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
<PropertyList>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D7(%rip),%rax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D7
datatype: string

tool: D7:     paved        = Paved (suitable for normal wheels)
    multilane    = Multiple lanes
    bridge       = Bridge
    tunnel       = Tunnel
    footroute    = A route marked for foot travel
    bicycleroute = A route marked for bicycle travel


assistant: ```c
const char *PropertyList(void) {
  return "    paved        = Paved (suitable for normal wheels)\n"
         "    multilane    = Multiple lanes\n"
         "    bridge       = Bridge\n"
         "    tunnel       = Tunnel\n"
         "    footroute    = A route marked for foot travel\n"
         "    bicycleroute = A route marked for bicycle travel\n";
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<PropertyList>:
  endbr64
  lea    D6(%rip),%rax
  ret
```

tool_calls: name: read_data
arguments: label: D6
type_infered: string

tool: D6:     paved        = Paved (suitable for normal wheels)
    multilane    = Multiple lanes
    bridge       = Bridge
    tunnel       = Tunnel
    footroute    = A route marked for foot travel
    bicycleroute = A route marked for bicycle travel


assistant: ```c
const char *PropertyList(void) {
  return "    paved        = Paved (suitable for normal wheels)\n"
         "    multilane    = Multiple lanes\n"
         "    bridge       = Bridge\n"
         "    tunnel       = Tunnel\n"
         "    footroute    = A route marked for foot travel\n"
         "    bicycleroute = A route marked for bicycle travel\n";
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<PropertyList>:
  endbr64
  lea    D5(%rip),%rax
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D5
datatype: string

tool: D5:     paved        = Paved (suitable for normal wheels)
    multilane    = Multiple lanes
    bridge       = Bridge
    tunnel       = Tunnel
    footroute    = A route marked for foot travel
    bicycleroute = A route marked for bicycle travel


assistant: ```c
const char *PropertyList(void) {
  return "    paved        = Paved (suitable for normal wheels)\n"
         "    multilane    = Multiple lanes\n"
         "    bridge       = Bridge\n"
         "    tunnel       = Tunnel\n"
         "    footroute    = A route marked for foot travel\n"
         "    bicycleroute = A route marked for bicycle travel\n";
}
```
user: What is the c source code of the assembly code below:

```asm
<PropertyList>:
  endbr64
  lea    D4(%rip),%rax
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D4
type_infered: string

tool: D4:     paved        = Paved (suitable for normal wheels)
    multilane    = Multiple lanes
    bridge       = Bridge
    tunnel       = Tunnel
    footroute    = A route marked for foot travel
    bicycleroute = A route marked for bicycle travel


assistant: ```c
const char *PropertyList(void) {
  return "    paved        = Paved (suitable for normal wheels)\n"
         "    multilane    = Multiple lanes\n"
         "    bridge       = Bridge\n"
         "    tunnel       = Tunnel\n"
         "    footroute    = A route marked for foot travel\n"
         "    bicycleroute = A route marked for bicycle travel\n";
}
```
user: Translate the below assembly code into C programming language:

```asm
<check_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x2,-0x8(%rbp)
  cmpl   $0x2,-0x14(%rbp)
  jne    L41
  mov    $0x1,%eax
  jmp    L9
L1:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x8(%rbp)
  mov    %edx,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L44
  mov    $0x0,%eax
  jmp    L9
L44:
  addl   $0x1,-0x8(%rbp)
L41:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L1
  mov    $0x1,%eax
L9:
  pop    %rbp
  ret
```

assistant: ```c
int check_prime(int num) {
  int j, i;
  i = 2;
  if (num == 2)
    return 1;
  while (i < num) {
    j = num % i;
    if (j == 0)
      return 0;
    i = i + 1;
  }
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<check_prime>:
  endbr64
  cmp    $0x2,%edi
  jle    L36
  test   $0x1,%dil
  je     L45
  mov    $0x2,%ecx
L5:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  je     L6
  mov    %edi,%eax
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %edx,%edx
  jne    L5
L40:
  ret
L6:
  mov    $0x1,%eax
  ret
L36:
  mov    $0x1,%eax
  ret
L45:
  mov    $0x0,%eax
  jmp    L40
```

assistant: ```c
int check_prime(int num) {
  int j, i;
  i = 2;
  if (num == 2)
    return 1;
  while (i < num) {
    j = num % i;
    if (j == 0)
      return 0;
    i = i + 1;
  }
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<check_prime>:
  endbr64
  cmp    $0x2,%edi
  jle    L58
  test   $0x1,%dil
  je     L49
  mov    $0x2,%ecx
  jmp    L30
  cs nopw 0x0(%rax,%rax,1)
L42:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L54
L30:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L42
L58:
  mov    $0x1,%edx
L54:
  mov    %edx,%eax
  ret
L49:
  xor    %edx,%edx
  jmp    L54
```

assistant: ```c
int check_prime(int num) {
  int j, i;
  i = 2;
  if (num == 2)
    return 1;
  while (i < num) {
    j = num % i;
    if (j == 0)
      return 0;
    i = i + 1;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<check_prime>:
  endbr64
  cmp    $0x2,%edi
  jle    L5
  test   $0x1,%dil
  je     L9
  mov    $0x2,%ecx
  jmp    L35
  cs nopw 0x0(%rax,%rax,1)
L55:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L14
L35:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L55
L5:
  mov    $0x1,%edx
L14:
  mov    %edx,%eax
  ret
L9:
  xor    %edx,%edx
  jmp    L14
```

assistant: ```c
int check_prime(int num) {
  int j, i;
  i = 2;
  if (num == 2)
    return 1;
  while (i < num) {
    j = num % i;
    if (j == 0)
      return 0;
    i = i + 1;
  }
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<id>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <getuid@plt>
  mov    %eax,%ebx
  mov    %eax,%esi
  lea    D71(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebx,%edi
  call   <getpwuid@plt>
  test   %rax,%rax
  je     L38
  mov    (%rax),%rsi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L38:
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  call   <geteuid@plt>
  mov    %eax,%ebx
  mov    %eax,%esi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebx,%edi
  call   <getpwuid@plt>
  test   %rax,%rax
  je     L32
  mov    (%rax),%rsi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L32:
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  call   <getgid@plt>
  mov    %eax,%ebx
  mov    %eax,%esi
  lea    D11(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebx,%edi
  call   <getgrgid@plt>
  test   %rax,%rax
  je     L14
  mov    (%rax),%rsi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L14:
  mov    $0x0,%eax
  call   <getegid@plt>
  mov    %eax,%ebx
  mov    %eax,%esi
  lea    D43(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebx,%edi
  call   <getgrgid@plt>
  test   %rax,%rax
  je     L5
  mov    (%rax),%rsi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L5:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%eax
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: label: D71
type: string
name: read_data
arguments: label: D3
type: string
name: read_data
arguments: label: D4
type: string
name: read_data
arguments: label: D11
type: string
name: read_data
arguments: label: D43
type: string

tool: D71: uid:%i

tool: D3: (%s)

tool: D4: euid:%i

tool: D11: gid:%i

tool: D43: egid:%i

assistant: ```c
int id(char **command, struct tm start) {
  struct passwd *pwd = NULL;
  struct group *gr = NULL;
  uid_t uid = getuid();
  printf("uid:%i", uid);
  pwd = getpwuid(uid);
  if (pwd)
    printf("(%s)", pwd->pw_name);
  printf(" ");
  uid_t euid = geteuid();
  printf("euid:%i", euid);
  pwd = getpwuid(euid);
  if (pwd)
    printf("(%s)", pwd->pw_name);
  printf(" ");
  gid_t gid = getgid();
  printf("gid:%i", gid);
  gr = getgrgid(gid);
  if (gr)
    printf("(%s)", gr->gr_name);
  gid_t egid = getegid();
  printf("egid:%i", egid);
  gr = getgrgid(egid);
  if (gr)
    printf("(%s)", gr->gr_name);
  printf("\n");
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<id>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <getuid@plt>
  lea    D33(%rip),%rdi
  mov    %eax,%ebp
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%edi
  call   <getpwuid@plt>
  test   %rax,%rax
  je     L39
  mov    (%rax),%rsi
  lea    D65(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L39:
  mov    $0x20,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  call   <geteuid@plt>
  lea    D54(%rip),%rdi
  mov    %eax,%ebp
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%edi
  call   <getpwuid@plt>
  test   %rax,%rax
  je     L8
  mov    (%rax),%rsi
  lea    D65(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L8:
  mov    $0x20,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  call   <getgid@plt>
  lea    D11(%rip),%rdi
  mov    %eax,%ebp
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%edi
  call   <getgrgid@plt>
  test   %rax,%rax
  je     L34
  mov    (%rax),%rsi
  lea    D65(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L34:
  xor    %eax,%eax
  call   <getegid@plt>
  lea    D40(%rip),%rdi
  mov    %eax,%ebp
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%edi
  call   <getgrgid@plt>
  test   %rax,%rax
  je     L32
  mov    (%rax),%rsi
  lea    D65(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L32:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D33
type: string
name: disassemble_data
arguments: data_tag: D65
type: string
name: disassemble_data
arguments: data_tag: D54
type: string
name: disassemble_data
arguments: data_tag: D11
type: string
name: disassemble_data
arguments: data_tag: D40
type: string

tool: D33: uid:%i

tool: D65: (%s)

tool: D54: euid:%i

tool: D11: gid:%i

tool: D40: egid:%i

assistant: ```c
int id(char **command, struct tm start) {
  struct passwd *pwd = NULL;
  struct group *gr = NULL;
  uid_t uid = getuid();
  printf("uid:%i", uid);
  pwd = getpwuid(uid);
  if (pwd)
    printf("(%s)", pwd->pw_name);
  printf(" ");
  uid_t euid = geteuid();
  printf("euid:%i", euid);
  pwd = getpwuid(euid);
  if (pwd)
    printf("(%s)", pwd->pw_name);
  printf(" ");
  gid_t gid = getgid();
  printf("gid:%i", gid);
  gr = getgrgid(gid);
  if (gr)
    printf("(%s)", gr->gr_name);
  gid_t egid = getegid();
  printf("egid:%i", egid);
  gr = getgrgid(egid);
  if (gr)
    printf("(%s)", gr->gr_name);
  printf("\n");
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<id>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <getuid@plt>
  lea    D13(%rip),%rdi
  mov    %eax,%ebp
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%edi
  call   <getpwuid@plt>
  test   %rax,%rax
  je     L5
  mov    (%rax),%rsi
  lea    D77(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L5:
  mov    $0x20,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  call   <geteuid@plt>
  lea    D65(%rip),%rdi
  mov    %eax,%ebp
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%edi
  call   <getpwuid@plt>
  test   %rax,%rax
  je     L32
  mov    (%rax),%rsi
  lea    D77(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L32:
  mov    $0x20,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  call   <getgid@plt>
  lea    D5(%rip),%rdi
  mov    %eax,%ebp
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%edi
  call   <getgrgid@plt>
  test   %rax,%rax
  je     L34
  mov    (%rax),%rsi
  lea    D77(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L34:
  xor    %eax,%eax
  call   <getegid@plt>
  lea    D48(%rip),%rdi
  mov    %eax,%ebp
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%edi
  call   <getgrgid@plt>
  test   %rax,%rax
  je     L35
  mov    (%rax),%rsi
  lea    D77(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L35:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D13
data_type: string
name: unpack_data
arguments: tag: D77
data_type: string
name: unpack_data
arguments: tag: D65
data_type: string
name: unpack_data
arguments: tag: D5
data_type: string
name: unpack_data
arguments: tag: D48
data_type: string

tool: D13: uid:%i

tool: D77: (%s)

tool: D65: euid:%i

tool: D5: gid:%i

tool: D48: egid:%i

assistant: ```c
int id(char **command, struct tm start) {
  struct passwd *pwd = NULL;
  struct group *gr = NULL;
  uid_t uid = getuid();
  printf("uid:%i", uid);
  pwd = getpwuid(uid);
  if (pwd)
    printf("(%s)", pwd->pw_name);
  printf(" ");
  uid_t euid = geteuid();
  printf("euid:%i", euid);
  pwd = getpwuid(euid);
  if (pwd)
    printf("(%s)", pwd->pw_name);
  printf(" ");
  gid_t gid = getgid();
  printf("gid:%i", gid);
  gr = getgrgid(gid);
  if (gr)
    printf("(%s)", gr->gr_name);
  gid_t egid = getegid();
  printf("egid:%i", egid);
  gr = getgrgid(egid);
  if (gr)
    printf("(%s)", gr->gr_name);
  printf("\n");
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<bwb_ON_ERROR_RETURN_NEXT>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    $0xffffffff,%edi
  call   <SetOnError@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <bwb_zline@plt>
  leave
  ret
```

assistant: ```c
struct bwb_line *bwb_ON_ERROR_RETURN_NEXT(struct bwb_line *l) {
  SetOnError(-1);
  return bwb_zline(l);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<bwb_ON_ERROR_RETURN_NEXT>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0xffffffff,%edi
  call   <SetOnError@plt>
  mov    %rbx,%rdi
  call   <bwb_zline@plt>
  pop    %rbx
  ret
```

assistant: ```c
struct bwb_line *bwb_ON_ERROR_RETURN_NEXT(struct bwb_line *l) {
  SetOnError(-1);
  return bwb_zline(l);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<bwb_ON_ERROR_RETURN_NEXT>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  mov    $0xffffffff,%edi
  call   <SetOnError@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <bwb_zline@plt>
```

assistant: ```c
struct bwb_line *bwb_ON_ERROR_RETURN_NEXT(struct bwb_line *l) {
  SetOnError(-1);
  return bwb_zline(l);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<bwb_ON_ERROR_RETURN_NEXT>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  mov    $0xffffffff,%edi
  call   <SetOnError@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <bwb_zline@plt>
```

assistant: ```c
struct bwb_line *bwb_ON_ERROR_RETURN_NEXT(struct bwb_line *l) {
  SetOnError(-1);
  return bwb_zline(l);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<xrop3_79_0_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x24(%rbp)
  mov    %al,-0x28(%rbp)
  movl   $0xff,-0x18(%rbp)
  movzbl -0x24(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x24(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x14(%rbp)
  movzbl -0x28(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x28(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  not    %eax
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  or     -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x10(%rbp),%eax
  and    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  notl   -0x4(%rbp)
  mov    -0x10(%rbp),%eax
  not    %eax
  or     -0x14(%rbp),%eax
  and    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  not    %eax
  and    -0x10(%rbp),%eax
  and    -0x18(%rbp),%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned xrop3_79_0_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = ~S;
  stk2 = D | stk3;
  stk1 = T & stk2;
  stk1 = ~stk1;
  return (stk1 & S) | (stk1 & (~T)) | (T & (~S) & D);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<xrop3_79_0_1>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  movzbl %sil,%eax
  mov    %eax,%esi
  shl    $0x8,%esi
  or     %eax,%esi
  mov    %edi,%ecx
  not    %ecx
  mov    %ecx,%edx
  or     $0xff,%dl
  and    %esi,%edx
  mov    %esi,%eax
  not    %eax
  or     %edi,%eax
  not    %edx
  and    %edx,%eax
  and    %ecx,%esi
  movzbl %sil,%esi
  or     %esi,%eax
  ret
```

assistant: ```c
unsigned xrop3_79_0_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = ~S;
  stk2 = D | stk3;
  stk1 = T & stk2;
  stk1 = ~stk1;
  return (stk1 & S) | (stk1 & (~T)) | (T & (~S) & D);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<xrop3_79_0_1>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  movzbl %sil,%eax
  mov    %eax,%esi
  mov    %edi,%ecx
  not    %ecx
  shl    $0x8,%esi
  or     %eax,%esi
  mov    %ecx,%edx
  or     $0xff,%dl
  mov    %esi,%eax
  and    %esi,%edx
  not    %eax
  and    %ecx,%esi
  or     %edi,%eax
  not    %edx
  movzbl %sil,%esi
  and    %edx,%eax
  or     %esi,%eax
  ret
```

assistant: ```c
unsigned xrop3_79_0_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = ~S;
  stk2 = D | stk3;
  stk1 = T & stk2;
  stk1 = ~stk1;
  return (stk1 & S) | (stk1 & (~T)) | (T & (~S) & D);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xrop3_79_0_1>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  movzbl %sil,%eax
  mov    %eax,%esi
  mov    %edi,%ecx
  not    %ecx
  shl    $0x8,%esi
  or     %eax,%esi
  mov    %ecx,%edx
  or     $0xff,%dl
  mov    %esi,%eax
  and    %esi,%edx
  not    %eax
  and    %ecx,%esi
  or     %edi,%eax
  not    %edx
  movzbl %sil,%esi
  and    %edx,%eax
  or     %esi,%eax
  ret
```

assistant: ```c
unsigned xrop3_79_0_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = ~S;
  stk2 = D | stk3;
  stk1 = T & stk2;
  stk1 = ~stk1;
  return (stk1 & S) | (stk1 & (~T)) | (T & (~S) & D);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<report_init>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <gg_loginit@plt>
  test   %rax,%rax
  jns    L17
  mov    $0xffffffff,%eax
  jmp    L8
L17:
  mov    $0x0,%eax
L8:
  pop    %rbp
  ret
```

assistant: ```c
int report_init(void) {
  if (gg_loginit() < 0)
    return -1;
  else
    return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<report_init>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <gg_loginit@plt>
  sar    $0x3f,%rax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int report_init(void) {
  if (gg_loginit() < 0)
    return -1;
  else
    return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<report_init>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <gg_loginit@plt>
  add    $0x8,%rsp
  sar    $0x3f,%rax
  ret
```

assistant: ```c
int report_init(void) {
  if (gg_loginit() < 0)
    return -1;
  else
    return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<report_init>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <gg_loginit@plt>
  add    $0x8,%rsp
  sar    $0x3f,%rax
  ret
```

assistant: ```c
int report_init(void) {
  if (gg_loginit() < 0)
    return -1;
  else
    return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x1,-0xc(%rbp)
  jmp    L4
L18:
  cmpl   $0x0,-0xc(%rbp)
  je     L38
  movl   $0x0,-0xc(%rbp)
  jmp    L3
L38:
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L3:
  mov    $0x0,%eax
  call   <test_case@plt>
L4:
  mov    -0x10(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  test   %eax,%eax
  jne    L18
  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: D8
type: string
name: extract_symbolic_data
arguments: data_tag: D5
type: string

tool: D8: %d

tool: D5: 


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  int first = 1;
  while (n--) {
    if (first)
      first = 0;
    else
      printf("\n");
    test_case();
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  lea    D0(%rip),%rbx
  test   %eax,%eax
  jne    L28
  jmp    L18
L38:
  mov    %rbx,%rdi
  call   <printf@plt>
L28:
  mov    $0x0,%eax
  call   <test_case@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  jne    L38
L18:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D16
datatype: string
name: unpack_data
arguments: data_label: D0
datatype: string

tool: D16: %d

tool: D0: 


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  int first = 1;
  while (n--) {
    if (first)
      first = 0;
    else
      printf("\n");
    test_case();
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbx
  lea    D10(%rip),%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  je     L37
  lea    D17(%rip),%rbx
  jmp    L20
  nopw   0x0(%rax,%rax,1)
L26:
  mov    %rbx,%rdi
  call   <printf@plt>
L20:
  xor    %eax,%eax
  call   <test_case@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  jne    L26
L37:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D10
type_guessed: string
name: unpack_data
arguments: label: D17
type_guessed: string

tool: D10: %d

tool: D17: 


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  int first = 1;
  while (n--) {
    if (first)
      first = 0;
    else
      printf("\n");
    test_case();
  }
  return 0;
}
```