ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<sb_getversion_s>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    $0x102,%eax
  sub    -0x4(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  js     L10
  mov    -0x4(%rbp),%eax
  cmp    $0x10,%eax
  ja     L10
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  lea    D18(%rip),%rax
  mov    (%rdx,%rax,1),%rax
  jmp    L9
L10:
  lea    D10(%rip),%rax
L9:
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D10
type_guessed: string

tool: D10: unknown

assistant: ```c
char *sb_getversion_s(int n) {
  static char *version_s[] = {"Justice", "CTF", "Assassin", "Summer", "Spring", "Gui", "Water", "Normalmap", "Sp", "Occlusion", "Shader", "Physics", "Mp", "", "Agc", "Quakecon", "Independence"};
  n = 258 - n;
  if (n >= 0 && (size_t)n < sizeof(version_s) / sizeof(version_s[0])) {
    return version_s[n];
  }
  return "unknown";
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sb_getversion_s>:
  endbr64
  mov    $0x102,%eax
  sub    %edi,%eax
  lea    D11(%rip),%rdx
  cmp    $0x10,%eax
  jbe    L18
L15:
  mov    %rdx,%rax
  ret
L18:
  cltq
  lea    D9(%rip),%rdx
  mov    (%rdx,%rax,8),%rdx
  jmp    L15
```

tool_calls: name: read_data
arguments: label: D11
datatype: string

tool: D11: unknown

assistant: ```c
char *sb_getversion_s(int n) {
  static char *version_s[] = {"Justice", "CTF", "Assassin", "Summer", "Spring", "Gui", "Water", "Normalmap", "Sp", "Occlusion", "Shader", "Physics", "Mp", "", "Agc", "Quakecon", "Independence"};
  n = 258 - n;
  if (n >= 0 && (size_t)n < sizeof(version_s) / sizeof(version_s[0])) {
    return version_s[n];
  }
  return "unknown";
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sb_getversion_s>:
  endbr64
  mov    $0x102,%eax
  lea    D1(%rip),%r8
  sub    %edi,%eax
  cmp    $0x10,%eax
  ja     L9
  cltq
  lea    D9(%rip),%rdx
  mov    (%rdx,%rax,8),%r8
L9:
  mov    %r8,%rax
  ret
```

tool_calls: name: read_data
arguments: label: D1
type_guessed: string

tool: D1: unknown

assistant: ```c
char *sb_getversion_s(int n) {
  static char *version_s[] = {"Justice", "CTF", "Assassin", "Summer", "Spring", "Gui", "Water", "Normalmap", "Sp", "Occlusion", "Shader", "Physics", "Mp", "", "Agc", "Quakecon", "Independence"};
  n = 258 - n;
  if (n >= 0 && (size_t)n < sizeof(version_s) / sizeof(version_s[0])) {
    return version_s[n];
  }
  return "unknown";
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sb_getversion_s>:
  endbr64
  mov    $0x102,%eax
  lea    D12(%rip),%r8
  sub    %edi,%eax
  cmp    $0x10,%eax
  ja     L2
  cltq
  lea    D15(%rip),%rdx
  mov    (%rdx,%rax,8),%r8
L2:
  mov    %r8,%rax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D12
datatype: string

tool: D12: unknown

assistant: ```c
char *sb_getversion_s(int n) {
  static char *version_s[] = {"Justice", "CTF", "Assassin", "Summer", "Spring", "Gui", "Water", "Normalmap", "Sp", "Occlusion", "Shader", "Physics", "Mp", "", "Agc", "Quakecon", "Independence"};
  n = 258 - n;
  if (n >= 0 && (size_t)n < sizeof(version_s) / sizeof(version_s[0])) {
    return version_s[n];
  }
  return "unknown";
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<rin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  mov    $0x0,%r12d
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  jmp    L15
L21:
  cmp    $0x2f,%ebx
  jle    L26
  cmp    $0x39,%ebx
  jle    L57
  jmp    L26
L83:
  test   %ebx,%ebx
  jne    L32
  mov    $0x0,%edi
  call   <exit@plt>
L32:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
L26:
  cmp    $0xa,%ebx
  jne    L83
  mov    $0x0,%eax
  jmp    L46
L57:
  mov    %r12d,%eax
  shl    $0x2,%eax
  add    %r12d,%eax
  add    %eax,%eax
  add    %ebx,%eax
  lea    -0x30(%rax),%r12d
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
L15:
  cmp    $0xa,%ebx
  je     L72
  cmp    $0x20,%ebx
  jne    L21
L72:
  mov    %r12d,%eax
L46:
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
int rin(void) {
  register int n, c;
  n = 0;
  c = getchar();
  while (c != '\n' && c != ' ') {
    if (c < '0' || c > '9') {
      while (c != '\n') {
        if (c == 0)
          exit(0);
        c = getchar();
      }
      return (0);
    }
    n = n * 10 + c - '0';
    c = getchar();
  }
  return (n);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rin>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  je     L88
  cmp    $0x20,%eax
  je     L88
  lea    -0x30(%rax),%edx
  cmp    $0x9,%edx
  ja     L93
  mov    $0x0,%ebx
L86:
  lea    (%rbx,%rbx,4),%edx
  lea    -0x30(%rax,%rdx,2),%ebx
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  je     L62
  cmp    $0x20,%eax
  je     L62
  lea    -0x30(%rax),%edx
  cmp    $0x9,%edx
  jbe    L86
  jmp    L93
L59:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  je     L32
L93:
  test   %eax,%eax
  jne    L59
  mov    $0x0,%edi
  call   <exit@plt>
  jmp    L59
L32:
  mov    $0x0,%ebx
L62:
  mov    %ebx,%eax
  pop    %rbx
  ret
L88:
  mov    $0x0,%ebx
  jmp    L62
```

assistant: ```c
int rin(void) {
  register int n, c;
  n = 0;
  c = getchar();
  while (c != '\n' && c != ' ') {
    if (c < '0' || c > '9') {
      while (c != '\n') {
        if (c == 0)
          exit(0);
        c = getchar();
      }
      return (0);
    }
    n = n * 10 + c - '0';
    c = getchar();
  }
  return (n);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rin>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0x20,%eax
  je     L58
  cmp    $0xa,%eax
  je     L58
  lea    -0x30(%rax),%edx
  cmp    $0x9,%edx
  ja     L86
  xor    %r12d,%r12d
  nopw   0x0(%rax,%rax,1)
L54:
  lea    (%r12,%r12,4),%edx
  lea    -0x30(%rax,%rdx,2),%r12d
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  je     L15
  cmp    $0x20,%eax
  je     L15
  lea    -0x30(%rax),%edx
  cmp    $0x9,%edx
  jbe    L54
  jmp    L86
  nopl   0x0(%rax)
L76:
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  je     L58
L86:
  test   %eax,%eax
  je     L76
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  jne    L86
L58:
  xor    %r12d,%r12d
L15:
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int rin(void) {
  register int n, c;
  n = 0;
  c = getchar();
  while (c != '\n' && c != ' ') {
    if (c < '0' || c > '9') {
      while (c != '\n') {
        if (c == 0)
          exit(0);
        c = getchar();
      }
      return (0);
    }
    n = n * 10 + c - '0';
    c = getchar();
  }
  return (n);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<rin>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0x20,%eax
  je     L25
  cmp    $0xa,%eax
  je     L25
  lea    -0x30(%rax),%edx
  cmp    $0x9,%edx
  ja     L46
  xor    %r12d,%r12d
  nopw   0x0(%rax,%rax,1)
L73:
  lea    (%r12,%r12,4),%edx
  lea    -0x30(%rax,%rdx,2),%r12d
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  je     L34
  cmp    $0x20,%eax
  je     L34
  lea    -0x30(%rax),%edx
  cmp    $0x9,%edx
  jbe    L73
  nopw   0x0(%rax,%rax,1)
L46:
  test   %eax,%eax
  je     L1
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  jne    L46
L25:
  xor    %r12d,%r12d
L34:
  mov    %r12d,%eax
  pop    %r12
  ret
  nopl   0x0(%rax)
L1:
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  jne    L46
  jmp    L25
```

assistant: ```c
int rin(void) {
  register int n, c;
  n = 0;
  c = getchar();
  while (c != '\n' && c != ' ') {
    if (c < '0' || c > '9') {
      while (c != '\n') {
        if (c == 0)
          exit(0);
        c = getchar();
      }
      return (0);
    }
    n = n * 10 + c - '0';
    c = getchar();
  }
  return (n);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <mtrace@plt>
  mov    $0x0,%eax
  call   <kmem_check_start@plt>
  mov    $0x0,%eax
  call   <test_echo_server@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  call   <kmem_check_leak@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  call   <kmem_check_stop@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  mtrace();
  kmem_check_start();
  test_echo_server();
  getchar();
  kmem_check_leak();
  getchar();
  kmem_check_stop();
  getchar();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <mtrace@plt>
  mov    $0x0,%eax
  call   <kmem_check_start@plt>
  mov    $0x0,%eax
  call   <test_echo_server@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  call   <kmem_check_leak@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  call   <kmem_check_stop@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  mtrace();
  kmem_check_start();
  test_echo_server();
  getchar();
  kmem_check_leak();
  getchar();
  kmem_check_stop();
  getchar();
  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
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <mtrace@plt>
  xor    %eax,%eax
  call   <kmem_check_start@plt>
  xor    %eax,%eax
  call   <test_echo_server@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  call   <kmem_check_leak@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  call   <kmem_check_stop@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  mtrace();
  kmem_check_start();
  test_echo_server();
  getchar();
  kmem_check_leak();
  getchar();
  kmem_check_stop();
  getchar();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <mtrace@plt>
  xor    %eax,%eax
  call   <kmem_check_start@plt>
  xor    %eax,%eax
  call   <test_echo_server@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  call   <kmem_check_leak@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  call   <kmem_check_stop@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  mtrace();
  kmem_check_start();
  test_echo_server();
  getchar();
  kmem_check_leak();
  getchar();
  kmem_check_stop();
  getchar();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mkvec_ret>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  sub    -0x14(%rbp),%eax
  add    $0x1,%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <smalloc_ret@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L15
  mov    $0x0,%eax
  jmp    L8
L15:
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x3,%rax
  neg    %rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
L8:
  leave
  ret
```

assistant: ```c
double *mkvec_ret(int nl, int nh) {
  double *v;
  v = smalloc_ret((nh - nl + 1) * sizeof(double));
  if (v == NULL) {
    return (NULL);
  } else {
    return (v - nl);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<mkvec_ret>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  sub    %edi,%esi
  lea    0x8(,%rsi,8),%edi
  call   <smalloc_ret@plt>
  test   %rax,%rax
  je     L2
  movslq %ebx,%rbx
  shl    $0x3,%rbx
  sub    %rbx,%rax
L2:
  pop    %rbx
  ret
```

assistant: ```c
double *mkvec_ret(int nl, int nh) {
  double *v;
  v = smalloc_ret((nh - nl + 1) * sizeof(double));
  if (v == NULL) {
    return (NULL);
  } else {
    return (v - nl);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<mkvec_ret>:
  endbr64
  push   %rbx
  movslq %edi,%rbx
  sub    %ebx,%esi
  lea    0x8(,%rsi,8),%edi
  call   <smalloc_ret@plt>
  test   %rax,%rax
  je     L8
  shl    $0x3,%rbx
  sub    %rbx,%rax
L8:
  pop    %rbx
  ret
```

assistant: ```c
double *mkvec_ret(int nl, int nh) {
  double *v;
  v = smalloc_ret((nh - nl + 1) * sizeof(double));
  if (v == NULL) {
    return (NULL);
  } else {
    return (v - nl);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<mkvec_ret>:
  endbr64
  push   %rbx
  movslq %edi,%rbx
  sub    %ebx,%esi
  lea    0x8(,%rsi,8),%edi
  call   <smalloc_ret@plt>
  test   %rax,%rax
  je     L7
  shl    $0x3,%rbx
  sub    %rbx,%rax
L7:
  pop    %rbx
  ret
```

assistant: ```c
double *mkvec_ret(int nl, int nh) {
  double *v;
  v = smalloc_ret((nh - nl + 1) * sizeof(double));
  if (v == NULL) {
    return (NULL);
  } else {
    return (v - nl);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<update_reject>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    %ecx,-0x14(%rbp)
  movsd  %xmm0,-0x20(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%ecx
  mov    -0x8(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <revert@plt>
  nop
  leave
  ret
```

assistant: ```c
void update_reject(design, iteration, which1, which2, current_eval) struct Design *design;
int iteration, which1, which2;
double current_eval;
{
  void revert(), write_step();
  revert(design, which1, which2);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<update_reject>:
  endbr64
  sub    $0x8,%rsp
  mov    %edx,%esi
  mov    %ecx,%edx
  mov    $0x0,%eax
  call   <revert@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void update_reject(design, iteration, which1, which2, current_eval) struct Design *design;
int iteration, which1, which2;
double current_eval;
{
  void revert(), write_step();
  revert(design, which1, which2);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<update_reject>:
  endbr64
  mov    %edx,%esi
  xor    %eax,%eax
  mov    %ecx,%edx
  jmp    1050 <revert@plt>
```

assistant: ```c
void update_reject(design, iteration, which1, which2, current_eval) struct Design *design;
int iteration, which1, which2;
double current_eval;
{
  void revert(), write_step();
  revert(design, which1, which2);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<symperms>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    $0x8,%eax
  jg     L75
  mov    $0x0,%eax
  jmp    L35
L75:
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L87:
  shll   $0x3,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x72,%al
  jne    L3
  mov    $0x4,%edx
  jmp    L107
L3:
  mov    $0x0,%edx
L107:
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x77,%al
  jne    L55
  mov    $0x2,%eax
  jmp    L103
L55:
  mov    $0x0,%eax
L103:
  add    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  cmp    $0x78,%al
  je     L102
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  cmp    $0x73,%al
  jne    L92
L102:
  mov    $0x1,%eax
  jmp    L91
L92:
  mov    $0x0,%eax
L91:
  add    %edx,%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
  addq   $0x3,-0x18(%rbp)
L13:
  cmpl   $0x2,-0x4(%rbp)
  jle    L87
  mov    -0x8(%rbp),%eax
L35:
  leave
  ret
```

assistant: ```c
int symperms(const char *s) {
  int perms = 0, i;
  if (strlen(s) < 9)
    return 0;
  for (i = 0; i < 3; i++, s += 3) {
    perms <<= 3;
    perms += (((s[0] == 'r') << 2) + ((s[1] == 'w') << 1) + (s[2] == 'x' || s[2] == 's'));
  }
  return perms;
}
```
user: Translate the below assembly code into C programming language:

```asm
<symperms>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    $0x0,%ecx
  cmp    $0x8,%eax
  jle    L19
  lea    0x9(%rbx),%rdi
L5:
  movzbl 0x2(%rbx),%esi
  cmpb   $0x77,0x1(%rbx)
  sete   %dl
  movzbl %dl,%edx
  cmpb   $0x72,(%rbx)
  sete   %al
  movzbl %al,%eax
  shl    $0x2,%eax
  lea    (%rax,%rdx,2),%edx
  cmp    $0x78,%sil
  sete   %al
  cmp    $0x73,%sil
  sete   %sil
  or     %esi,%eax
  movzbl %al,%eax
  add    %edx,%eax
  lea    (%rax,%rcx,8),%ecx
  add    $0x3,%rbx
  cmp    %rbx,%rdi
  jne    L5
L19:
  mov    %ecx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int symperms(const char *s) {
  int perms = 0, i;
  if (strlen(s) < 9)
    return 0;
  for (i = 0; i < 3; i++, s += 3) {
    perms <<= 3;
    perms += (((s[0] == 'r') << 2) + ((s[1] == 'w') << 1) + (s[2] == 'x' || s[2] == 's'));
  }
  return perms;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<symperms>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  xor    %r8d,%r8d
  cmp    $0x8,%eax
  jle    L14
  lea    0x9(%rbx),%rsi
L12:
  xor    %edx,%edx
  cmpb   $0x72,(%rbx)
  movzbl 0x2(%rbx),%ecx
  sete   %dl
  xor    %eax,%eax
  cmpb   $0x77,0x1(%rbx)
  sete   %al
  add    %eax,%eax
  cmp    $0x78,%cl
  lea    (%rax,%rdx,4),%edx
  sete   %al
  cmp    $0x73,%cl
  sete   %cl
  add    $0x3,%rbx
  or     %ecx,%eax
  movzbl %al,%eax
  add    %edx,%eax
  lea    (%rax,%r8,8),%r8d
  cmp    %rsi,%rbx
  jne    L12
L14:
  mov    %r8d,%eax
  pop    %rbx
  ret
```

assistant: ```c
int symperms(const char *s) {
  int perms = 0, i;
  if (strlen(s) < 9)
    return 0;
  for (i = 0; i < 3; i++, s += 3) {
    perms <<= 3;
    perms += (((s[0] == 'r') << 2) + ((s[1] == 'w') << 1) + (s[2] == 'x' || s[2] == 's'));
  }
  return perms;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<symperms>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%r8d
  xor    %eax,%eax
  cmp    $0x8,%r8d
  jle    L5
  xor    %ecx,%ecx
  cmpb   $0x72,(%rbx)
  movzbl 0x2(%rbx),%edx
  sete   %cl
  xor    %eax,%eax
  cmpb   $0x77,0x1(%rbx)
  movzbl 0x5(%rbx),%esi
  sete   %al
  add    %eax,%eax
  cmp    $0x78,%dl
  lea    (%rax,%rcx,4),%ecx
  sete   %al
  cmp    $0x73,%dl
  sete   %dl
  or     %edx,%eax
  xor    %edx,%edx
  movzbl %al,%eax
  add    %eax,%ecx
  cmpb   $0x72,0x3(%rbx)
  sete   %dl
  xor    %eax,%eax
  cmpb   $0x77,0x4(%rbx)
  sete   %al
  add    %eax,%eax
  cmp    $0x78,%sil
  lea    (%rax,%rdx,4),%edx
  sete   %al
  cmp    $0x73,%sil
  sete   %sil
  or     %esi,%eax
  movzbl %al,%eax
  add    %edx,%eax
  xor    %edx,%edx
  cmpb   $0x72,0x6(%rbx)
  lea    (%rax,%rcx,8),%esi
  sete   %dl
  movzbl 0x8(%rbx),%ecx
  xor    %eax,%eax
  cmpb   $0x77,0x7(%rbx)
  sete   %al
  add    %eax,%eax
  cmp    $0x78,%cl
  lea    (%rax,%rdx,4),%edx
  sete   %al
  cmp    $0x73,%cl
  sete   %cl
  or     %ecx,%eax
  movzbl %al,%eax
  add    %edx,%eax
  lea    (%rax,%rsi,8),%eax
L5:
  pop    %rbx
  ret
```

assistant: ```c
int symperms(const char *s) {
  int perms = 0, i;
  if (strlen(s) < 9)
    return 0;
  for (i = 0; i < 3; i++, s += 3) {
    perms <<= 3;
    perms += (((s[0] == 'r') << 2) + ((s[1] == 'w') << 1) + (s[2] == 'x' || s[2] == 's'));
  }
  return perms;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jle    L2
  mov    -0x20(%rbp),%rax
  mov    0x8(%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <chdir@plt>
  cmp    $0xffffffff,%eax
  jne    L2
  mov    $0x12,%esi
  mov    $0x14,%edi
  call   <gmove@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <gprintf@plt>
  mov    $0x0,%eax
  call   <ggetch@plt>
  mov    $0x0,%eax
  call   <gend@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L2:
  mov    $0x0,%eax
  call   <ginit@plt>
  mov    $0x0,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <reader@plt>
  mov    $0x0,%eax
  call   <gend@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D16
datatype: string
name: retrieve_data
arguments: data_label: D10
datatype: string

tool: D16: Error Setting Directory %s


tool: D10: news

assistant: ```c
int main(int argc, char *argv[]) {
  struct stat st_buf;
  char *gamedir, *p;
  if (argc > 1) {
    gamedir = argv[1];
    if (chdir(gamedir) == -1) {
      gmove(20, 18);
      gprintf("Error Setting Directory %s\n", gamedir);
      ggetch();
      gend();
      exit(1);
    }
  }
  ginit();
  reader("news", 0);
  gend();
  exit(0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  cmp    $0x1,%edi
  jg     L18
L4:
  mov    $0x0,%eax
  call   <ginit@plt>
  mov    $0x0,%esi
  lea    D10(%rip),%rdi
  call   <reader@plt>
  mov    $0x0,%eax
  call   <gend@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
L18:
  mov    0x8(%rsi),%rbx
  mov    %rbx,%rdi
  call   <chdir@plt>
  cmp    $0xffffffff,%eax
  jne    L4
  mov    $0x12,%esi
  mov    $0x14,%edi
  call   <gmove@plt>
  mov    %rbx,%rsi
  lea    D17(%rip),%rdi
  call   <gprintf@plt>
  mov    $0x0,%eax
  call   <ggetch@plt>
  mov    $0x0,%eax
  call   <gend@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L4
```

tool_calls: name: retrieve_data
arguments: data_tag: D10
type_guessed: string
name: retrieve_data
arguments: data_tag: D17
type_guessed: string

tool: D10: news

tool: D17: Error Setting Directory %s


assistant: ```c
int main(int argc, char *argv[]) {
  struct stat st_buf;
  char *gamedir, *p;
  if (argc > 1) {
    gamedir = argv[1];
    if (chdir(gamedir) == -1) {
      gmove(20, 18);
      gprintf("Error Setting Directory %s\n", gamedir);
      ggetch();
      gend();
      exit(1);
    }
  }
  ginit();
  reader("news", 0);
  gend();
  exit(0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  cmp    $0x1,%edi
  jg     L8
L24:
  xor    %eax,%eax
  call   <ginit@plt>
  xor    %esi,%esi
  lea    D9(%rip),%rdi
  call   <reader@plt>
  xor    %eax,%eax
  call   <gend@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
L8:
  mov    0x8(%rsi),%rbp
  mov    %rbp,%rdi
  call   <chdir@plt>
  add    $0x1,%eax
  jne    L24
  mov    $0x12,%esi
  mov    $0x14,%edi
  call   <gmove@plt>
  mov    %rbp,%rsi
  lea    D16(%rip),%rdi
  call   <gprintf@plt>
  xor    %eax,%eax
  call   <ggetch@plt>
  xor    %eax,%eax
  call   <gend@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L24
```

tool_calls: name: retrieve_data
arguments: label: D9
type: string
name: retrieve_data
arguments: label: D16
type: string

tool: D9: news

tool: D16: Error Setting Directory %s


assistant: ```c
int main(int argc, char *argv[]) {
  struct stat st_buf;
  char *gamedir, *p;
  if (argc > 1) {
    gamedir = argv[1];
    if (chdir(gamedir) == -1) {
      gmove(20, 18);
      gprintf("Error Setting Directory %s\n", gamedir);
      ggetch();
      gend();
      exit(1);
    }
  }
  ginit();
  reader("news", 0);
  gend();
  exit(0);
}
```