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
<PRStringConstructorInt>:
  endbr64
  sub    $0x28,%rsp
  mov    %edi,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  js     L90
  jne    L42
  movb   $0x30,(%rsp)
  mov    $0x1,%esi
  jmp    L2
L90:
  neg    %edx
  mov    $0x1,%r10d
L21:
  mov    $0x1,%ecx
  lea    -0x1(%rsp),%r9
L9:
  mov    %ecx,%esi
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  lea    (%rax,%rax,4),%r8d
  add    %r8d,%r8d
  mov    %edx,%edi
  sub    %r8d,%edi
  add    $0x30,%edi
  mov    %dil,(%r9,%rcx,1)
  mov    %edx,%edi
  mov    %eax,%edx
  add    $0x1,%rcx
  cmp    $0x9,%edi
  jg     L9
  test   %r10d,%r10d
  je     L109
  movslq %esi,%rax
  movb   $0x2d,(%rsp,%rax,1)
  lea    0x1(%rsi),%esi
L109:
  lea    -0x1(%rsi),%ecx
  test   %ecx,%ecx
  jle    L2
  mov    %rsp,%rdx
  movslq %esi,%rax
  add    %rdx,%rax
L84:
  movzbl (%rdx),%edi
  movzbl -0x1(%rax),%r8d
  mov    %r8b,(%rdx)
  mov    %dil,-0x1(%rax)
  mov    %ecx,%r8d
  sub    $0x1,%ecx
  add    $0x1,%rdx
  sub    $0x1,%rax
  mov    %esi,%edi
  sub    %r8d,%edi
  cmp    %edi,%ecx
  jg     L84
L2:
  movslq %esi,%rsi
  movb   $0x0,(%rsp,%rsi,1)
  mov    %rsp,%rdi
  call   <PRStringConstructorCStr@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L75
  add    $0x28,%rsp
  ret
L42:
  jle    L93
  mov    $0x0,%r10d
  jmp    L21
L93:
  mov    $0x0,%esi
  jmp    L2
L75:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
struct String *PRStringConstructorInt(int num) {
  char str[20], tmp = 0;
  int len = 0, isNeg = 0, i = 0, j = 0;
  ;
  if (num < 0) {
    isNeg = 1;
    num = -num;
  }
  if (num == 0) {
    str[len++] = '0';
  } else {
    while (num > 0) {
      str[len++] = (char)(num % 10) + '0';
      num /= 10;
    }
  }
  if (isNeg)
    str[len++] = '-';
  for (i = 0, j = len - 1; i < j; ++i, --j) {
    tmp = str[i];
    str[i] = str[j];
    str[j] = tmp;
  }
  str[len] = 0;
  ;
  return PRStringConstructorCStr((char *)str);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<PRStringConstructorInt>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  js     L36
  je     L24
  xor    %ebp,%ebp
L12:
  mov    %rsp,%r9
  xor    %r8d,%r8d
  mov    $0xcccccccd,%ebx
  mov    %r9,%rcx
  mov    %r9,%r11
  nopl   0x0(%rax,%rax,1)
L94:
  mov    %edi,%eax
  mov    %r8d,%r10d
  add    $0x1,%r11
  add    $0x1,%r8d
  imul   %rbx,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  lea    0x30(%rdi),%esi
  mov    %eax,%edi
  mov    %sil,-0x1(%r11)
  test   %eax,%eax
  jne    L94
  test   %ebp,%ebp
  je     L39
  movslq %r8d,%rdx
  lea    0x2(%r10),%eax
  mov    $0x2d,%esi
  mov    %r8d,%r10d
  movb   $0x2d,(%rsp,%rdx,1)
  mov    %eax,%r8d
L31:
  movslq %r10d,%rax
  mov    %r10d,%edx
  add    %r9,%rax
  jmp    L56
  nopw   0x0(%rax,%rax,1)
L89:
  movzbl (%rax),%esi
L56:
  movzbl (%rcx),%edi
  sub    $0x1,%edx
  mov    %sil,(%rcx)
  mov    %r10d,%esi
  sub    %edx,%esi
  add    $0x1,%rcx
  sub    $0x1,%rax
  mov    %dil,0x1(%rax)
  cmp    %edx,%esi
  jl     L89
L109:
  movslq %r8d,%r8
  mov    %r9,%rdi
  movb   $0x0,(%rsp,%r8,1)
  call   <PRStringConstructorCStr@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L43
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L24:
  movb   $0x30,(%rsp)
  mov    $0x1,%r8d
  mov    %rsp,%r9
  jmp    L109
  nop
L36:
  neg    %edi
  mov    $0x1,%ebp
  jmp    L12
  nopl   0x0(%rax)
L39:
  test   %r10d,%r10d
  jne    L31
  jmp    L109
L43:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
struct String *PRStringConstructorInt(int num) {
  char str[20], tmp = 0;
  int len = 0, isNeg = 0, i = 0, j = 0;
  ;
  if (num < 0) {
    isNeg = 1;
    num = -num;
  }
  if (num == 0) {
    str[len++] = '0';
  } else {
    while (num > 0) {
      str[len++] = (char)(num % 10) + '0';
      num /= 10;
    }
  }
  if (isNeg)
    str[len++] = '-';
  for (i = 0, j = len - 1; i < j; ++i, --j) {
    tmp = str[i];
    str[i] = str[j];
    str[j] = tmp;
  }
  str[len] = 0;
  ;
  return PRStringConstructorCStr((char *)str);
}
```
user: Translate the below assembly code into C programming language:

```asm
<copy_quaddobl_Laurent_start_system_to_container>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%r8d
  mov    %rax,%rsi
  mov    $0x311,%edi
  call   <_ada_use_c2phc@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  leave
  ret
```

assistant: ```c
int copy_quaddobl_Laurent_start_system_to_container(void) {
  int *a, *b, fail;
  double *c;
  fail = _ada_use_c2phc(785, a, b, c, 0);
  return fail;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<copy_quaddobl_Laurent_start_system_to_container>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x311,%edi
  call   <_ada_use_c2phc@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int copy_quaddobl_Laurent_start_system_to_container(void) {
  int *a, *b, fail;
  double *c;
  fail = _ada_use_c2phc(785, a, b, c, 0);
  return fail;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<copy_quaddobl_Laurent_start_system_to_container>:
  endbr64
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0x311,%edi
  jmp    1050 <_ada_use_c2phc@plt>
```

assistant: ```c
int copy_quaddobl_Laurent_start_system_to_container(void) {
  int *a, *b, fail;
  double *c;
  fail = _ada_use_c2phc(785, a, b, c, 0);
  return fail;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<copy_quaddobl_Laurent_start_system_to_container>:
  endbr64
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0x311,%edi
  jmp    1050 <_ada_use_c2phc@plt>
```

assistant: ```c
int copy_quaddobl_Laurent_start_system_to_container(void) {
  int *a, *b, fail;
  double *c;
  fail = _ada_use_c2phc(785, a, b, c, 0);
  return fail;
}
```
user: What is the c source code of the assembly code below?

```asm
<get_time_resource>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <get_string_resource@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L13
  mov    $0x0,%eax
  jmp    L17
L13:
  mov    D7(%rip),%rax
  mov    (%rax),%edx
  mov    -0x2c(%rbp),%ecx
  mov    -0x8(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <parse_time@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0xc(%rbp),%eax
  mov    $0x0,%edx
  test   %eax,%eax
  cmovs  %edx,%eax
L17:
  leave
  ret
```

assistant: ```c
unsigned int get_time_resource(Display *dpy, char *res_name, char *res_class, Bool sec_p) {
  int val;
  char *s = get_string_resource(dpy, res_name, res_class);
  if (!s)
    return 0;
  val = parse_time(s, sec_p, False);
  free(s);
  return (val < 0 ? 0 : val);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<get_time_resource>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %ecx,%ebx
  call   <get_string_resource@plt>
  mov    %rax,%rbp
  mov    $0x0,%eax
  test   %rbp,%rbp
  je     L1
  mov    D3(%rip),%rax
  mov    (%rax),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <parse_time@plt>
  mov    %eax,%ebx
  mov    %rbp,%rdi
  call   <free@plt>
  test   %ebx,%ebx
  mov    $0x0,%eax
  cmovns %ebx,%eax
L1:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
unsigned int get_time_resource(Display *dpy, char *res_name, char *res_class, Bool sec_p) {
  int val;
  char *s = get_string_resource(dpy, res_name, res_class);
  if (!s)
    return 0;
  val = parse_time(s, sec_p, False);
  free(s);
  return (val < 0 ? 0 : val);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_time_resource>:
  endbr64
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  push   %rbx
  call   <get_string_resource@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  test   %rbp,%rbp
  je     L7
  mov    D0(%rip),%rax
  mov    %r12d,%esi
  mov    %rbp,%rdi
  mov    (%rax),%edx
  call   <parse_time@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <free@plt>
  xor    %eax,%eax
  test   %ebx,%ebx
  cmovns %ebx,%eax
L7:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
unsigned int get_time_resource(Display *dpy, char *res_name, char *res_class, Bool sec_p) {
  int val;
  char *s = get_string_resource(dpy, res_name, res_class);
  if (!s)
    return 0;
  val = parse_time(s, sec_p, False);
  free(s);
  return (val < 0 ? 0 : val);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<get_time_resource>:
  endbr64
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  push   %rbx
  call   <get_string_resource@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  test   %rbp,%rbp
  je     L8
  mov    D7(%rip),%rax
  mov    %r12d,%esi
  mov    %rbp,%rdi
  mov    (%rax),%edx
  call   <parse_time@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <free@plt>
  xor    %eax,%eax
  test   %ebx,%ebx
  cmovns %ebx,%eax
L8:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
unsigned int get_time_resource(Display *dpy, char *res_name, char *res_class, Bool sec_p) {
  int val;
  char *s = get_string_resource(dpy, res_name, res_class);
  if (!s)
    return 0;
  val = parse_time(s, sec_p, False);
  free(s);
  return (val < 0 ? 0 : val);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<printHexBuf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L6
L11:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L6:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L11
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
type: string

tool: D9: %02x 

assistant: ```c
void printHexBuf(unsigned char *buf, int n) {
  int i;
  for (i = 0; i < n; i++) {
    printf("%02x ", buf[i]);
  }
  printf("\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<printHexBuf>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L19
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%r12
  lea    D8(%rip),%rbp
L17:
  movzbl (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L17
L19:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D8
type_infered: string

tool: D8: %02x 

assistant: ```c
void printHexBuf(unsigned char *buf, int n) {
  int i;
  for (i = 0; i < n; i++) {
    printf("%02x ", buf[i]);
  }
  printf("\n");
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<printHexBuf>:
  endbr64
  test   %esi,%esi
  jle    L9
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x1(%rdi,%rax,1),%r12
  push   %rbp
  lea    D3(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  xchg   %ax,%ax
L10:
  movzbl (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L10
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
  nopl   0x0(%rax)
L9:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

tool_calls: name: read_data
arguments: tag: D3
data_type: string

tool: D3: %02x 

assistant: ```c
void printHexBuf(unsigned char *buf, int n) {
  int i;
  for (i = 0; i < n; i++) {
    printf("%02x ", buf[i]);
  }
  printf("\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<printHexBuf>:
  endbr64
  test   %esi,%esi
  jle    L13
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x1(%rdi,%rax,1),%r12
  push   %rbp
  lea    D7(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  xchg   %ax,%ax
L7:
  movzbl (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L7
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
  nopl   0x0(%rax)
L13:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D7
type: string

tool: D7: %02x 

assistant: ```c
void printHexBuf(unsigned char *buf, int n) {
  int i;
  for (i = 0; i < n; i++) {
    printf("%02x ", buf[i]);
  }
  printf("\n");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<dexists>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    D24(%rip),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L8:
  cmp    %rdx,%rsp
  je     L79
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L8
L79:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L74
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L74:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <realpath@plt>
  lea    -0x2c(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <lstat@plt>
  test   %rax,%rax
  je     L11
  mov    D30(%rip),%rax
  mov    (%rax),%rdx
  mov    D12(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L66
  mov    D30(%rip),%rax
  mov    (%rax),%rdx
  mov    D47(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L11
L66:
  mov    $0x0,%eax
  jmp    L37
L11:
  mov    -0x2c(%rbp),%eax
  mov    %eax,%edi
  call   <S_ISDIR@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
L37:
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L32
  call   <__stack_chk_fail@plt>
L32:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int dexists(const char *path) {
  char buf[PATH_MAX];
  struct stat info;
  realpath(path, &buf[0]);
  if (lstat(buf, &info) != 0) {
    if (errno == ENOENT || errno == EACCES)
      return 0;
  }
  return S_ISDIR(info.st_mode) ? 1 : 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dexists>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D1(%rip),%rax
  movslq (%rax),%rax
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L31:
  cmp    %rdx,%rsp
  je     L39
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L31
L39:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L11
  orq    $0x0,-0x8(%rsp,%rax,1)
L11:
  mov    %rsp,%rsi
  call   <realpath@plt>
  lea    -0xc(%rbp),%rsi
  mov    %rsp,%rdi
  call   <lstat@plt>
  test   %rax,%rax
  je     L49
  mov    D39(%rip),%rax
  mov    (%rax),%rdx
  mov    $0x0,%eax
  mov    D2(%rip),%rcx
  cmp    (%rcx),%rdx
  je     L57
  mov    D15(%rip),%rcx
  cmp    (%rcx),%rdx
  je     L57
L49:
  mov    -0xc(%rbp),%edi
  call   <S_ISDIR@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
L57:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  leave
  ret
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dexists(const char *path) {
  char buf[PATH_MAX];
  struct stat info;
  realpath(path, &buf[0]);
  if (lstat(buf, &info) != 0) {
    if (errno == ENOENT || errno == EACCES)
      return 0;
  }
  return S_ISDIR(info.st_mode) ? 1 : 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dexists>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  mov    D19(%rip),%rax
  mov    %rsp,%rcx
  movslq (%rax),%rax
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L57
L63:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L63
L57:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L51
L66:
  mov    %rsp,%rsi
  call   <realpath@plt>
  lea    -0xc(%rbp),%rsi
  mov    %rsp,%rdi
  call   <lstat@plt>
  test   %rax,%rax
  je     L32
  mov    D29(%rip),%rax
  mov    D27(%rip),%rcx
  mov    (%rax),%rdx
  xor    %eax,%eax
  cmp    (%rcx),%rdx
  je     L18
  mov    D0(%rip),%rcx
  cmp    (%rcx),%rdx
  je     L18
L32:
  mov    -0xc(%rbp),%edi
  call   <S_ISDIR@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
L18:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L54
  leave
  ret
  nopl   0x0(%rax)
L51:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L66
L54:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dexists(const char *path) {
  char buf[PATH_MAX];
  struct stat info;
  realpath(path, &buf[0]);
  if (lstat(buf, &info) != 0) {
    if (errno == ENOENT || errno == EACCES)
      return 0;
  }
  return S_ISDIR(info.st_mode) ? 1 : 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<dexists>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  mov    D31(%rip),%rax
  mov    %rsp,%rcx
  movslq (%rax),%rax
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L34
L49:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L49
L34:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L19
L71:
  mov    %rsp,%rsi
  call   <realpath@plt>
  lea    -0xc(%rbp),%rsi
  mov    %rsp,%rdi
  call   <lstat@plt>
  test   %rax,%rax
  je     L7
  mov    D17(%rip),%rax
  mov    D39(%rip),%rcx
  mov    (%rax),%rdx
  xor    %eax,%eax
  cmp    (%rcx),%rdx
  je     L47
  mov    D29(%rip),%rcx
  cmp    (%rcx),%rdx
  je     L47
L7:
  mov    -0xc(%rbp),%edi
  call   <S_ISDIR@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
L47:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L64
  leave
  ret
  nopl   0x0(%rax)
L19:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L71
L64:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dexists(const char *path) {
  char buf[PATH_MAX];
  struct stat info;
  realpath(path, &buf[0]);
  if (lstat(buf, &info) != 0) {
    if (errno == ENOENT || errno == EACCES)
      return 0;
  }
  return S_ISDIR(info.st_mode) ? 1 : 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<part2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0x18(%rbp)
  mov    $0x1,%esi
  mov    $0x9,%edi
  call   <calloc@plt>
  mov    %rax,-0x10(%rbp)
  jmp    L29
L79:
  mov    -0x18(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x18(%rbp)
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <hash@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <interesting@plt>
  test   %rax,%rax
  je     L29
  mov    -0x8(%rbp),%rax
  movzbl 0x5(%rax),%eax
  mov    %al,-0x19(%rbp)
  cmpb   $0x2f,-0x19(%rbp)
  jle    L78
  cmpb   $0x37,-0x19(%rbp)
  jle    L25
L78:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  jmp    L29
L25:
  movsbl -0x19(%rbp),%eax
  sub    $0x30,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L12
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  jmp    L29
L12:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  movzbl 0x6(%rax),%eax
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L29:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    $0x7,%eax
  jle    L79
  mov    -0x10(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *part2(char *doorid) {
  unsigned int n = 0;
  char *password = calloc(8 + 1, sizeof(char));
  while (strlen(password) < 8) {
    char *s = hash(doorid, n++);
    if (interesting(s)) {
      char pos = s[5];
      if (pos < '0' || pos > '7') {
        free(s);
        continue;
      }
      int idx = (int)(pos - '0');
      if (password[idx] != '\0') {
        free(s);
        continue;
      }
      password[idx] = s[6];
      free(s);
    }
  }
  return password;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<part2>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    $0x1,%esi
  mov    $0x9,%edi
  call   <calloc@plt>
  mov    %rax,%r12
  mov    $0x0,%ebp
  jmp    L23
L8:
  mov    %rbx,%rdi
  call   <free@plt>
  jmp    L35
L60:
  mov    %rbx,%rdi
  call   <free@plt>
L35:
  add    $0x1,%ebp
L23:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    $0x7,%eax
  jg     L13
  mov    %ebp,%esi
  mov    %r13,%rdi
  call   <hash@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <interesting@plt>
  test   %rax,%rax
  je     L35
  movzbl 0x5(%rbx),%eax
  lea    -0x30(%rax),%edx
  cmp    $0x7,%dl
  ja     L8
  movsbl %al,%eax
  sub    $0x30,%eax
  cltq
  add    %r12,%rax
  cmpb   $0x0,(%rax)
  jne    L60
  movzbl 0x6(%rbx),%edx
  mov    %dl,(%rax)
  mov    %rbx,%rdi
  call   <free@plt>
  jmp    L35
L13:
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *part2(char *doorid) {
  unsigned int n = 0;
  char *password = calloc(8 + 1, sizeof(char));
  while (strlen(password) < 8) {
    char *s = hash(doorid, n++);
    if (interesting(s)) {
      char pos = s[5];
      if (pos < '0' || pos > '7') {
        free(s);
        continue;
      }
      int idx = (int)(pos - '0');
      if (password[idx] != '\0') {
        free(s);
        continue;
      }
      password[idx] = s[6];
      free(s);
    }
  }
  return password;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<part2>:
  endbr64
  push   %r13
  mov    $0x1,%esi
  mov    %rdi,%r13
  mov    $0x9,%edi
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <calloc@plt>
  mov    %rax,%r12
  jmp    L27
  nopw   0x0(%rax,%rax,1)
L20:
  mov    %r13,%rdi
  mov    %ebx,%esi
  call   <hash@plt>
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <interesting@plt>
  test   %rax,%rax
  je     L14
  movzbl 0x5(%rbp),%eax
  sub    $0x30,%eax
  cmp    $0x7,%al
  ja     L33
  movsbq %al,%rax
  add    %r12,%rax
  cmpb   $0x0,(%rax)
  jne    L33
  movzbl 0x6(%rbp),%edx
  mov    %dl,(%rax)
L33:
  mov    %rbp,%rdi
  call   <free@plt>
L14:
  add    $0x1,%ebx
L27:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    $0x7,%eax
  jle    L20
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *part2(char *doorid) {
  unsigned int n = 0;
  char *password = calloc(8 + 1, sizeof(char));
  while (strlen(password) < 8) {
    char *s = hash(doorid, n++);
    if (interesting(s)) {
      char pos = s[5];
      if (pos < '0' || pos > '7') {
        free(s);
        continue;
      }
      int idx = (int)(pos - '0');
      if (password[idx] != '\0') {
        free(s);
        continue;
      }
      password[idx] = s[6];
      free(s);
    }
  }
  return password;
}
```