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
<strstr>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <__tokencap_is_ro@plt>
  test   %rax,%rax
  jne    L6
L78:
  mov    %rbx,%rdi
  call   <__tokencap_is_ro@plt>
  test   %rax,%rax
  jne    L54
L24:
  movzbl (%rbx),%esi
  nopl   0x0(%rax,%rax,1)
L57:
  test   %sil,%sil
  je     L1
  mov    %esi,%ecx
  xor    %eax,%eax
  jmp    L62
  nopl   0x0(%rax,%rax,1)
L37:
  test   %dl,%dl
  je     L44
  add    $0x1,%rax
  movzbl (%rbx,%rax,1),%ecx
  test   %cl,%cl
  je     L1
L62:
  movzbl (%r12,%rax,1),%edx
  cmp    %cl,%dl
  je     L37
L44:
  add    $0x1,%r12
  cmpb   $0x0,-0x1(%r12)
  jne    L57
  xor    %r12d,%r12d
L1:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
L54:
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    $0x1,%edx
  mov    %rbx,%rdi
  mov    %eax,%esi
  call   <__tokencap_dump@plt>
  jmp    L24
L6:
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    $0x1,%edx
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <__tokencap_dump@plt>
  jmp    L78
```

assistant: ```c
char *strstr(const char *haystack, const char *needle) {
  if (__tokencap_is_ro(haystack))
    __tokencap_dump(haystack, strlen(haystack), 1);
  if (__tokencap_is_ro(needle))
    __tokencap_dump(needle, strlen(needle), 1);
  do {
    const char *n = needle;
    const char *h = haystack;
    while (*n && *h && *n == *h)
      n++, h++;
    if (!*n)
      return (char *)haystack;
  } while (*(haystack++));
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strstr>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <__tokencap_is_ro@plt>
  test   %rax,%rax
  jne    L72
L11:
  mov    %rbx,%rdi
  call   <__tokencap_is_ro@plt>
  test   %rax,%rax
  jne    L53
L58:
  movzbl (%rbx),%esi
  nopl   0x0(%rax,%rax,1)
L24:
  test   %sil,%sil
  je     L74
  mov    %esi,%ecx
  xor    %eax,%eax
  jmp    L50
  nopl   0x0(%rax,%rax,1)
L35:
  test   %dl,%dl
  je     L82
  add    $0x1,%rax
  movzbl (%rbx,%rax,1),%ecx
  test   %cl,%cl
  je     L74
L50:
  movzbl (%r12,%rax,1),%edx
  cmp    %cl,%dl
  je     L35
L82:
  add    $0x1,%r12
  cmpb   $0x0,-0x1(%r12)
  jne    L24
  xor    %r12d,%r12d
L74:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
L53:
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    $0x1,%edx
  mov    %rbx,%rdi
  mov    %eax,%esi
  call   <__tokencap_dump@plt>
  jmp    L58
L72:
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    $0x1,%edx
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <__tokencap_dump@plt>
  jmp    L11
```

assistant: ```c
char *strstr(const char *haystack, const char *needle) {
  if (__tokencap_is_ro(haystack))
    __tokencap_dump(haystack, strlen(haystack), 1);
  if (__tokencap_is_ro(needle))
    __tokencap_dump(needle, strlen(needle), 1);
  do {
    const char *n = needle;
    const char *h = haystack;
    while (*n && *h && *n == *h)
      n++, h++;
    if (!*n)
      return (char *)haystack;
  } while (*(haystack++));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<largest_range_pos_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jne    L17
  mov    $0xffffffff,%eax
  jmp    L21
L17:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  addq   $0x4,-0x18(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L35
L46:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0xc(%rbp)
  jge    L44
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L44:
  addq   $0x4,-0x18(%rbp)
  addl   $0x1,-0x4(%rbp)
L35:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jne    L46
  mov    -0x8(%rbp),%eax
L21:
  pop    %rbp
  ret
```

assistant: ```c
int largest_range_pos_int(int *range, int length) {
  if (length == 0)
    return -1;
  int largest_num;
  int largest_pos;
  largest_num = *range;
  largest_pos = 0;
  ++range;
  int i;
  for (i = 1; i != length; ++i) {
    if (*range > largest_num) {
      largest_num = *range;
      largest_pos = i;
    }
    ++range;
  }
  return largest_pos;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<largest_range_pos_int>:
  endbr64
  test   %esi,%esi
  je     L49
  mov    (%rdi),%ecx
  add    $0x4,%rdi
  cmp    $0x1,%esi
  je     L17
  mov    $0x1,%eax
  mov    $0x0,%r8d
  jmp    L9
L15:
  add    $0x4,%rdi
  add    $0x1,%eax
  cmp    %eax,%esi
  je     L68
L9:
  mov    (%rdi),%edx
  cmp    %ecx,%edx
  jle    L15
  mov    %eax,%r8d
  mov    %edx,%ecx
  jmp    L15
L49:
  mov    $0xffffffff,%r8d
  jmp    L68
L17:
  mov    $0x0,%r8d
L68:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int largest_range_pos_int(int *range, int length) {
  if (length == 0)
    return -1;
  int largest_num;
  int largest_pos;
  largest_num = *range;
  largest_pos = 0;
  ++range;
  int i;
  for (i = 1; i != length; ++i) {
    if (*range > largest_num) {
      largest_num = *range;
      largest_pos = i;
    }
    ++range;
  }
  return largest_pos;
}
```
user: What is the c source code of the assembly code below?

```asm
<largest_range_pos_int>:
  endbr64
  test   %esi,%esi
  je     L1
  mov    (%rdi),%ecx
  add    $0x4,%rdi
  cmp    $0x1,%esi
  je     L6
  mov    $0x1,%eax
  xor    %r8d,%r8d
  nopl   0x0(%rax,%rax,1)
L42:
  mov    (%rdi),%edx
  cmp    %ecx,%edx
  jle    L43
  mov    %eax,%r8d
  mov    %edx,%ecx
L43:
  add    $0x1,%eax
  add    $0x4,%rdi
  cmp    %eax,%esi
  jne    L42
L11:
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
L1:
  mov    $0xffffffff,%r8d
  jmp    L11
```

assistant: ```c
int largest_range_pos_int(int *range, int length) {
  if (length == 0)
    return -1;
  int largest_num;
  int largest_pos;
  largest_num = *range;
  largest_pos = 0;
  ++range;
  int i;
  for (i = 1; i != length; ++i) {
    if (*range > largest_num) {
      largest_num = *range;
      largest_pos = i;
    }
    ++range;
  }
  return largest_pos;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<largest_range_pos_int>:
  endbr64
  test   %esi,%esi
  je     L9
  mov    (%rdi),%ecx
  add    $0x4,%rdi
  cmp    $0x1,%esi
  je     L37
  mov    $0x1,%eax
  xor    %r8d,%r8d
  nopl   0x0(%rax,%rax,1)
L21:
  mov    (%rdi),%edx
  cmp    %edx,%ecx
  jge    L39
  mov    %eax,%r8d
  mov    %edx,%ecx
L39:
  add    $0x1,%eax
  add    $0x4,%rdi
  cmp    %eax,%esi
  jne    L21
L26:
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L37:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
L9:
  mov    $0xffffffff,%r8d
  jmp    L26
```

assistant: ```c
int largest_range_pos_int(int *range, int length) {
  if (length == 0)
    return -1;
  int largest_num;
  int largest_pos;
  largest_num = *range;
  largest_pos = 0;
  ++range;
  int i;
  for (i = 1; i != length; ++i) {
    if (*range > largest_num) {
      largest_num = *range;
      largest_pos = i;
    }
    ++range;
  }
  return largest_pos;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<formata_rg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L12
L16:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L25
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  lea    0x30(%rax),%ecx
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L12
L25:
  addq   $0x1,-0x20(%rbp)
L12:
  cmpl   $0x8,-0x4(%rbp)
  jle    L16
  nop
  nop
  leave
  ret
```

assistant: ```c
void formata_rg(int *n, char *rg) {
  int i = 0;
  while (i < 9)
    if (isspace(*rg))
      *rg++ = n[i++] + '0';
    else
      rg++;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<formata_rg>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  lea    0x1(%rsi),%rbx
  mov    $0x0,%ebp
  jmp    L31
L37:
  add    $0x1,%rbx
  cmp    $0x8,%ebp
  jg     L6
L31:
  movsbl -0x1(%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L37
  movslq %ebp,%rax
  mov    (%r12,%rax,4),%eax
  add    $0x30,%eax
  mov    %al,-0x1(%rbx)
  lea    0x1(%rbp),%ebp
  jmp    L37
L6:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void formata_rg(int *n, char *rg) {
  int i = 0;
  while (i < 9)
    if (isspace(*rg))
      *rg++ = n[i++] + '0';
    else
      rg++;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<formata_rg>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rsi,%rbx
L1:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <isspace@plt>
  test   %rax,%rax
  je     L2
  movslq %ebp,%rax
  add    $0x1,%ebp
  mov    (%r12,%rax,4),%eax
  add    $0x30,%eax
  mov    %al,-0x1(%rbx)
L2:
  cmp    $0x9,%ebp
  jne    L1
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void formata_rg(int *n, char *rg) {
  int i = 0;
  while (i < 9)
    if (isspace(*rg))
      *rg++ = n[i++] + '0';
    else
      rg++;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<formata_rg>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rsi,%rbx
L37:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <isspace@plt>
  test   %rax,%rax
  je     L16
  movslq %ebp,%rax
  lea    0x1(%rbp),%edx
  mov    (%r12,%rax,4),%eax
  add    $0x30,%eax
  mov    %al,-0x1(%rbx)
  cmp    $0x8,%ebp
  je     L22
  mov    %edx,%ebp
  jmp    L37
  nopw   0x0(%rax,%rax,1)
L16:
  cmp    $0x9,%ebp
  jne    L37
L22:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void formata_rg(int *n, char *rg) {
  int i = 0;
  while (i < 9)
    if (isspace(*rg))
      *rg++ = n[i++] + '0';
    else
      rg++;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x14(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x10(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0xc(%rbp)
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movss  -0x14(%rbp),%xmm1
  movss  -0x10(%rbp),%xmm0
  addss  %xmm1,%xmm0
  movss  D54(%rip),%xmm1
  divss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  pxor   %xmm2,%xmm2
  cvtss2sd -0xc(%rbp),%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movss  -0xc(%rbp),%xmm0
  comiss D15(%rip),%xmm0
  jb     L8
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L3
L8:
  lea    D67(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L3:
  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: data_tag: D2
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D4
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D58
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D54
type_infered: byte[4]
name: extract_symbolic_data
arguments: data_tag: D20
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D15
type_infered: byte[4]
name: extract_symbolic_data
arguments: data_tag: D0
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D67
type_infered: string

tool: D2: Informe a nota da Av1:

tool: D4: %f

tool: D58: Informe a nota da Av2:

tool: {"D54": [0x00,0x00,0x00,0x40]}

tool: D20: A media e igual a %.2f

tool: {"D15": [0x00,0x00,0xc0,0x40]}

tool: D0: aprovado

tool: D67: Reprovado

assistant: ```c
int main() {
  float av1 = 0, av2 = 0, media = 0;
  printf("Informe a nota da Av1:");
  scanf("%f", &av1);
  printf("Informe a nota da Av2:");
  scanf("%f", &av2);
  media = (av1 + av2) / 2;
  printf("A media e igual a %.2f", media);
  if (media >= 6) {
    printf("aprovado");
  } else
    printf("Reprovado");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x10(%rsp)
  movl   $0x0,0x14(%rsp)
  lea    D36(%rip),%rdi
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  lea    D22(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D20(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movss  0x10(%rsp),%xmm0
  addss  0x14(%rsp),%xmm0
  mulss  D8(%rip),%xmm0
  movss  %xmm0,0xc(%rsp)
  cvtss2sd %xmm0,%xmm0
  lea    D0(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movss  0xc(%rsp),%xmm0
  comiss D29(%rip),%xmm0
  jb     L21
  lea    D11(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L11:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L21:
  lea    D58(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L11
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D36
type: string
name: unpack_data
arguments: data_tag: D22
type: string
name: unpack_data
arguments: data_tag: D20
type: string
name: unpack_data
arguments: data_tag: D8
type: byte[4]
name: unpack_data
arguments: data_tag: D0
type: string
name: unpack_data
arguments: data_tag: D29
type: byte[4]
name: unpack_data
arguments: data_tag: D11
type: string
name: unpack_data
arguments: data_tag: D58
type: string

tool: D36: Informe a nota da Av1:

tool: D22: %f

tool: D20: Informe a nota da Av2:

tool: {"D8": [0x00,0x00,0x00,0x3f]}

tool: D0: A media e igual a %.2f

tool: {"D29": [0x00,0x00,0xc0,0x40]}

tool: D11: aprovado

tool: D58: Reprovado

assistant: ```c
int main() {
  float av1 = 0, av2 = 0, media = 0;
  printf("Informe a nota da Av1:");
  scanf("%f", &av1);
  printf("Informe a nota da Av2:");
  scanf("%f", &av2);
  media = (av1 + av2) / 2;
  printf("A media e igual a %.2f", media);
  if (media >= 6) {
    printf("aprovado");
  } else
    printf("Reprovado");
  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
  lea    D68(%rip),%rdi
  lea    D16(%rip),%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x10(%rsp)
  movl   $0x0,0x14(%rsp)
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D35(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movss  0x10(%rsp),%xmm1
  addss  0x14(%rsp),%xmm1
  pxor   %xmm0,%xmm0
  mulss  D15(%rip),%xmm1
  lea    D48(%rip),%rdi
  mov    $0x1,%eax
  movss  %xmm1,0xc(%rsp)
  cvtss2sd %xmm1,%xmm0
  call   <printf@plt>
  movss  0xc(%rsp),%xmm1
  comiss D45(%rip),%xmm1
  jb     L17
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L26:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L17:
  lea    D2(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L26
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D68
type_infered: string
name: parse_data
arguments: tag: D16
type_infered: string
name: parse_data
arguments: tag: D35
type_infered: string
name: parse_data
arguments: tag: D15
type_infered: byte[4]
name: parse_data
arguments: tag: D48
type_infered: string
name: parse_data
arguments: tag: D45
type_infered: byte[4]
name: parse_data
arguments: tag: D7
type_infered: string
name: parse_data
arguments: tag: D2
type_infered: string

tool: D68: Informe a nota da Av1:

tool: D16: %f

tool: D35: Informe a nota da Av2:

tool: {"D15": [0x00,0x00,0x00,0x3f]}

tool: D48: A media e igual a %.2f

tool: {"D45": [0x00,0x00,0xc0,0x40]}

tool: D7: aprovado

tool: D2: Reprovado

assistant: ```c
int main() {
  float av1 = 0, av2 = 0, media = 0;
  printf("Informe a nota da Av1:");
  scanf("%f", &av1);
  printf("Informe a nota da Av2:");
  scanf("%f", &av2);
  media = (av1 + av2) / 2;
  printf("A media e igual a %.2f", media);
  if (media >= 6) {
    printf("aprovado");
  } else
    printf("Reprovado");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D55(%rip),%rdi
  lea    D49(%rip),%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x10(%rsp)
  movl   $0x0,0x14(%rsp)
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D53(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movss  0x10(%rsp),%xmm1
  addss  0x14(%rsp),%xmm1
  pxor   %xmm0,%xmm0
  mulss  D60(%rip),%xmm1
  lea    D43(%rip),%rdi
  mov    $0x1,%eax
  movss  %xmm1,0xc(%rsp)
  cvtss2sd %xmm1,%xmm0
  call   <printf@plt>
  movss  0xc(%rsp),%xmm1
  comiss D23(%rip),%xmm1
  jb     L28
  lea    D21(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L7:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L28:
  lea    D76(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L7
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D55
type_infered: string
name: parse_data
arguments: data_label: D49
type_infered: string
name: parse_data
arguments: data_label: D53
type_infered: string
name: parse_data
arguments: data_label: D60
type_infered: byte[4]
name: parse_data
arguments: data_label: D43
type_infered: string
name: parse_data
arguments: data_label: D23
type_infered: byte[4]
name: parse_data
arguments: data_label: D21
type_infered: string
name: parse_data
arguments: data_label: D76
type_infered: string

tool: D55: Informe a nota da Av1:

tool: D49: %f

tool: D53: Informe a nota da Av2:

tool: {"D60": [0x00,0x00,0x00,0x3f]}

tool: D43: A media e igual a %.2f

tool: {"D23": [0x00,0x00,0xc0,0x40]}

tool: D21: aprovado

tool: D76: Reprovado

assistant: ```c
int main() {
  float av1 = 0, av2 = 0, media = 0;
  printf("Informe a nota da Av1:");
  scanf("%f", &av1);
  printf("Informe a nota da Av2:");
  scanf("%f", &av2);
  media = (av1 + av2) / 2;
  printf("A media e igual a %.2f", media);
  if (media >= 6) {
    printf("aprovado");
  } else
    printf("Reprovado");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<_ftok>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <stat@plt>
  cmp    $0xffffffff,%eax
  jne    L11
  mov    $0xffffffff,%eax
  jmp    L6
L11:
  movl   $0x0,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0x2c(%rbp),%eax
  movzbl %al,%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x10(%rbp),%eax
  and    $0xff,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0xc(%rbp),%eax
  and    $0xffff,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x20(%rbp),%eax
  or     %eax,-0x14(%rbp)
  shll   $0x8,-0x14(%rbp)
  mov    -0x1c(%rbp),%eax
  or     %eax,-0x14(%rbp)
  shll   $0x10,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  or     %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
L6:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

assistant: ```c
key_t _ftok(char *pathname, int proj) {
  struct stat info;
  if (stat(pathname, &info) == -1)
    return -1;
  int projbits = 0, devbits = 0, inodebits = 0;
  key_t key = 0;
  projbits = proj & 0xff;
  devbits = info.st_dev & 0xff;
  inodebits = info.st_ino & 0xffff;
  key |= projbits;
  key <<= 8;
  key |= devbits;
  key <<= 16;
  key |= inodebits;
  return key;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<_ftok>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %esi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <stat@plt>
  cmp    $0xffffffff,%eax
  je     L0
  shl    $0x8,%ebx
  mov    %ebx,%esi
  movzbl (%rsp),%eax
  or     %eax,%esi
  shl    $0x10,%esi
  movzwl 0x4(%rsp),%eax
  or     %esi,%eax
L0:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x10,%rsp
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
key_t _ftok(char *pathname, int proj) {
  struct stat info;
  if (stat(pathname, &info) == -1)
    return -1;
  int projbits = 0, devbits = 0, inodebits = 0;
  key_t key = 0;
  projbits = proj & 0xff;
  devbits = info.st_dev & 0xff;
  inodebits = info.st_ino & 0xffff;
  key |= projbits;
  key <<= 8;
  key |= devbits;
  key <<= 16;
  key |= inodebits;
  return key;
}
```
user: What is the c source code of the assembly code below:

```asm
<_ftok>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <stat@plt>
  cmp    $0xffffffff,%eax
  je     L7
  movzbl (%rsp),%eax
  shl    $0x8,%ebx
  mov    %ebx,%esi
  or     %eax,%esi
  movzwl 0x4(%rsp),%eax
  shl    $0x10,%esi
  or     %esi,%eax
L7:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x10,%rsp
  pop    %rbx
  ret
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
key_t _ftok(char *pathname, int proj) {
  struct stat info;
  if (stat(pathname, &info) == -1)
    return -1;
  int projbits = 0, devbits = 0, inodebits = 0;
  key_t key = 0;
  projbits = proj & 0xff;
  devbits = info.st_dev & 0xff;
  inodebits = info.st_ino & 0xffff;
  key |= projbits;
  key <<= 8;
  key |= devbits;
  key <<= 16;
  key |= inodebits;
  return key;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_ftok>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <stat@plt>
  cmp    $0xffffffff,%eax
  je     L6
  movzbl (%rsp),%eax
  shl    $0x8,%ebx
  mov    %ebx,%esi
  or     %eax,%esi
  movzwl 0x4(%rsp),%eax
  shl    $0x10,%esi
  or     %esi,%eax
L6:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x10,%rsp
  pop    %rbx
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
key_t _ftok(char *pathname, int proj) {
  struct stat info;
  if (stat(pathname, &info) == -1)
    return -1;
  int projbits = 0, devbits = 0, inodebits = 0;
  key_t key = 0;
  projbits = proj & 0xff;
  devbits = info.st_dev & 0xff;
  inodebits = info.st_ino & 0xffff;
  key |= projbits;
  key <<= 8;
  key |= devbits;
  key <<= 16;
  key |= inodebits;
  return key;
}
```