ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<mnemonic2words>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  lea    D4(%rip),%r12
  push   %rbp
  mov    %r12,%rbp
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L2
L48:
  cmp    $0x5f,%dil
  je     L49
  call   <toupper@plt>
  lea    0x1(%rbx),%r14
  lea    0x1(%rbp),%r13
  mov    %al,0x0(%rbp)
  movsbl 0x1(%rbx),%edi
  test   %dil,%dil
  je     L36
L52:
  cmp    $0x5f,%dil
  je     L34
  call   <tolower@plt>
  add    $0x1,%r14
  add    $0x1,%r13
  mov    %al,-0x1(%r13)
  movsbl (%r14),%edi
  test   %dil,%dil
  jne    L52
L36:
  mov    %r13,%rbp
L2:
  movb   $0x0,0x0(%rbp)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L34:
  mov    %r13,%rbp
  mov    %r14,%rbx
L49:
  add    $0x1,%rbx
  movb   $0x20,0x0(%rbp)
  add    $0x1,%rbp
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L48
  movb   $0x0,0x0(%rbp)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *mnemonic2words(char *mnemonic) {
  static char a[1024 + 1];
  char *ptr = a;
  int first_letter_of_word = 1;
  while (*mnemonic) {
    if (*mnemonic == '_') {
      *ptr++ = ' ';
      mnemonic++;
      first_letter_of_word = 1;
    } else if (first_letter_of_word) {
      *ptr++ = toupper(*mnemonic++);
      first_letter_of_word = 0;
    } else {
      *ptr++ = tolower(*mnemonic++);
    }
  }
  *ptr = '\0';
  return a;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getLocalPort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  movl   $0x4,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  lea    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <bzero@plt>
  lea    -0xc(%rbp),%rdx
  lea    -0x10(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <getsockname@plt>
  mov    -0x10(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
uint16_t getLocalPort(int fd) {
  struct sockaddr_in sin = {};
  socklen_t slen = sizeof(struct sockaddr);
  bzero(&sin, slen);
  getsockname(fd, (struct sockaddr *)&sin, &slen);
  return sin.sin_port;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getLocalPort>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rsp)
  movl   $0x4,0x4(%rsp)
  mov    %rsp,%rbp
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <bzero@plt>
  lea    0x4(%rsp),%rdx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <getsockname@plt>
  mov    (%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint16_t getLocalPort(int fd) {
  struct sockaddr_in sin = {};
  socklen_t slen = sizeof(struct sockaddr);
  bzero(&sin, slen);
  getsockname(fd, (struct sockaddr *)&sin, &slen);
  return sin.sin_port;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<getLocalPort>:
  endbr64
  push   %r12
  mov    $0x4,%esi
  push   %rbp
  mov    %edi,%ebp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  movl   $0x0,(%rsp)
  mov    %r12,%rdi
  movl   $0x4,0x4(%rsp)
  call   <bzero@plt>
  lea    0x4(%rsp),%rdx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <getsockname@plt>
  mov    (%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint16_t getLocalPort(int fd) {
  struct sockaddr_in sin = {};
  socklen_t slen = sizeof(struct sockaddr);
  bzero(&sin, slen);
  getsockname(fd, (struct sockaddr *)&sin, &slen);
  return sin.sin_port;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<getLocalPort>:
  endbr64
  push   %r12
  mov    $0x4,%esi
  push   %rbp
  mov    %edi,%ebp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  movl   $0x0,(%rsp)
  mov    %r12,%rdi
  movl   $0x4,0x4(%rsp)
  call   <bzero@plt>
  lea    0x4(%rsp),%rdx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <getsockname@plt>
  mov    (%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint16_t getLocalPort(int fd) {
  struct sockaddr_in sin = {};
  socklen_t slen = sizeof(struct sockaddr);
  bzero(&sin, slen);
  getsockname(fd, (struct sockaddr *)&sin, &slen);
  return sin.sin_port;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<safe_fprintf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xd0(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L8
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L8:
  mov    -0xd0(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb4(%rbp),%edx
  mov    -0xd0(%rbp),%rcx
  mov    -0xc8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <safe_vfprintf@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  nop
  leave
  ret
```

assistant: ```c
void safe_fprintf(FILE *f, char *m, ...) {
  va_list ap;
  va_start(ap, m);
  safe_vfprintf(f, m, ap);
  va_end(ap);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<safe_fprintf>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <safe_vfprintf@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void safe_fprintf(FILE *f, char *m, ...) {
  va_list ap;
  va_start(ap, m);
  safe_vfprintf(f, m, ap);
  va_end(ap);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<safe_fprintf>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %r12d,%edi
  mov    %rsi,%rbp
  call   <va_start@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <safe_vfprintf@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void safe_fprintf(FILE *f, char *m, ...) {
  va_list ap;
  va_start(ap, m);
  safe_vfprintf(f, m, ap);
  va_end(ap);
}
```
user: What is the c source code of the assembly code below?

```asm
<os_alloc_assign_pid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x28(%rbp)
  mov    %rcx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1c(%rbp)
  lea    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <_pqos_cap_get@plt>
  mov    -0x10(%rbp),%rax
  lea    -0x1c(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <resctrl_alloc_get_grps_num@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L10
  mov    -0x14(%rbp),%eax
  jmp    L43
L10:
  mov    -0x1c(%rbp),%eax
  test   %eax,%eax
  jne    L56
  mov    $0x1,%eax
  jmp    L43
L56:
  mov    $0x0,%eax
  call   <resctrl_lock_exclusive@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L78
  mov    -0x14(%rbp),%eax
  jmp    L43
L78:
  mov    -0x1c(%rbp),%eax
  mov    -0x38(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <resctrl_alloc_get_unused_group@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L96
  movl   $0x0,-0x18(%rbp)
  jmp    L124
L22:
  mov    -0x18(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <resctrl_alloc_task_write@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L30
  addl   $0x1,-0x18(%rbp)
L124:
  mov    -0x18(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jb     L22
  jmp    L67
L96:
  nop
  jmp    L67
L30:
  nop
L67:
  mov    $0x0,%eax
  call   <resctrl_lock_release@plt>
  mov    -0x14(%rbp),%eax
L43:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
int os_alloc_assign_pid(const unsigned technology, const pid_t *task_array, const unsigned task_num, unsigned *class_id) {
  unsigned i, num_rctl_grps = 0;
  int ret;
  const struct pqos_cap *cap;
  ;
  ;
  ;
  (void)(technology);
  _pqos_cap_get(&cap, NULL);
  ret = resctrl_alloc_get_grps_num(cap, &num_rctl_grps);
  if (ret != 0)
    return ret;
  if (num_rctl_grps == 0)
    return 1;
  ret = resctrl_lock_exclusive();
  if (ret != 0)
    return ret;
  ret = resctrl_alloc_get_unused_group(num_rctl_grps, class_id);
  if (ret != 0)
    goto os_alloc_assign_pid_unlock;
  for (i = 0; i < task_num; i++) {
    ret = resctrl_alloc_task_write(*class_id, task_array[i]);
    if (ret != 0)
      goto os_alloc_assign_pid_unlock;
  }
os_alloc_assign_pid_unlock:
  resctrl_lock_release();
  return ret;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<os_alloc_assign_pid>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rsi,%r13
  mov    %edx,%r12d
  mov    %rcx,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0xc(%rsp)
  lea    0x10(%rsp),%rdi
  mov    $0x0,%esi
  call   <_pqos_cap_get@plt>
  lea    0xc(%rsp),%rsi
  mov    0x10(%rsp),%rdi
  call   <resctrl_alloc_get_grps_num@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L57
  mov    $0x1,%ebx
  cmpl   $0x0,0xc(%rsp)
  jne    L3
L57:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  mov    %ebx,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L3:
  mov    $0x0,%eax
  call   <resctrl_lock_exclusive@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L57
  mov    %rbp,%rsi
  mov    0xc(%rsp),%edi
  call   <resctrl_alloc_get_unused_group@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L16
  test   %r12d,%r12d
  je     L16
  mov    %r13,%r14
  lea    -0x1(%r12),%eax
  lea    0x4(%r13,%rax,4),%r12
L1:
  mov    (%r14),%esi
  mov    0x0(%rbp),%edi
  call   <resctrl_alloc_task_write@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L16
  add    $0x4,%r14
  cmp    %r12,%r14
  jne    L1
L16:
  mov    $0x0,%eax
  call   <resctrl_lock_release@plt>
  jmp    L57
L31:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int os_alloc_assign_pid(const unsigned technology, const pid_t *task_array, const unsigned task_num, unsigned *class_id) {
  unsigned i, num_rctl_grps = 0;
  int ret;
  const struct pqos_cap *cap;
  ;
  ;
  ;
  (void)(technology);
  _pqos_cap_get(&cap, NULL);
  ret = resctrl_alloc_get_grps_num(cap, &num_rctl_grps);
  if (ret != 0)
    return ret;
  if (num_rctl_grps == 0)
    return 1;
  ret = resctrl_lock_exclusive();
  if (ret != 0)
    return ret;
  ret = resctrl_alloc_get_unused_group(num_rctl_grps, class_id);
  if (ret != 0)
    goto os_alloc_assign_pid_unlock;
  for (i = 0; i < task_num; i++) {
    ret = resctrl_alloc_task_write(*class_id, task_array[i]);
    if (ret != 0)
      goto os_alloc_assign_pid_unlock;
  }
os_alloc_assign_pid_unlock:
  resctrl_lock_release();
  return ret;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<os_alloc_assign_pid>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  movl   $0x0,0xc(%rsp)
  call   <_pqos_cap_get@plt>
  mov    0x10(%rsp),%rdi
  lea    0xc(%rsp),%rsi
  call   <resctrl_alloc_get_grps_num@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L45
  mov    0xc(%rsp),%eax
  mov    $0x1,%r12d
  test   %eax,%eax
  jne    L8
L45:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L83
  add    $0x28,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L8:
  xor    %eax,%eax
  call   <resctrl_lock_exclusive@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L45
  mov    0xc(%rsp),%edi
  mov    %rbp,%rsi
  call   <resctrl_alloc_get_unused_group@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L65
  test   %r13d,%r13d
  je     L65
  lea    -0x1(%r13),%eax
  lea    0x4(%rbx,%rax,4),%r13
  jmp    L79
  nopl   0x0(%rax)
L5:
  add    $0x4,%rbx
  cmp    %r13,%rbx
  je     L65
L79:
  mov    (%rbx),%esi
  mov    0x0(%rbp),%edi
  call   <resctrl_alloc_task_write@plt>
  test   %eax,%eax
  je     L5
  mov    %eax,%r12d
L65:
  xor    %eax,%eax
  call   <resctrl_lock_release@plt>
  jmp    L45
L83:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int os_alloc_assign_pid(const unsigned technology, const pid_t *task_array, const unsigned task_num, unsigned *class_id) {
  unsigned i, num_rctl_grps = 0;
  int ret;
  const struct pqos_cap *cap;
  ;
  ;
  ;
  (void)(technology);
  _pqos_cap_get(&cap, NULL);
  ret = resctrl_alloc_get_grps_num(cap, &num_rctl_grps);
  if (ret != 0)
    return ret;
  if (num_rctl_grps == 0)
    return 1;
  ret = resctrl_lock_exclusive();
  if (ret != 0)
    return ret;
  ret = resctrl_alloc_get_unused_group(num_rctl_grps, class_id);
  if (ret != 0)
    goto os_alloc_assign_pid_unlock;
  for (i = 0; i < task_num; i++) {
    ret = resctrl_alloc_task_write(*class_id, task_array[i]);
    if (ret != 0)
      goto os_alloc_assign_pid_unlock;
  }
os_alloc_assign_pid_unlock:
  resctrl_lock_release();
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<os_alloc_assign_pid>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  movl   $0x0,0xc(%rsp)
  call   <_pqos_cap_get@plt>
  mov    0x10(%rsp),%rdi
  lea    0xc(%rsp),%rsi
  call   <resctrl_alloc_get_grps_num@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L90
  mov    0xc(%rsp),%eax
  mov    $0x1,%r12d
  test   %eax,%eax
  jne    L82
L90:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  add    $0x28,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L82:
  xor    %eax,%eax
  call   <resctrl_lock_exclusive@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L90
  mov    0xc(%rsp),%edi
  mov    %rbp,%rsi
  call   <resctrl_alloc_get_unused_group@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L23
  test   %r13d,%r13d
  je     L23
  lea    -0x1(%r13),%eax
  lea    0x4(%rbx,%rax,4),%r13
  jmp    L30
  nopl   0x0(%rax)
L32:
  add    $0x4,%rbx
  cmp    %r13,%rbx
  je     L23
L30:
  mov    (%rbx),%esi
  mov    0x0(%rbp),%edi
  call   <resctrl_alloc_task_write@plt>
  test   %eax,%eax
  je     L32
  mov    %eax,%r12d
L23:
  xor    %eax,%eax
  call   <resctrl_lock_release@plt>
  jmp    L90
L53:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int os_alloc_assign_pid(const unsigned technology, const pid_t *task_array, const unsigned task_num, unsigned *class_id) {
  unsigned i, num_rctl_grps = 0;
  int ret;
  const struct pqos_cap *cap;
  ;
  ;
  ;
  (void)(technology);
  _pqos_cap_get(&cap, NULL);
  ret = resctrl_alloc_get_grps_num(cap, &num_rctl_grps);
  if (ret != 0)
    return ret;
  if (num_rctl_grps == 0)
    return 1;
  ret = resctrl_lock_exclusive();
  if (ret != 0)
    return ret;
  ret = resctrl_alloc_get_unused_group(num_rctl_grps, class_id);
  if (ret != 0)
    goto os_alloc_assign_pid_unlock;
  for (i = 0; i < task_num; i++) {
    ret = resctrl_alloc_task_write(*class_id, task_array[i]);
    if (ret != 0)
      goto os_alloc_assign_pid_unlock;
  }
os_alloc_assign_pid_unlock:
  resctrl_lock_release();
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<align_malloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rcx
  lea    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <posix_memalign@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L9
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <strerror@plt>
  mov    %rax,%rdx
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <fprintf@plt>
  mov    $0x0,%eax
  call   <abort@plt>
L9:
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
void *align_malloc(size_t align, size_t size) {
  void *ptr;
  int ret = posix_memalign(&ptr, align, size);
  if (ret != 0) {
    fprintf(stderr, strerror(ret));
    abort();
  }
  return ptr;
}
```
user: Translate the below assembly code into C programming language:

```asm
<align_malloc>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rsi,%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rdi,%rsi
  mov    %rax,%rdi
  call   <posix_memalign@plt>
  test   %eax,%eax
  jne    L9
  mov    (%rsp),%rbx
L23:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  mov    %rbx,%rax
  add    $0x10,%rsp
  pop    %rbx
  ret
L9:
  mov    %eax,%edi
  call   <strerror@plt>
  mov    %rax,%rsi
  mov    D6(%rip),%rax
  mov    (%rax),%edi
  call   <fprintf@plt>
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L23
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *align_malloc(size_t align, size_t size) {
  void *ptr;
  int ret = posix_memalign(&ptr, align, size);
  if (ret != 0) {
    fprintf(stderr, strerror(ret));
    abort();
  }
  return ptr;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<align_malloc>:
  endbr64
  push   %r12
  mov    %rdi,%r8
  mov    %rsi,%rdx
  mov    %r8,%rsi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <posix_memalign@plt>
  test   %eax,%eax
  je     L6
  mov    %eax,%edi
  call   <strerror@plt>
  mov    %rax,%rsi
  mov    D2(%rip),%rax
  mov    (%rax),%edi
  call   <fprintf@plt>
  xor    %eax,%eax
  call   <abort@plt>
L23:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x10,%rsp
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L6:
  mov    (%rsp),%r12
  jmp    L23
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *align_malloc(size_t align, size_t size) {
  void *ptr;
  int ret = posix_memalign(&ptr, align, size);
  if (ret != 0) {
    fprintf(stderr, strerror(ret));
    abort();
  }
  return ptr;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<align_malloc>:
  endbr64
  push   %r12
  mov    %rdi,%r8
  mov    %rsi,%rdx
  mov    %r8,%rsi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <posix_memalign@plt>
  test   %eax,%eax
  je     L22
  mov    %eax,%edi
  call   <strerror@plt>
  mov    %rax,%rsi
  mov    D4(%rip),%rax
  mov    (%rax),%edi
  call   <fprintf@plt>
  xor    %eax,%eax
  call   <abort@plt>
L2:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x10,%rsp
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L22:
  mov    (%rsp),%r12
  jmp    L2
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *align_malloc(size_t align, size_t size) {
  void *ptr;
  int ret = posix_memalign(&ptr, align, size);
  if (ret != 0) {
    fprintf(stderr, strerror(ret));
    abort();
  }
  return ptr;
}
```
user: What would the C source look like for the assembly code provided here?

```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
  mov    $0x0,%eax
  call   <megaInit@plt>
  cmpl   $0x2,-0x34(%rbp)
  je     L42
  mov    $0x1,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
L42:
  movl   $0x0,-0x28(%rbp)
  jmp    L25
L37:
  mov    -0x28(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtoul@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x28(%rbp),%eax
  cltq
  mov    -0x20(%rbp),%rdx
  mov    %rdx,-0x18(%rbp,%rax,8)
  addl   $0x1,-0x28(%rbp)
L25:
  cmpl   $0x0,-0x28(%rbp)
  jle    L37
  lea    -0x10(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <RandomFunc@plt>
  mov    -0x10(%rbp),%rax
  movabs $0x2e27ef92e8a,%rdx
  cmp    %rdx,%rax
  jne    L16
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L16:
  movl   $0x0,-0x24(%rbp)
  jmp    L33
L52:
  mov    -0x24(%rbp),%eax
  cltq
  mov    -0x10(%rbp,%rax,8),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x24(%rbp)
L33:
  cmpl   $0x0,-0x24(%rbp)
  jle    L52
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L56
  call   <__stack_chk_fail@plt>
L56:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D19
type_guessed: string
name: extract_symbolic_data
arguments: tag: D16
type_guessed: string
name: extract_symbolic_data
arguments: tag: D3
type_guessed: string

tool: D19: Call this program with %i arguments


tool: D16: You win!

tool: D3: %lu


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long input[1];
  unsigned long output[1];
  int randomFuns_i5;
  unsigned long randomFuns_value6;
  int randomFuns_main_i7;
  {
    megaInit();
    if (argc != 2) {
      printf("Call this program with %i arguments\n", 1);
      exit(-1);
    } else {
    }
    randomFuns_i5 = 0;
    while (randomFuns_i5 < 1) {
      randomFuns_value6 = strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 3171816124042UL) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%lu\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```