ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x4(%rsp)
  jle    L1
  mov    $0x1,%ebx
  lea    D3(%rip),%rbp
L11:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,0x4(%rsp)
  jge    L11
L1:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D13
type_guessed: string
name: retrieve_data
arguments: label: D3
type_guessed: string

tool: D13: %d

tool: D3: %d 

assistant: ```c
int main() {
  int num;
  scanf("%d", &num);
  for (int i = 1; i <= num; i++) {
    printf("%d ", i);
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D8(%rip),%rdi
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L24
  mov    $0x1,%ebx
  lea    D14(%rip),%rbp
  nop
L7:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jge    L7
L24:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D8
data_type: string
name: parse_data
arguments: data_tag: D14
data_type: string

tool: D8: %d

tool: D14: %d 

assistant: ```c
int main() {
  int num;
  scanf("%d", &num);
  for (int i = 1; i <= num; i++) {
    printf("%d ", i);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D14(%rip),%rdi
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L8
  mov    $0x1,%ebx
  lea    D18(%rip),%rbp
  nop
L23:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jge    L23
L8:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D14
datatype: string
name: parse_data
arguments: label: D18
datatype: string

tool: D14: %d

tool: D18: %d 

assistant: ```c
int main() {
  int num;
  scanf("%d", &num);
  for (int i = 1; i <= num; i++) {
    printf("%d ", i);
  }
  return 0;
}
```
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    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L58
L35:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x14(%rbp)
  mov    -0x1c(%rbp),%eax
  test   %eax,%eax
  jle    L24
  movl   $0x1,-0x10(%rbp)
  jmp    L44
L69:
  mov    -0x1c(%rbp),%eax
  cltd
  idivl  -0x10(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L73
  addl   $0x1,-0x14(%rbp)
L73:
  addl   $0x1,-0x10(%rbp)
L44:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jle    L69
  cmpl   $0x2,-0x14(%rbp)
  jne    L24
  addl   $0x1,-0xc(%rbp)
L24:
  addl   $0x1,-0x18(%rbp)
L58:
  cmpl   $0xa,-0x18(%rbp)
  jle    L35
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L42
  call   <__stack_chk_fail@plt>
L42:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D17
datatype: string
name: unpack_data
arguments: data_tag: D28
datatype: string
name: unpack_data
arguments: data_tag: D7
datatype: string

tool: D17: Digite o numero inteiro positivo #%d: 

tool: D28:  %d

tool: D7: 
Numero de valores primos entrados: %d.


assistant: ```c
void main() {
  int i, num, cont, j, primo = 0;
  for (i = 1; i <= 10; i++) {
    printf("Digite o numero inteiro positivo #%d: ", i);
    scanf(" %d", &num);
    cont = 0;
    if (num >= 1) {
      for (j = 1; j <= num; j++)
        if (num % j == 0)
          cont++;
      if (cont == 2)
        primo++;
    }
  }
  printf("\nNumero de valores primos entrados: %d.\n", primo);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x0,%r14d
  mov    $0x1,%ebx
  lea    D9(%rip),%r13
  lea    0x4(%rsp),%r12
  lea    D3(%rip),%rbp
  jmp    L20
L39:
  cmp    $0x2,%edi
  sete   %al
  movzbl %al,%eax
  add    %eax,%r14d
L54:
  add    $0x1,%ebx
  cmp    $0xb,%ebx
  je     L47
L20:
  mov    %ebx,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L54
  lea    0x1(%rsi),%r8d
  mov    $0x1,%ecx
  mov    $0x0,%edi
L25:
  mov    %esi,%eax
  cltd
  idiv   %ecx
  cmp    $0x1,%edx
  adc    $0x0,%edi
  add    $0x1,%ecx
  cmp    %ecx,%r8d
  jne    L25
  jmp    L39
L47:
  mov    %r14d,%esi
  lea    D23(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L38:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D9
type: string
name: unpack_data
arguments: data_label: D3
type: string
name: unpack_data
arguments: data_label: D23
type: string

tool: D9: Digite o numero inteiro positivo #%d: 

tool: D3:  %d

tool: D23: 
Numero de valores primos entrados: %d.


assistant: ```c
void main() {
  int i, num, cont, j, primo = 0;
  for (i = 1; i <= 10; i++) {
    printf("Digite o numero inteiro positivo #%d: ", i);
    scanf(" %d", &num);
    cont = 0;
    if (num >= 1) {
      for (j = 1; j <= num; j++)
        if (num % j == 0)
          cont++;
      if (cont == 2)
        primo++;
    }
  }
  printf("\nNumero de valores primos entrados: %d.\n", primo);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x1,%r14d
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  lea    D28(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D11(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  jmp    L21
  xchg   %ax,%ax
L10:
  add    $0x1,%r14d
  cmp    $0xb,%r14d
  je     L39
L21:
  mov    %r14d,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L10
  lea    0x1(%rsi),%r8d
  mov    $0x1,%ecx
  xor    %edi,%edi
  nopl   0x0(%rax,%rax,1)
L49:
  mov    %esi,%eax
  cltd
  idiv   %ecx
  cmp    $0x1,%edx
  adc    $0x0,%edi
  add    $0x1,%ecx
  cmp    %ecx,%r8d
  jne    L49
  xor    %eax,%eax
  cmp    $0x2,%edi
  sete   %al
  add    $0x1,%r14d
  add    %eax,%r13d
  cmp    $0xb,%r14d
  jne    L21
L39:
  mov    %r13d,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L33:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D28
data_type: string
name: unpack_data
arguments: label: D11
data_type: string
name: unpack_data
arguments: label: D9
data_type: string

tool: D28: Digite o numero inteiro positivo #%d: 

tool: D11:  %d

tool: D9: 
Numero de valores primos entrados: %d.


assistant: ```c
void main() {
  int i, num, cont, j, primo = 0;
  for (i = 1; i <= 10; i++) {
    printf("Digite o numero inteiro positivo #%d: ", i);
    scanf(" %d", &num);
    cont = 0;
    if (num >= 1) {
      for (j = 1; j <= num; j++)
        if (num % j == 0)
          cont++;
      if (cont == 2)
        primo++;
    }
  }
  printf("\nNumero de valores primos entrados: %d.\n", primo);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x1,%r14d
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  lea    D0(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D8(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  jmp    L28
  xchg   %ax,%ax
L29:
  add    $0x1,%r14d
  cmp    $0xb,%r14d
  je     L10
L28:
  mov    %r14d,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L29
  lea    0x1(%rsi),%r8d
  mov    $0x1,%ecx
  xor    %edi,%edi
  nopl   0x0(%rax,%rax,1)
L51:
  mov    %esi,%eax
  cltd
  idiv   %ecx
  cmp    $0x1,%edx
  adc    $0x0,%edi
  add    $0x1,%ecx
  cmp    %r8d,%ecx
  jne    L51
  xor    %eax,%eax
  cmp    $0x2,%edi
  sete   %al
  add    $0x1,%r14d
  add    %eax,%r13d
  cmp    $0xb,%r14d
  jne    L28
L10:
  mov    %r13d,%esi
  lea    D25(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D0
type: string
name: extract_symbolic_data
arguments: tag: D8
type: string
name: extract_symbolic_data
arguments: tag: D25
type: string

tool: D0: Digite o numero inteiro positivo #%d: 

tool: D8:  %d

tool: D25: 
Numero de valores primos entrados: %d.


assistant: ```c
void main() {
  int i, num, cont, j, primo = 0;
  for (i = 1; i <= 10; i++) {
    printf("Digite o numero inteiro positivo #%d: ", i);
    scanf(" %d", &num);
    cont = 0;
    if (num >= 1) {
      for (j = 1; j <= num; j++)
        if (num % j == 0)
          cont++;
      if (cont == 2)
        primo++;
    }
  }
  printf("\nNumero de valores primos entrados: %d.\n", primo);
}
```
user: What is the c source code of the assembly code below?

```asm
<get_highest_bit_position>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L46
  mov    $0x0,%eax
  jmp    L17
L46:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  and    $0xffff0000,%eax
  test   %rax,%rax
  jne    L7
  addl   $0x10,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  shl    $0x10,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
L7:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  and    $0xff000000,%eax
  test   %rax,%rax
  jne    L28
  addl   $0x8,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  shl    $0x8,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
L28:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  and    $0xf0000000,%eax
  test   %rax,%rax
  jne    L6
  addl   $0x4,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
L6:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  and    $0xc0000000,%eax
  test   %rax,%rax
  jne    L36
  addl   $0x2,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
L36:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  and    $0x80000000,%eax
  test   %rax,%rax
  je     L19
  subl   $0x1,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  shr    %rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
L19:
  mov    -0x4(%rbp),%eax
L17:
  pop    %rbp
  ret
```

assistant: ```c
int get_highest_bit_position(unsigned long *value) {
  int position;
  position = 0;
  if (*value == 0)
    return 0;
  if ((*value & 0xFFFF0000) == 0) {
    position += 16;
    *value = *value << 16;
  }
  if ((*value & 0xFF000000) == 0) {
    position += 8;
    *value = *value << 8;
  }
  if ((*value & 0xF0000000) == 0) {
    position += 4;
    *value = *value << 4;
  }
  if ((*value & 0xC0000000) == 0) {
    position += 2;
    *value = *value << 2;
  }
  if ((*value & 0x80000000)) {
    position -= 1;
    *value = *value >> 1;
  }
  return position;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<get_highest_bit_position>:
  endbr64
  mov    (%rdi),%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  je     L1
  test   $0xffff0000,%edx
  jne    L30
  shl    $0x10,%rdx
  mov    %rdx,(%rdi)
  mov    $0x10,%eax
L30:
  mov    (%rdi),%rdx
  test   $0xff000000,%edx
  jne    L28
  add    $0x8,%eax
  shl    $0x8,%rdx
  mov    %rdx,(%rdi)
L28:
  mov    (%rdi),%rdx
  test   $0xf0000000,%edx
  jne    L55
  add    $0x4,%eax
  shl    $0x4,%rdx
  mov    %rdx,(%rdi)
L55:
  mov    (%rdi),%rdx
  test   $0xc0000000,%edx
  jne    L57
  add    $0x2,%eax
  shl    $0x2,%rdx
  mov    %rdx,(%rdi)
L57:
  mov    (%rdi),%rdx
  test   $0x80000000,%edx
  je     L1
  sub    $0x1,%eax
  shr    %rdx
  mov    %rdx,(%rdi)
L1:
  ret
```

assistant: ```c
int get_highest_bit_position(unsigned long *value) {
  int position;
  position = 0;
  if (*value == 0)
    return 0;
  if ((*value & 0xFFFF0000) == 0) {
    position += 16;
    *value = *value << 16;
  }
  if ((*value & 0xFF000000) == 0) {
    position += 8;
    *value = *value << 8;
  }
  if ((*value & 0xF0000000) == 0) {
    position += 4;
    *value = *value << 4;
  }
  if ((*value & 0xC0000000) == 0) {
    position += 2;
    *value = *value << 2;
  }
  if ((*value & 0x80000000)) {
    position -= 1;
    *value = *value >> 1;
  }
  return position;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<get_highest_bit_position>:
  endbr64
  mov    (%rdi),%rax
  xor    %r8d,%r8d
  test   %rax,%rax
  je     L15
  test   $0xffff0000,%eax
  jne    L19
  shl    $0x10,%rax
  mov    $0x10,%r8d
  mov    %rax,(%rdi)
L19:
  test   $0xff000000,%eax
  jne    L9
  shl    $0x8,%rax
  add    $0x8,%r8d
  mov    %rax,(%rdi)
L9:
  test   $0xf0000000,%eax
  jne    L26
  shl    $0x4,%rax
  add    $0x4,%r8d
  mov    %rax,(%rdi)
L26:
  test   $0xc0000000,%eax
  jne    L10
  shl    $0x2,%rax
  add    $0x2,%r8d
  mov    %rax,(%rdi)
L10:
  test   $0x80000000,%eax
  je     L15
  shr    %rax
  sub    $0x1,%r8d
  mov    %rax,(%rdi)
L15:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int get_highest_bit_position(unsigned long *value) {
  int position;
  position = 0;
  if (*value == 0)
    return 0;
  if ((*value & 0xFFFF0000) == 0) {
    position += 16;
    *value = *value << 16;
  }
  if ((*value & 0xFF000000) == 0) {
    position += 8;
    *value = *value << 8;
  }
  if ((*value & 0xF0000000) == 0) {
    position += 4;
    *value = *value << 4;
  }
  if ((*value & 0xC0000000) == 0) {
    position += 2;
    *value = *value << 2;
  }
  if ((*value & 0x80000000)) {
    position -= 1;
    *value = *value >> 1;
  }
  return position;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_highest_bit_position>:
  endbr64
  mov    (%rdi),%rax
  xor    %r8d,%r8d
  test   %rax,%rax
  je     L13
  mov    $0x8,%edx
  test   $0xffff0000,%eax
  jne    L46
  shl    $0x10,%rax
  mov    $0x18,%edx
  mov    $0x10,%r8d
  mov    %rax,(%rdi)
L46:
  test   $0xff000000,%eax
  jne    L22
  shl    $0x8,%rax
  mov    %edx,%r8d
  mov    %rax,(%rdi)
L22:
  test   $0xf0000000,%eax
  jne    L41
  shl    $0x4,%rax
  add    $0x4,%r8d
  mov    %rax,(%rdi)
L41:
  test   $0xc0000000,%eax
  jne    L36
  shl    $0x2,%rax
  add    $0x2,%r8d
  mov    %rax,(%rdi)
L36:
  test   $0x80000000,%eax
  je     L13
  shr    %rax
  sub    $0x1,%r8d
  mov    %rax,(%rdi)
L13:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int get_highest_bit_position(unsigned long *value) {
  int position;
  position = 0;
  if (*value == 0)
    return 0;
  if ((*value & 0xFFFF0000) == 0) {
    position += 16;
    *value = *value << 16;
  }
  if ((*value & 0xFF000000) == 0) {
    position += 8;
    *value = *value << 8;
  }
  if ((*value & 0xF0000000) == 0) {
    position += 4;
    *value = *value << 4;
  }
  if ((*value & 0xC0000000) == 0) {
    position += 2;
    *value = *value << 2;
  }
  if ((*value & 0x80000000)) {
    position -= 1;
    *value = *value >> 1;
  }
  return position;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x430,%rsp
  mov    %edi,-0x424(%rbp)
  mov    %rsi,-0x430(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D13(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x418(%rbp)
  jmp    L5
L13:
  mov    -0x414(%rbp),%edx
  lea    -0x410(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <arrayToUpper@plt>
  mov    -0x414(%rbp),%edx
  lea    -0x410(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <printArray@plt>
L5:
  lea    -0x410(%rbp),%rcx
  mov    -0x418(%rbp),%eax
  mov    $0x400,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x414(%rbp)
  cmpl   $0x0,-0x414(%rbp)
  jg     L13
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int fd_in = STDIN_FILENO;
  ssize_t num_read;
  char buf[1024];
  while ((num_read = read(fd_in, buf, 1024)) > 0) {
    arrayToUpper(buf, num_read);
    printArray(buf, num_read);
  }
  return EXIT_SUCCESS;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    D6(%rip),%rax
  mov    (%rax),%r12d
  mov    %rsp,%rbp
  jmp    L9
L14:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <arrayToUpper@plt>
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printArray@plt>
L9:
  mov    $0x400,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jg     L14
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  mov    0x408(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int fd_in = STDIN_FILENO;
  ssize_t num_read;
  char buf[1024];
  while ((num_read = read(fd_in, buf, 1024)) > 0) {
    arrayToUpper(buf, num_read);
    printArray(buf, num_read);
  }
  return EXIT_SUCCESS;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  mov    D14(%rip),%rax
  mov    %rsp,%rbx
  mov    (%rax),%r12d
  jmp    L29
  nop
L28:
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <arrayToUpper@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <printArray@plt>
L29:
  mov    $0x400,%edx
  mov    %rbx,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  mov    %eax,%ebp
  test   %eax,%eax
  jg     L28
  mov    D4(%rip),%rax
  mov    (%rax),%eax
  mov    0x408(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int fd_in = STDIN_FILENO;
  ssize_t num_read;
  char buf[1024];
  while ((num_read = read(fd_in, buf, 1024)) > 0) {
    arrayToUpper(buf, num_read);
    printArray(buf, num_read);
  }
  return EXIT_SUCCESS;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  mov    D11(%rip),%rax
  mov    %rsp,%rbx
  mov    (%rax),%r12d
  jmp    L7
  nop
L16:
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <arrayToUpper@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <printArray@plt>
L7:
  mov    $0x400,%edx
  mov    %rbx,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  mov    %eax,%ebp
  test   %eax,%eax
  jg     L16
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  mov    0x408(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int fd_in = STDIN_FILENO;
  ssize_t num_read;
  char buf[1024];
  while ((num_read = read(fd_in, buf, 1024)) > 0) {
    arrayToUpper(buf, num_read);
    printArray(buf, num_read);
  }
  return EXIT_SUCCESS;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ppMem>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L12
L13:
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x3,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%esi
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x2,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%ecx
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdi
  mov    -0x18(%rbp),%rax
  add    %rdi,%rax
  mov    (%rax),%eax
  mov    %esi,%r8d
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x2,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x3,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  cmpl   $0x2,-0x4(%rbp)
  jne    L5
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0xffffffff,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x2,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x3,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
L5:
  addl   $0x4,-0x4(%rbp)
L12:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L13
  nop
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D10
data_type: string
name: retrieve_data
arguments: label: D2
data_type: string

tool: D10: MEM1:

tool: D2: 0x%x, 0x%x, 0x%x, 0x%x


assistant: ```c
void ppMem(unsigned int *m, int len) {
  int i;
  printf("MEM1:\n");
  for (i = 0; i < len; i += 4) {
    printf("0x%x, 0x%x, 0x%x, 0x%x\n", m[i], m[i + 1], m[i + 2], m[i + 3]);
    m[i] = 0;
    m[i + 1] = 0;
    m[i + 2] = 0;
    m[i + 3] = 0;
    if (i == 2) {
      m[i] = 0xffffffff;
      m[i + 1] = 0;
      m[i + 2] = 0;
      m[i + 3] = 0;
    }
  }
}
```