
  mov    $0x0,%r14d
  mov    $0x1,%r13d
  mov    %rsp,%r15
  jmp    L18
L82:
  add    $0x1,%ebp
  mov    %ebp,%eax
  imul   %ebp,%eax
  add    %eax,%ebx
  cmp    $0x5f5e100,%ebx
  jg     L61
L16:
  mov    %ebx,%edi
  call   <reverse@plt>
  cmp    %ebx,%eax
  jne    L82
  mov    %ebx,%esi
  mov    (%rsp),%rdi
  call   <seen@plt>
  test   %eax,%eax
  jne    L82
  movslq %ebx,%rax
  add    %rax,%r14
  mov    %ebx,%esi
  mov    %r15,%rdi
  call   <append@plt>
  jmp    L82
L61:
  cmp    $0x5f5e0ff,%r13d
  jg     L27
L18:
  add    $0x1,%r12d
  mov    %r13d,%ebx
  mov    %r12d,%r13d
  imul   %r12d,%r13d
  add    %r13d,%ebx
  cmp    $0x5f5e100,%ebx
  jg     L61
  mov    %r12d,%ebp
  jmp    L16
L27:
  mov    %r14,%rsi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D5
data_type: string

tool: D5: %ld


assistant: ```c
int main(int argc, char **argv) {
  const int limit = 100000000;
  long sum = 0;
  struct node *results = 0;
  for (int i = 1; i * i < limit; i++) {
    int candidate = i * i;
    for (int j = i + 1;; j++) {
      candidate += j * j;
      if (candidate > limit) {
        break;
      }
      if (candidate == reverse(candidate) && !seen(results, candidate)) {
        sum += candidate;
        append(&results, candidate);
      }
    }
  }
  printf("%ld\n", sum);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsp,%r14
  nopl   0x0(%rax,%rax,1)
L70:
  add    $0x1,%ebp
  mov    %r12d,%ebx
  mov    %ebp,%r12d
  imul   %ebp,%r12d
  add    %r12d,%ebx
  cmp    $0x5f5e100,%ebx
  jg     L68
  mov    %ebp,%r15d
  jmp    L52
  nopl   (%rax)
L27:
  add    $0x1,%r15d
  mov    %r15d,%eax
  imul   %r15d,%eax
  add    %eax,%ebx
  cmp    $0x5f5e100,%ebx
  jg     L68
L52:
  mov    %ebx,%edi
  call   <reverse@plt>
  cmp    %ebx,%eax
  jne    L27
  mov    (%rsp),%rdi
  mov    %ebx,%esi
  call   <seen@plt>
  test   %eax,%eax
  jne    L27
  movslq %ebx,%rax
  mov    %ebx,%esi
  mov    %r14,%rdi
  add    %rax,%r13
  call   <append@plt>
  jmp    L27
  nopl   0x0(%rax)
L68:
  cmp    $0x5f5e0ff,%r12d
  jle    L70
  mov    %r13,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D2
datatype: string

tool: D2: %ld


assistant: ```c
int main(int argc, char **argv) {
  const int limit = 100000000;
  long sum = 0;
  struct node *results = 0;
  for (int i = 1; i * i < limit; i++) {
    int candidate = i * i;
    for (int j = i + 1;; j++) {
      candidate += j * j;
      if (candidate > limit) {
        break;
      }
      if (candidate == reverse(candidate) && !seen(results, candidate)) {
        sum += candidate;
        append(&results, candidate);
      }
    }
  }
  printf("%ld\n", sum);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsp,%r14
  nopl   0x0(%rax,%rax,1)
L49:
  add    $0x1,%ebp
  mov    %r12d,%ebx
  mov    %ebp,%r12d
  imul   %ebp,%r12d
  add    %r12d,%ebx
  cmp    $0x5f5e100,%ebx
  jg     L18
  mov    %ebp,%r15d
  jmp    L1
  nopl   (%rax)
L47:
  add    $0x1,%r15d
  mov    %r15d,%eax
  imul   %r15d,%eax
  add    %eax,%ebx
  cmp    $0x5f5e100,%ebx
  jg     L18
L1:
  mov    %ebx,%edi
  call   <reverse@plt>
  cmp    %ebx,%eax
  jne    L47
  mov    (%rsp),%rdi
  mov    %ebx,%esi
  call   <seen@plt>
  test   %eax,%eax
  jne    L47
  movslq %ebx,%rax
  mov    %ebx,%esi
  mov    %r14,%rdi
  add    %rax,%r13
  call   <append@plt>
  jmp    L47
  nopl   0x0(%rax)
L18:
  cmp    $0x5f5e0ff,%r12d
  jle    L49
  mov    %r13,%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L59
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L59:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D1
datatype: string

tool: D1: %ld


assistant: ```c
int main(int argc, char **argv) {
  const int limit = 100000000;
  long sum = 0;
  struct node *results = 0;
  for (int i = 1; i * i < limit; i++) {
    int candidate = i * i;
    for (int j = i + 1;; j++) {
      candidate += j * j;
      if (candidate > limit) {
        break;
      }
      if (candidate == reverse(candidate) && !seen(results, candidate)) {
        sum += candidate;
        append(&results, candidate);
      }
    }
  }
  printf("%ld\n", sum);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L31:
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L26
L98:
  movl   $0x2,-0xc(%rbp)
  jmp    L63
L80:
  mov    -0x10(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  je     L36
  addl   $0x1,-0xc(%rbp)
L63:
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jl     L80
  jmp    L51
L36:
  nop
L51:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jne    L109
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L28
L109:
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L28:
  addl   $0x1,-0x10(%rbp)
L26:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jle    L98
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x19(%rbp),%rax
  mov    %rax,%rsi
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movzbl -0x19(%rbp),%eax
  cmp    $0x73,%al
  je     L31
  movzbl -0x19(%rbp),%eax
  cmp    $0x53,%al
  je     L31
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D0
datatype: string
name: extract_symbolic_data
arguments: tag: D29
datatype: string
name: extract_symbolic_data
arguments: tag: D12
datatype: string
name: extract_symbolic_data
arguments: tag: D43
datatype: string
name: extract_symbolic_data
arguments: tag: D18
datatype: string
name: extract_symbolic_data
arguments: tag: D54
datatype: string

tool: D0: 
Informe um numero: 

tool: D29: %i

tool: D12: 
%i - Primos!

tool: D43: 
%i - Nao primos!

tool: D18: 
Deseja calcular outra sequencia de numeros? 

tool: D54: %s

assistant: ```c
int main() {
  int num1, num2, i, j;
  char op;
  do {
    printf("\nInforme um numero: ");
    scanf("%i", &num1);
    printf("\nInforme um numero: ");
    scanf("%i", &num2);
    for (i = num1; i <= num2; i++) {
      for (j = 2; j < i; j++)
        if (i % j == 0)
          break;
      if (i == j)
        printf("\n%i - Primos!", i);
      else
        printf("\n%i - Nao primos!", i);
    }
    printf("\nDeseja calcular outra sequencia de numeros? ");
    scanf("%s", &op);
  } while (op == 's' || op == 'S');
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D50(%rip),%r14
  lea    D7(%rip),%r13
  lea    D19(%rip),%rbp
  jmp    L33
L73:
  je     L39
L1:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L31:
  add    $0x1,%ebx
  cmp    %ebx,0x14(%rsp)
  jl     L78
L0:
  cmp    $0x2,%ebx
  jle    L73
  test   $0x1,%bl
  je     L1
  mov    $0x2,%ecx
L32:
  add    $0x1,%ecx
  cmp    %ebx,%ecx
  je     L39
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L32
  jmp    L1
L39:
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L31
L78:
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xf(%rsp),%rsi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movzbl 0xf(%rsp),%eax
  and    $0xffffffdf,%eax
  cmp    $0x53,%al
  jne    L118
L33:
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%ebx
  lea    D15(%rip),%r12
  cmp    0x14(%rsp),%ebx
  jle    L0
  jmp    L78
L118:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L50
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L50:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D50
type_guessed: string
name: unpack_data
arguments: label: D7
type_guessed: string
name: unpack_data
arguments: label: D19
type_guessed: string
name: unpack_data
arguments: label: D4
type_guessed: string
name: unpack_data
arguments: label: D1
type_guessed: string
name: unpack_data
arguments: label: D15
type_guessed: string

tool: D50: 
Informe um numero: 

tool: D7: %i

tool: D19: 
%i - Nao primos!

tool: D4: 
Deseja calcular outra sequencia de numeros? 

tool: D1: %s

tool: D15: 
%i - Primos!

assistant: ```c
int main() {
  int num1, num2, i, j;
  char op;
  do {
    printf("\nInforme um numero: ");
    scanf("%i", &num1);
    printf("\nInforme um numero: ");
    scanf("%i", &num2);
    for (i = num1; i <= num2; i++) {
      for (j = 2; j < i; j++)
        if (i % j == 0)
          break;
      if (i == j)
        printf("\n%i - Primos!", i);
      else
        printf("\n%i - Nao primos!", i);
    }
    printf("\nDeseja calcular outra sequencia de numeros? ");
    scanf("%s", &op);
  } while (op == 's' || op == 'S');
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  lea    D56(%rip),%r13
  push   %r12
  lea    D7(%rip),%r12
  push   %rbp
  lea    D4(%rip),%rbp
  push   %rbx
  lea    D6(%rip),%rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x24(%rsp),%rax
  lea    0x20(%rsp),%r15
  mov    %rax,0x8(%rsp)
  lea    0x1f(%rsp),%rax
  mov    %rax,(%rsp)
  cs nopw 0x0(%rax,%rax,1)
L87:
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r15,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x20(%rsp),%r14d
  cmp    0x24(%rsp),%r14d
  jg     L72
  nopl   0x0(%rax)
L103:
  cmp    $0x2,%r14d
  jle    L15
  test   $0x1,%r14b
  je     L91
  mov    $0x2,%ecx
  jmp    L65
  nopl   0x0(%rax,%rax,1)
L4:
  mov    %r14d,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L91
L65:
  add    $0x1,%ecx
  cmp    %r14d,%ecx
  jne    L4
L26:
  mov    %r14d,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L41:
  add    $0x1,%r14d
  cmp    %r14d,0x24(%rsp)
  jge    L103
L72:
  lea    D37(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rsp),%rsi
  lea    D58(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  movzbl 0x1f(%rsp),%eax
  and    $0xffffffdf,%eax
  cmp    $0x53,%al
  je     L87
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L58
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L15:
  je     L26
L91:
  mov    %r14d,%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L41
L58:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D56
type_infered: string
name: extract_symbolic_data
arguments: tag: D7
type_infered: string
name: extract_symbolic_data
arguments: tag: D4
type_infered: string
name: extract_symbolic_data
arguments: tag: D6
type_infered: string
name: extract_symbolic_data
arguments: tag: D37
type_infered: string
name: extract_symbolic_data
arguments: tag: D58
type_infered: string

tool: D56: 
Informe um numero: 

tool: D7: %i

tool: D4: 
%i - Primos!

tool: D6: 
%i - Nao primos!

tool: D37: 
Deseja calcular outra sequencia de numeros? 

tool: D58: %s

assistant: ```c
int main() {
  int num1, num2, i, j;
  char op;
  do {
    printf("\nInforme um numero: ");
    scanf("%i", &num1);
    printf("\nInforme um numero: ");
    scanf("%i", &num2);
    for (i = num1; i <= num2; i++) {
      for (j = 2; j < i; j++)
        if (i % j == 0)
          break;
      if (i == j)
        printf("\n%i - Primos!", i);
      else
        printf("\n%i - Nao primos!", i);
    }
    printf("\nDeseja calcular outra sequencia de numeros? ");
    scanf("%s", &op);
  } while (op == 's' || op == 'S');
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  lea    D55(%rip),%r13
  push   %r12
  lea    D58(%rip),%r12
  push   %rbp
  lea    D32(%rip),%rbp
  push   %rbx
  lea    D47(%rip),%rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x24(%rsp),%rax
  lea    0x20(%rsp),%r15
  mov    %rax,0x8(%rsp)
  lea    0x1f(%rsp),%rax
  mov    %rax,(%rsp)
  cs nopw 0x0(%rax,%rax,1)
L78:
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r15,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x20(%rsp),%r14d
  cmp    0x24(%rsp),%r14d
  jg     L94
  nopl   0x0(%rax)
L22:
  cmp    $0x2,%r14d
  jle    L101
  test   $0x1,%r14b
  je     L42
  mov    $0x2,%ecx
  jmp    L48
  nopl   0x0(%rax,%rax,1)
L40:
  mov    %r14d,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L42
L48:
  add    $0x1,%ecx
  cmp    %r14d,%ecx
  jne    L40
L4:
  mov    %r14d,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L15:
  add    $0x1,%r14d
  cmp    %r14d,0x24(%rsp)
  jge    L22
L94:
  lea    D8(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rsp),%rsi
  lea    D16(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  movzbl 0x1f(%rsp),%eax
  and    $0xffffffdf,%eax
  cmp    $0x53,%al
  je     L78
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L101:
  je     L4
L42:
  mov    %r14d,%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L15
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D55
data_type: string
name: unpack_data
arguments: data_label: D58
data_type: string
name: unpack_data
arguments: data_label: D32
data_type: string
name: unpack_data
arguments: data_label: D47
data_type: string
name: unpack_data
arguments: data_label: D8
data_type: string
name: unpack_data
arguments: data_label: D16
data_type: string

tool: D55: 
Informe um numero: 

tool: D58: %i

tool: D32: 
%i - Primos!

tool: D47: 
%i - Nao primos!

tool: D8: 
Deseja calcular outra sequencia de numeros? 

tool: D16: %s

assistant: ```c
int main() {
  int num1, num2, i, j;
  char op;
  do {
    printf("\nInforme um numero: ");
    scanf("%i", &num1);
    printf("\nInforme um numero: ");
    scanf("%i", &num2);
    for (i = num1; i <= num2; i++) {
      for (j = 2; j < i; j++)
        if (i % j == 0)
          break;
      if (i == j)
        printf("\n%i - Primos!", i);
      else
        printf("\n%i - Nao primos!", i);
    }
    printf("\nDeseja calcular outra sequencia de numeros? ");
    scanf("%s", &op);
  } while (op == 's' || op == 'S');
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<get_destroy_root>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x228,%rsp
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x218(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <stat@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  js     L70
  mov    0x8(%rsp),%r14
  mov    %rbp,%rdi
  call   <strlen@plt>
  movslq %eax,%r12
  add    %rbp,%r12
  lea    0x10(%rsp),%r13
  jmp    L139
L83:
  cmp    %rbx,%rbp
  jae    L143
L111:
  cmpb   $0x2f,(%rbx)
  je     L135
  sub    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L111
L143:
  sub    %rbp,%r12
  mov    %r12d,%edx
  mov    $0x0,%eax
  test   %r12d,%r12d
  jne    L75
L70:
  mov    0x218(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L156
  add    $0x228,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L135:
  cmp    %rbx,%rbp
  jae    L143
  mov    %rbx,%r15
  sub    %rbp,%r15
  add    $0x1,%r15d
  mov    %r15d,%edx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <strncpy@plt>
  movslq %r15d,%r15
  movb   $0x0,0x10(%rsp,%r15,1)
  lea    0x8(%rsp),%rsi
  mov    %r13,%rdi
  call   <stat@plt>
  test   %rax,%rax
  js     L109
  cmp    %r14,0x8(%rsp)
  jne    L143
  mov    %rbx,%r12
L139:
  cmp    %rbp,%r12
  jbe    L143
  mov    %r12,%rbx
L145:
  movzbl (%rbx),%eax
  sub    $0x2e,%eax
  cmp    $0x1,%al
  ja     L83
  sub    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L145
  jmp    L143
L75:
  lea    0x10(%rsp),%rbx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strncpy@plt>
  movslq %r12d,%r12
  movb   $0x0,0x10(%rsp,%r12,1)
  mov    %rbx,%rdi
  call   <strdup@plt>
  jmp    L70
L109:
  mov    $0x0,%eax
  jmp    L70
L156:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *get_destroy_root(const char *dir) {
  struct stat st;
  dev_t id;
  int len;
  const char *p, *prev;
  char tmp[512];
  if (stat(dir, &st) < 0)
    return NULL;
  id = st.st_dev;
  p = dir + strlen(dir);
  prev = p;
  while (p > dir) {
    while (p > dir && (*p == '/' || *p == '.'))
      p--;
    while (p > dir && *p != '/')
      p--;
    if (p <= dir)
      break;
    len = p - dir + 1;
    strncpy(tmp, dir, len);
    tmp[len] = 0;
    if (stat(tmp, &st) < 0)
      return NULL;
    if (id != st.st_dev)
      break;
    prev = p;
  }
  len = prev - dir;
  if (len) {
    strncpy(tmp, dir, len);
    tmp[len] = 0;
    return strdup(tmp);
  }
  return NULL;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<get_destroy_root>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x238,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x228(%rsp)
  xor    %eax,%eax
  lea    0x18(%rsp),%r13
  mov    %r13,%rsi
  call   <stat@plt>
  test   %rax,%rax
  js     L1
  mov    0x18(%rsp),%rax
  mov    %rbp,%rdi
  lea    0x20(%rsp),%r12
  mov    %rax,0x8(%rsp)
  call   <strlen@plt>
  movslq %eax,%r15
  add    %rbp,%r15
  nopw   0x0(%rax,%rax,1)
L40:
  cmp    %rbp,%r15
  jbe    L113
  mov    %r15,%rbx
  nopl   0x0(%rax,%rax,1)
L44:
  movzbl (%rbx),%eax
  sub    $0x2e,%eax
  cmp    $0x1,%al
  ja     L65
  sub    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L44
L113:
  sub    %rbp,%r15
  mov    %r15d,%edx
  test   %r15d,%r15d
  jne    L2
L1:
  xor    %eax,%eax
L135:
  mov    0x228(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L42
  add    $0x238,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L65:
  cmp    %rbx,%rbp
  jae    L113
  nopl   (%rax)
L46:
  sub    $0x1,%rbx
  cmp    %rbx,%rbp
  je     L113
  cmpb   $0x2f,(%rbx)
  jne    L46
  cmp    %rbx,%rbp
  jae    L113
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  sub    %rbp,%rdx
  lea    0x1(%rdx),%r14d
  mov    %r14d,%edx
  call   <strncpy@plt>
  movslq %r14d,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  movb   $0x0,0x20(%rsp,%rdx,1)
  call   <stat@plt>
  test   %rax,%rax
  js     L1
  mov    0x8(%rsp),%rax
  cmp    %rax,0x18(%rsp)
  jne    L113
  mov    %rbx,%r15
  jmp    L40
  nopl   0x0(%rax,%rax,1)
L2:
  lea    0x20(%rsp),%r12
  mov    %rbp,%rsi
  movslq %r15d,%r15
  mov    %r12,%rdi
  call   <strncpy@plt>
  mov    %r12,%rdi
  movb   $0x0,0x20(%rsp,%r15,1)
  call   <strdup@plt>
  jmp    L135
L42:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *get_destroy_root(const char *dir) {
  struct stat st;
  dev_t id;
  int len;
  const char *p, *prev;
  char tmp[512];
  if (stat(dir, &st) < 0)
    return NULL;
  id = st.st_dev;
  p = dir + strlen(dir);
  prev = p;
  while (p > dir) {
    while (p > dir && (*p == '/' || *p == '.'))
      p--;
    while (p > dir && *p != '/')
      p--;
    if (p <= dir)
      break;
    len = p - dir + 1;
    strncpy(tmp, dir, len);
    tmp[len] = 0;
    if (stat(tmp, &st) < 0)
      return NULL;
    if (id != st.st_dev)
      break;
    prev = p;
  }
  len = prev - dir;
  if (len) {
    strncpy(tmp, dir, len);
    tmp[len] = 0;
    return strdup(tmp);
  }
  return NULL;
}
```