ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D3(%rip),%rdi
  push   %r14
  lea    D28(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x278,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x268(%rsp)
  xor    %eax,%eax
  lea    0x190(%rsp),%rbp
  lea    0x200(%rsp),%r13
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D10(%rip),%rdi
  call   <scanf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <strcpy@plt>
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  je     L19
  mov    %rsp,%rbx
  mov    %rax,%rsi
  xor    %ebp,%ebp
  mov    %rbx,%r15
  nopl   0x0(%rax)
L59:
  mov    %r15,%rdi
  movslq %ebp,%r12
  add    $0x14,%r15
  add    $0x1,%ebp
  call   <strcpy@plt>
  mov    %r14,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L59
L68:
  mov    %r12d,%eax
  shr    $0x1f,%eax
  add    %r12d,%eax
  sar    %eax
  cltq
  lea    (%rax,%rax,4),%rax
  lea    0x14(%rsp,%rax,4),%r14
  lea    (%r12,%r12,4),%rax
  xor    %r12d,%r12d
  lea    (%rbx,%rax,2),%rbp
  add    %rbp,%rbp
  jmp    L88
  nopw   0x0(%rax,%rax,1)
L75:
  add    $0x14,%rbx
  mov    $0x1,%r12d
  cmp    %r14,%rbx
  je     L74
L88:
  mov    %rbp,%rdi
  mov    %rbx,%rsi
  sub    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L75
  sub    $0x1,%r12d
  je     L74
  mov    %r13,%rsi
  lea    D41(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L58:
  mov    0x268(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x278,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L74:
  mov    %r13,%rsi
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L58
L19:
  or     $0xffffffffffffffff,%r12
  mov    %rsp,%rbx
  jmp    L68
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D3
type: string
name: unpack_data
arguments: label: D28
type: string
name: unpack_data
arguments: label: D10
type: string
name: unpack_data
arguments: label: D41
type: string
name: unpack_data
arguments: label: D38
type: string

tool: D3: Enter a sentence : 

tool: D28:  ,.?!

tool: D10: %[^
]s

tool: D41: "%s" is NOT a Palindrome paragraph.


tool: D38: "%s" is a Palindrome paragraph.


assistant: ```c
int main() {
  printf("Enter a sentence : ");
  char s[100];
  scanf("%[^\n]s", s);
  char str[100];
  strcpy(str, s);
  char *word;
  word = strtok(s, " ,.?!");
  char arr[20][20];
  int i = 0;
  while (word != NULL) {
    strcpy(arr[i], word);
    i++;
    word = strtok(NULL, " ,.?!");
  }
  int count = i - 1;
  int flag = 0;
  for (i = 0; i <= count / 2; i++) {
    if (strcmp(arr[count - i], arr[i]) != 0)
      break;
    else
      flag = 1;
  }
  if (flag == 1)
    printf("\"%s\" is a Palindrome paragraph.\n", str);
  else
    printf("\"%s\" is NOT a Palindrome paragraph.\n", str);
  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   %r15
  lea    D17(%rip),%rdi
  push   %r14
  lea    D16(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x278,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x268(%rsp)
  xor    %eax,%eax
  lea    0x190(%rsp),%rbp
  lea    0x200(%rsp),%r13
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <strcpy@plt>
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  je     L21
  mov    %rsp,%rbx
  mov    %rax,%rsi
  xor    %ebp,%ebp
  mov    %rbx,%r15
  nopl   0x0(%rax)
L5:
  mov    %r15,%rdi
  movslq %ebp,%r12
  add    $0x14,%r15
  add    $0x1,%ebp
  call   <strcpy@plt>
  mov    %r14,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L5
L57:
  mov    %r12d,%eax
  shr    $0x1f,%eax
  add    %r12d,%eax
  sar    %eax
  cltq
  lea    (%rax,%rax,4),%rax
  lea    0x14(%rsp,%rax,4),%r14
  lea    (%r12,%r12,4),%rax
  xor    %r12d,%r12d
  lea    (%rbx,%rax,2),%rbp
  add    %rbp,%rbp
  jmp    L24
  nopw   0x0(%rax,%rax,1)
L75:
  add    $0x14,%rbx
  mov    $0x1,%r12d
  cmp    %rbx,%r14
  je     L13
L24:
  mov    %rbp,%rdi
  mov    %rbx,%rsi
  sub    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L75
  sub    $0x1,%r12d
  je     L13
  mov    %r13,%rsi
  lea    D12(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L20:
  mov    0x268(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L64
  add    $0x278,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L13:
  mov    %r13,%rsi
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L20
L21:
  or     $0xffffffffffffffff,%r12
  mov    %rsp,%rbx
  jmp    L57
L64:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D17
type: string
name: extract_symbolic_data
arguments: data_tag: D16
type: string
name: extract_symbolic_data
arguments: data_tag: D14
type: string
name: extract_symbolic_data
arguments: data_tag: D12
type: string
name: extract_symbolic_data
arguments: data_tag: D15
type: string

tool: D17: Enter a sentence : 

tool: D16:  ,.?!

tool: D14: %[^
]s

tool: D12: "%s" is NOT a Palindrome paragraph.


tool: D15: "%s" is a Palindrome paragraph.


assistant: ```c
int main() {
  printf("Enter a sentence : ");
  char s[100];
  scanf("%[^\n]s", s);
  char str[100];
  strcpy(str, s);
  char *word;
  word = strtok(s, " ,.?!");
  char arr[20][20];
  int i = 0;
  while (word != NULL) {
    strcpy(arr[i], word);
    i++;
    word = strtok(NULL, " ,.?!");
  }
  int count = i - 1;
  int flag = 0;
  for (i = 0; i <= count / 2; i++) {
    if (strcmp(arr[count - i], arr[i]) != 0)
      break;
    else
      flag = 1;
  }
  if (flag == 1)
    printf("\"%s\" is a Palindrome paragraph.\n", str);
  else
    printf("\"%s\" is NOT a Palindrome paragraph.\n", str);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    D13(%rip),%rax
  movl   $0x50,(%rax)
  mov    D13(%rip),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    D13(%rip),%rax
  mov    %edx,(%rax)
  mov    D13(%rip),%rax
  mov    (%rax),%eax
  lea    -0x2(%rax),%edx
  mov    D27(%rip),%rax
  mov    %edx,(%rax)
  movl   $0x0,-0x30(%rbp)
  mov    D13(%rip),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  movslq %eax,%rdx
  mov    %rdx,%rsi
  mov    $0x0,%edi
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%edi
  mov    $0x0,%edx
  div    %rdi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L2:
  cmp    %rdx,%rsp
  je     L46
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L2
L46:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L42
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L42:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x20(%rbp)
  jmp    L78
L20:
  cmpl   $0xa,-0x2c(%rbp)
  je     L18
  mov    D27(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x30(%rbp)
  jle    L24
L18:
  mov    -0x20(%rbp),%rdx
  mov    -0x30(%rbp),%eax
  cltq
  movb   $0x0,(%rdx,%rax,1)
  mov    -0x2c(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <println@plt>
  mov    %eax,-0x30(%rbp)
  jmp    L78
L24:
  mov    -0x30(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x30(%rbp)
  mov    -0x2c(%rbp),%edx
  mov    %edx,%ecx
  mov    -0x20(%rbp),%rdx
  cltq
  mov    %cl,(%rdx,%rax,1)
L78:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x2c(%rbp)
  mov    D38(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x2c(%rbp)
  jne    L20
  mov    $0x0,%eax
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L69
  call   <__stack_chk_fail@plt>
L69:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int main(void) {
  extern int maxstr;
  extern int endarr;
  maxstr = 80;
  ++maxstr;
  endarr = maxstr - 2;
  int c;
  int col = 0;
  char line[maxstr];
  while ((c = getchar()) != EOF) {
    if (c == '\n' || col > endarr) {
      line[col] = '\0';
      col = println(line, c);
    } else {
      line[col++] = c;
    }
  }
  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
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    D26(%rip),%rax
  movl   $0x51,(%rax)
  mov    D16(%rip),%rax
  movl   $0x4f,(%rax)
  mov    $0x0,%ebx
  mov    %rsp,%rbp
  jmp    L24
L14:
  movslq %ebx,%rbx
  movb   $0x0,(%rsp,%rbx,1)
  mov    %rbp,%rdi
  call   <println@plt>
  mov    %eax,%ebx
L24:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%esi
  mov    D31(%rip),%rax
  cmp    %esi,(%rax)
  je     L16
  cmp    $0xa,%esi
  je     L14
  mov    D16(%rip),%rax
  cmp    %ebx,(%rax)
  jl     L14
  movslq %ebx,%rax
  mov    %sil,(%rsp,%rax,1)
  lea    0x1(%rbx),%ebx
  jmp    L24
L16:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  mov    $0x0,%eax
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  ret
L29:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  extern int maxstr;
  extern int endarr;
  maxstr = 80;
  ++maxstr;
  endarr = maxstr - 2;
  int c;
  int col = 0;
  char line[maxstr];
  while ((c = getchar()) != EOF) {
    if (c == '\n' || col > endarr) {
      line[col] = '\0';
      col = println(line, c);
    } else {
      line[col++] = c;
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x68,%rsp
  mov    D22(%rip),%r12
  mov    D10(%rip),%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  mov    D24(%rip),%rax
  mov    %rsp,%r13
  movl   $0x4f,(%r12)
  movl   $0x51,(%rax)
  nopl   0x0(%rax)
L3:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%esi
  cmp    %eax,0x0(%rbp)
  je     L49
L38:
  cmp    $0xa,%esi
  je     L16
  cmp    %ebx,(%r12)
  jl     L16
  movslq %ebx,%rax
  add    $0x1,%ebx
  mov    %sil,(%rsp,%rax,1)
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%esi
  cmp    %eax,0x0(%rbp)
  jne    L38
L49:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  add    $0x68,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L16:
  movslq %ebx,%rbx
  mov    %r13,%rdi
  movb   $0x0,(%rsp,%rbx,1)
  call   <println@plt>
  mov    %eax,%ebx
  jmp    L3
L48:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  extern int maxstr;
  extern int endarr;
  maxstr = 80;
  ++maxstr;
  endarr = maxstr - 2;
  int c;
  int col = 0;
  char line[maxstr];
  while ((c = getchar()) != EOF) {
    if (c == '\n' || col > endarr) {
      line[col] = '\0';
      col = println(line, c);
    } else {
      line[col++] = c;
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x68,%rsp
  mov    D28(%rip),%r12
  mov    D6(%rip),%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  mov    D11(%rip),%rax
  mov    %rsp,%r13
  movl   $0x4f,(%r12)
  movl   $0x51,(%rax)
  nopl   0x0(%rax)
L10:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%esi
  cmp    %eax,0x0(%rbp)
  je     L49
L31:
  cmp    $0xa,%esi
  je     L55
  cmp    %ebx,(%r12)
  jl     L55
  movslq %ebx,%rax
  add    $0x1,%ebx
  mov    %sil,(%rsp,%rax,1)
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%esi
  cmp    %eax,0x0(%rbp)
  jne    L31
L49:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x68,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L55:
  movslq %ebx,%rbx
  mov    %r13,%rdi
  movb   $0x0,(%rsp,%rbx,1)
  call   <println@plt>
  mov    %eax,%ebx
  jmp    L10
L26:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  extern int maxstr;
  extern int endarr;
  maxstr = 80;
  ++maxstr;
  endarr = maxstr - 2;
  int c;
  int col = 0;
  char line[maxstr];
  while ((c = getchar()) != EOF) {
    if (c == '\n' || col > endarr) {
      line[col] = '\0';
      col = println(line, c);
    } else {
      line[col++] = c;
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<breakupPacket>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    -0x20(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x28(%rbp),%rax
  movq   $0x0,(%rax)
  mov    -0x30(%rbp),%rax
  movq   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <ntohs@plt>
  mov    D18(%rip),%rdx
  mov    (%rdx),%rdx
  cmp    %rdx,%rax
  jne    L7
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  lea    0x4(%rax),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    (%rax),%rdx
  mov    D19(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L39
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    0x8(%rax),%eax
  shl    $0x2,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    %rax,%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  jmp    L22
L7:
  nop
  jmp    L22
L39:
  nop
L22:
  leave
  ret
```

assistant: ```c
void breakupPacket(void *buf, struct ether_header **e, struct ip **i, struct tcphdr **t) {
  int ip_hl;
  *e = (struct ether_header *)buf;
  *i = NULL;
  *t = NULL;
  if (ETHERTYPE_IP != ntohs((*e)->ether_type))
    return;
  *i = (struct ip *)(*e + 1);
  if (IPPROTO_TCP != (*i)->ip_p)
    return;
  ip_hl = (*i)->ip_hl << 2;
  *t = (struct tcphdr *)((char *)*i + ip_hl);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<breakupPacket>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    %rdx,%rbp
  mov    %rcx,%r12
  mov    %rdi,(%rsi)
  movq   $0x0,(%rdx)
  movq   $0x0,(%rcx)
  mov    (%rsi),%rax
  mov    (%rax),%edi
  call   <ntohs@plt>
  mov    D1(%rip),%rdx
  cmp    (%rdx),%rax
  jne    L15
  mov    (%rbx),%rdx
  lea    0x4(%rdx),%rax
  mov    %rax,0x0(%rbp)
  mov    D16(%rip),%rcx
  mov    (%rcx),%rsi
  cmp    %rsi,0x4(%rdx)
  jne    L15
  mov    0xc(%rdx),%edx
  shl    $0x2,%edx
  movslq %edx,%rdx
  add    %rdx,%rax
  mov    %rax,(%r12)
L15:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void breakupPacket(void *buf, struct ether_header **e, struct ip **i, struct tcphdr **t) {
  int ip_hl;
  *e = (struct ether_header *)buf;
  *i = NULL;
  *t = NULL;
  if (ETHERTYPE_IP != ntohs((*e)->ether_type))
    return;
  *i = (struct ip *)(*e + 1);
  if (IPPROTO_TCP != (*i)->ip_p)
    return;
  ip_hl = (*i)->ip_hl << 2;
  *t = (struct tcphdr *)((char *)*i + ip_hl);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<breakupPacket>:
  endbr64
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  mov    %rdi,(%rsi)
  mov    (%rdi),%edi
  movq   $0x0,(%rdx)
  movq   $0x0,(%rcx)
  call   <ntohs@plt>
  mov    D8(%rip),%rdx
  cmp    (%rdx),%rax
  jne    L15
  mov    0x0(%rbp),%rdx
  mov    D1(%rip),%rcx
  lea    0x4(%rdx),%rax
  mov    %rax,(%rbx)
  mov    (%rcx),%rbx
  cmp    %rbx,0x4(%rdx)
  je     L9
L15:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L9:
  mov    0xc(%rdx),%edx
  pop    %rbx
  pop    %rbp
  shl    $0x2,%edx
  movslq %edx,%rdx
  add    %rdx,%rax
  mov    %rax,(%r12)
  pop    %r12
  ret
```

assistant: ```c
void breakupPacket(void *buf, struct ether_header **e, struct ip **i, struct tcphdr **t) {
  int ip_hl;
  *e = (struct ether_header *)buf;
  *i = NULL;
  *t = NULL;
  if (ETHERTYPE_IP != ntohs((*e)->ether_type))
    return;
  *i = (struct ip *)(*e + 1);
  if (IPPROTO_TCP != (*i)->ip_p)
    return;
  ip_hl = (*i)->ip_hl << 2;
  *t = (struct tcphdr *)((char *)*i + ip_hl);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<breakupPacket>:
  endbr64
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  mov    %rdi,(%rsi)
  mov    (%rdi),%edi
  movq   $0x0,(%rdx)
  movq   $0x0,(%rcx)
  call   <ntohs@plt>
  mov    D6(%rip),%rdx
  cmp    (%rdx),%rax
  jne    L7
  mov    0x0(%rbp),%rdx
  mov    D15(%rip),%rcx
  lea    0x4(%rdx),%rax
  mov    %rax,(%rbx)
  mov    (%rcx),%rbx
  cmp    %rbx,0x4(%rdx)
  je     L14
L7:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L14:
  mov    0xc(%rdx),%edx
  pop    %rbx
  pop    %rbp
  shl    $0x2,%edx
  movslq %edx,%rdx
  add    %rdx,%rax
  mov    %rax,(%r12)
  pop    %r12
  ret
```

assistant: ```c
void breakupPacket(void *buf, struct ether_header **e, struct ip **i, struct tcphdr **t) {
  int ip_hl;
  *e = (struct ether_header *)buf;
  *i = NULL;
  *t = NULL;
  if (ETHERTYPE_IP != ntohs((*e)->ether_type))
    return;
  *i = (struct ip *)(*e + 1);
  if (IPPROTO_TCP != (*i)->ip_p)
    return;
  ip_hl = (*i)->ip_hl << 2;
  *t = (struct tcphdr *)((char *)*i + ip_hl);
}
```
user: Translate the below assembly code into C programming language:

```asm
<max_heapify>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <left@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <right@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L3
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L3
  mov    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L22
L3:
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L22:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L12
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L12
  mov    -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L12:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  je     L25
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <exchange_num@plt>
  mov    -0x20(%rbp),%edx
  mov    -0xc(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <max_heapify>
L25:
  nop
  leave
  ret
```

assistant: ```c
void max_heapify(int *num, int index, int num_length) {
  int left_index, right_index, max_index;
  left_index = left(index);
  right_index = right(index);
  if ((left_index < num_length) && (num[left_index] > num[index]))
    max_index = left_index;
  else
    max_index = index;
  if ((right_index < num_length) && (num[right_index] > num[max_index]))
    max_index = right_index;
  if (max_index != index) {
    exchange_num(num + index, num + max_index);
    max_heapify(num, max_index, num_length);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<max_heapify>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%ebx
  mov    %edx,%r13d
  mov    %esi,%edi
  call   <left@plt>
  mov    %eax,%r14d
  mov    %ebx,%edi
  call   <right@plt>
  mov    %eax,%r12d
  cmp    %r13d,%r14d
  jge    L36
  movslq %r14d,%rax
  movslq %ebx,%rdx
  mov    0x0(%rbp,%rdx,4),%ecx
  cmp    %ecx,0x0(%rbp,%rax,4)
  jle    L36
  cmp    %r13d,%r12d
  jge    L26
  movslq %r12d,%rdx
  mov    0x0(%rbp,%rax,4),%eax
  cmp    %eax,0x0(%rbp,%rdx,4)
  cmovle %r14d,%r12d
  jmp    L7
L26:
  mov    %r14d,%r12d
  jmp    L7
L36:
  cmp    %r13d,%r12d
  jge    L6
  movslq %r12d,%rax
  movslq %ebx,%rdx
  mov    0x0(%rbp,%rdx,4),%ecx
  cmp    %ecx,0x0(%rbp,%rax,4)
  jle    L6
L7:
  cmp    %ebx,%r12d
  je     L6
  movslq %r12d,%rax
  lea    0x0(%rbp,%rax,4),%rsi
  movslq %ebx,%rbx
  lea    0x0(%rbp,%rbx,4),%rdi
  call   <exchange_num@plt>
  mov    %r13d,%edx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <max_heapify>
L6:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void max_heapify(int *num, int index, int num_length) {
  int left_index, right_index, max_index;
  left_index = left(index);
  right_index = right(index);
  if ((left_index < num_length) && (num[left_index] > num[index]))
    max_index = left_index;
  else
    max_index = index;
  if ((right_index < num_length) && (num[right_index] > num[max_index]))
    max_index = right_index;
  if (max_index != index) {
    exchange_num(num + index, num + max_index);
    max_heapify(num, max_index, num_length);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<max_heapify>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  jmp    L33
  cs nopw 0x0(%rax,%rax,1)
L20:
  movslq %ebp,%rax
  movslq %r12d,%rdx
  mov    (%r14,%rax,4),%ecx
  cmp    %ecx,(%r14,%rdx,4)
  jle    L5
  cmp    %r13d,%ebx
  jl     L28
  mov    %r12d,%ebx
L7:
  cmp    %ebp,%ebx
  je     L23
  movslq %ebx,%rdx
  lea    (%r14,%rax,4),%rdi
  mov    %ebx,%ebp
  lea    (%r14,%rdx,4),%rsi
  call   <exchange_num@plt>
L33:
  mov    %ebp,%edi
  call   <left@plt>
  mov    %ebp,%edi
  mov    %eax,%r12d
  call   <right@plt>
  mov    %eax,%ebx
  cmp    %r13d,%r12d
  jl     L20
L5:
  cmp    %r13d,%ebx
  jge    L23
  movslq %ebp,%rax
  movslq %ebx,%rdx
  mov    (%r14,%rax,4),%ecx
  cmp    %ecx,(%r14,%rdx,4)
  jg     L7
L23:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L28:
  movslq %ebx,%rcx
  mov    (%r14,%rdx,4),%esi
  cmp    %esi,(%r14,%rcx,4)
  cmovle %r12d,%ebx
  jmp    L7
```

assistant: ```c
void max_heapify(int *num, int index, int num_length) {
  int left_index, right_index, max_index;
  left_index = left(index);
  right_index = right(index);
  if ((left_index < num_length) && (num[left_index] > num[index]))
    max_index = left_index;
  else
    max_index = index;
  if ((right_index < num_length) && (num[right_index] > num[max_index]))
    max_index = right_index;
  if (max_index != index) {
    exchange_num(num + index, num + max_index);
    max_heapify(num, max_index, num_length);
  }
}
```