ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5a,%esi
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x5a,%ecx
  xor    %eax,%eax
  mov    $0x1e,%edx
  mov    $0x3,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D0
type: string
name: extract_symbolic_data
arguments: data_label: D9
type: string

tool: D0: ticket %d


tool: D9: %d X %d = %d


assistant: ```c
int main() {
  int shuzi = 3, jiage = 30;
  printf("ticket %d\n", shuzi * jiage);
  printf("%d X %d = %d\n", shuzi, jiage, shuzi * jiage);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5a,%esi
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x5a,%ecx
  xor    %eax,%eax
  mov    $0x1e,%edx
  mov    $0x3,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D13
type: string
name: read_data
arguments: data_tag: D1
type: string

tool: D13: ticket %d


tool: D1: %d X %d = %d


assistant: ```c
int main() {
  int shuzi = 3, jiage = 30;
  printf("ticket %d\n", shuzi * jiage);
  printf("%d X %d = %d\n", shuzi, jiage, shuzi * jiage);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<read_line_fd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x50,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x50,-0x18(%rbp)
L40:
  lea    -0x19(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <READ@plt>
  cmp    $0x1,%eax
  je     L51
  cmpl   $0x0,-0x14(%rbp)
  jne    L24
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  movq   $0x0,-0x10(%rbp)
L24:
  mov    -0x10(%rbp),%rax
  jmp    L41
L51:
  lea    -0x18(%rbp),%rdx
  mov    -0x14(%rbp),%esi
  lea    -0x10(%rbp),%rax
  mov    $0x1,%ecx
  mov    %rax,%rdi
  call   <resize@plt>
  movzbl -0x19(%rbp),%eax
  cmp    $0xa,%al
  je     L31
  movzbl -0x19(%rbp),%edx
  mov    D5(%rip),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L31
  movzbl -0x19(%rbp),%eax
  cmp    $0xd,%al
  je     L94
  mov    -0x10(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x14(%rbp)
  cltq
  lea    (%rcx,%rax,1),%rdx
  movzbl -0x19(%rbp),%eax
  mov    %al,(%rdx)
  jmp    L40
L31:
  lea    -0x18(%rbp),%rdx
  mov    -0x14(%rbp),%esi
  lea    -0x10(%rbp),%rax
  mov    $0x1,%ecx
  mov    %rax,%rdi
  call   <resize@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  cltq
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
  jmp    L41
L94:
  nop
  jmp    L40
L41:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
char *read_line_fd(int fd) {
  char *str = malloc(80);
  int n_in = 0, size = 80;
  for (;;) {
    char c;
    if (READ(fd, &c, 1) != 1) {
      if (n_in == 0) {
        free((void *)str);
        ;
        str = NULL;
      }
      return str;
    }
    resize((void **)&str, n_in, &size, sizeof(char));
    if (c == 10 || c == EOF)
      break;
    if (c == 13)
      continue;
    str[n_in++] = c;
  }
  resize((void **)&str, n_in, &size, sizeof(char));
  str[n_in] = 0x00;
  return str;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<read_line_fd>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %edi,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    $0x50,%edi
  call   <malloc@plt>
  mov    %rax,0x10(%rsp)
  movl   $0x50,0xc(%rsp)
  mov    $0x0,%ebx
  lea    0xb(%rsp),%r12
L89:
  mov    $0x1,%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <READ@plt>
  cmp    $0x1,%eax
  jne    L2
  lea    0xc(%rsp),%rdx
  lea    0x10(%rsp),%rdi
  mov    $0x1,%ecx
  mov    %ebx,%esi
  call   <resize@plt>
  movzbl 0xb(%rsp),%eax
  cmp    $0xa,%al
  je     L74
  mov    D3(%rip),%rdx
  cmp    (%rdx),%al
  je     L74
  cmp    $0xd,%al
  je     L89
  movslq %ebx,%rdx
  mov    0x10(%rsp),%rcx
  mov    %al,(%rcx,%rdx,1)
  lea    0x1(%rbx),%ebx
  jmp    L89
L2:
  test   %ebx,%ebx
  je     L80
L51:
  mov    0x10(%rsp),%rax
  jmp    L46
L80:
  mov    0x10(%rsp),%rdi
  call   <free@plt>
  movq   $0x0,0x10(%rsp)
  jmp    L51
L74:
  lea    0xc(%rsp),%rdx
  lea    0x10(%rsp),%rdi
  mov    $0x1,%ecx
  mov    %ebx,%esi
  call   <resize@plt>
  movslq %ebx,%rbx
  mov    0x10(%rsp),%rax
  movb   $0x0,(%rax,%rbx,1)
  mov    0x10(%rsp),%rax
L46:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L71
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L71:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *read_line_fd(int fd) {
  char *str = malloc(80);
  int n_in = 0, size = 80;
  for (;;) {
    char c;
    if (READ(fd, &c, 1) != 1) {
      if (n_in == 0) {
        free((void *)str);
        ;
        str = NULL;
      }
      return str;
    }
    resize((void **)&str, n_in, &size, sizeof(char));
    if (c == 10 || c == EOF)
      break;
    if (c == 13)
      continue;
    str[n_in++] = c;
  }
  resize((void **)&str, n_in, &size, sizeof(char));
  str[n_in] = 0x00;
  return str;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<read_line_fd>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  mov    $0x50,%edi
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xb(%rsp),%r13
  lea    0xc(%rsp),%r12
  call   <malloc@plt>
  movl   $0x50,0xc(%rsp)
  mov    %rax,0x10(%rsp)
  nopl   (%rax)
L35:
  mov    $0x1,%edx
  mov    %r13,%rsi
  mov    %ebp,%edi
  call   <READ@plt>
  cmp    $0x1,%eax
  jne    L75
  lea    0x10(%rsp),%r14
  mov    $0x1,%ecx
  mov    %r12,%rdx
  mov    %ebx,%esi
  mov    %r14,%rdi
  call   <resize@plt>
  movzbl 0xb(%rsp),%eax
  cmp    $0xa,%al
  je     L27
  mov    D3(%rip),%rdx
  cmp    (%rdx),%al
  je     L27
  cmp    $0xd,%al
  je     L35
  mov    0x10(%rsp),%rcx
  movslq %ebx,%rdx
  add    $0x1,%ebx
  mov    %al,(%rcx,%rdx,1)
  jmp    L35
  nop
L27:
  mov    %ebx,%esi
  mov    $0x1,%ecx
  mov    %r12,%rdx
  mov    %r14,%rdi
  call   <resize@plt>
  mov    0x10(%rsp),%rax
  movslq %ebx,%rbx
  movb   $0x0,(%rax,%rbx,1)
  mov    0x10(%rsp),%rax
L3:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L31
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L75:
  mov    0x10(%rsp),%rax
  test   %ebx,%ebx
  jne    L3
  mov    %rax,%rdi
  call   <free@plt>
  xor    %eax,%eax
  jmp    L3
L31:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *read_line_fd(int fd) {
  char *str = malloc(80);
  int n_in = 0, size = 80;
  for (;;) {
    char c;
    if (READ(fd, &c, 1) != 1) {
      if (n_in == 0) {
        free((void *)str);
        ;
        str = NULL;
      }
      return str;
    }
    resize((void **)&str, n_in, &size, sizeof(char));
    if (c == 10 || c == EOF)
      break;
    if (c == 13)
      continue;
    str[n_in++] = c;
  }
  resize((void **)&str, n_in, &size, sizeof(char));
  str[n_in] = 0x00;
  return str;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<read_line_fd>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  mov    $0x50,%edi
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xb(%rsp),%r13
  lea    0xc(%rsp),%r12
  call   <malloc@plt>
  movl   $0x50,0xc(%rsp)
  mov    %rax,0x10(%rsp)
  nopl   (%rax)
L15:
  mov    $0x1,%edx
  mov    %r13,%rsi
  mov    %ebp,%edi
  call   <READ@plt>
  cmp    $0x1,%eax
  jne    L49
  lea    0x10(%rsp),%r14
  mov    $0x1,%ecx
  mov    %r12,%rdx
  mov    %ebx,%esi
  mov    %r14,%rdi
  call   <resize@plt>
  movzbl 0xb(%rsp),%eax
  cmp    $0xa,%al
  je     L63
  mov    D5(%rip),%rdx
  cmp    (%rdx),%al
  je     L63
  cmp    $0xd,%al
  je     L15
  mov    0x10(%rsp),%rcx
  movslq %ebx,%rdx
  add    $0x1,%ebx
  mov    %al,(%rcx,%rdx,1)
  jmp    L15
  nop
L63:
  mov    %ebx,%esi
  mov    $0x1,%ecx
  mov    %r12,%rdx
  mov    %r14,%rdi
  call   <resize@plt>
  mov    0x10(%rsp),%rax
  movslq %ebx,%rbx
  movb   $0x0,(%rax,%rbx,1)
  mov    0x10(%rsp),%rax
L74:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L49:
  mov    0x10(%rsp),%rax
  test   %ebx,%ebx
  jne    L74
  mov    %rax,%rdi
  call   <free@plt>
  xor    %eax,%eax
  jmp    L74
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *read_line_fd(int fd) {
  char *str = malloc(80);
  int n_in = 0, size = 80;
  for (;;) {
    char c;
    if (READ(fd, &c, 1) != 1) {
      if (n_in == 0) {
        free((void *)str);
        ;
        str = NULL;
      }
      return str;
    }
    resize((void **)&str, n_in, &size, sizeof(char));
    if (c == 10 || c == EOF)
      break;
    if (c == 13)
      continue;
    str[n_in++] = c;
  }
  resize((void **)&str, n_in, &size, sizeof(char));
  str[n_in] = 0x00;
  return str;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<aaip_encode_acl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8d,-0x20(%rbp)
  mov    -0x30(%rbp),%rax
  movq   $0x0,(%rax)
  mov    -0x28(%rbp),%rax
  movq   $0x0,(%rax)
  mov    -0x20(%rbp),%eax
  and    $0xe,%eax
  or     $0x1,%eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %edx,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    %rax,%rdi
  call   <aaip_encode_acl_text@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xfffffffe,-0x4(%rbp)
  jge    L54
  mov    -0x4(%rbp),%eax
  jmp    L58
L54:
  cmpl   $0x0,-0x4(%rbp)
  jns    L59
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  jmp    L58
L59:
  mov    -0x20(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L81
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,(%rax)
  mov    $0x1,%eax
  jmp    L58
L81:
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    -0x30(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L87
  mov    $0xffffffff,%eax
  jmp    L58
L87:
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x20(%rbp),%eax
  and    $0xe,%eax
  mov    %eax,%edi
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rcx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <aaip_encode_acl_text@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xfffffffe,-0x4(%rbp)
  jge    L86
  mov    -0x4(%rbp),%eax
  jmp    L58
L86:
  cmpl   $0x0,-0x4(%rbp)
  jns    L79
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  jmp    L58
L79:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L117
  mov    -0x28(%rbp),%rax
  movq   $0x0,(%rax)
  mov    $0xfffffffe,%eax
  jmp    L58
L117:
  mov    $0x1,%eax
L58:
  leave
  ret
```

assistant: ```c
int aaip_encode_acl(char *acl_text, mode_t st_mode, size_t *result_len, unsigned char **result, int flag) {
  ssize_t bytes;
  *result = NULL;
  *result_len = 0;
  bytes = aaip_encode_acl_text(acl_text, st_mode, (size_t)0, NULL, 1 | (flag & (2 | 4 | 8)));
  if (bytes < -2)
    return (bytes);
  if (bytes < 0)
    return ((int)bytes - 1);
  if (flag & 1) {
    *result_len = bytes;
    return (1);
  }
  *result = calloc(bytes + 1, 1);
  if (*result == NULL)
    return (-1);
  (*result)[bytes] = 0;
  *result_len = bytes;
  bytes = aaip_encode_acl_text(acl_text, st_mode, *result_len, *result, (flag & (2 | 4 | 8)));
  if (bytes < -2)
    return (bytes);
  if (bytes < 0)
    return ((int)bytes - 1);
  if ((size_t)bytes != *result_len) {
    *result_len = 0;
    return (-2);
  }
  return (1);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<aaip_encode_acl>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %esi,%r14d
  mov    %rdx,%r12
  mov    %rcx,%r13
  mov    %r8d,%ebp
  movq   $0x0,(%rcx)
  movq   $0x0,(%rdx)
  mov    %r8d,%r15d
  and    $0xe,%r15d
  mov    %r15d,%r8d
  or     $0x1,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  call   <aaip_encode_acl_text@plt>
  mov    %eax,%ebx
  cmp    $0xfffffffe,%eax
  jl     L23
  test   %eax,%eax
  js     L111
  and    $0x1,%ebp
  je     L22
  movslq %eax,%rbx
  mov    %rbx,(%r12)
  mov    %ebp,%ebx
L23:
  mov    %ebx,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L111:
  sub    $0x1,%ebx
  jmp    L23
L22:
  lea    0x1(%rax),%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,0x0(%r13)
  test   %rax,%rax
  je     L105
  movslq %ebx,%rdx
  movb   $0x0,(%rax,%rdx,1)
  mov    %rdx,(%r12)
  mov    %r15d,%r8d
  mov    0x0(%r13),%rcx
  mov    %r14d,%esi
  mov    0x8(%rsp),%rdi
  call   <aaip_encode_acl_text@plt>
  mov    %eax,%ebx
  cmp    $0xfffffffe,%eax
  jl     L23
  test   %eax,%eax
  js     L42
  movslq %eax,%rbx
  cmp    (%r12),%rbx
  jne    L95
  mov    $0x1,%ebx
  jmp    L23
L42:
  sub    $0x1,%ebx
  jmp    L23
L95:
  movq   $0x0,(%r12)
  mov    $0xfffffffe,%ebx
  jmp    L23
L105:
  mov    $0xffffffff,%ebx
  jmp    L23
```

assistant: ```c
int aaip_encode_acl(char *acl_text, mode_t st_mode, size_t *result_len, unsigned char **result, int flag) {
  ssize_t bytes;
  *result = NULL;
  *result_len = 0;
  bytes = aaip_encode_acl_text(acl_text, st_mode, (size_t)0, NULL, 1 | (flag & (2 | 4 | 8)));
  if (bytes < -2)
    return (bytes);
  if (bytes < 0)
    return ((int)bytes - 1);
  if (flag & 1) {
    *result_len = bytes;
    return (1);
  }
  *result = calloc(bytes + 1, 1);
  if (*result == NULL)
    return (-1);
  (*result)[bytes] = 0;
  *result_len = bytes;
  bytes = aaip_encode_acl_text(acl_text, st_mode, *result_len, *result, (flag & (2 | 4 | 8)));
  if (bytes < -2)
    return (bytes);
  if (bytes < 0)
    return ((int)bytes - 1);
  if ((size_t)bytes != *result_len) {
    *result_len = 0;
    return (-2);
  }
  return (1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<aaip_encode_acl>:
  endbr64
  push   %r15
  mov    %r8d,%eax
  mov    %rdi,%r15
  push   %r14
  and    $0xe,%eax
  mov    %r8d,%r14d
  push   %r13
  mov    %eax,%r8d
  mov    %esi,%r13d
  push   %r12
  or     $0x1,%r8d
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x18,%rsp
  movq   $0x0,(%rcx)
  xor    %ecx,%ecx
  movq   $0x0,(%rdx)
  xor    %edx,%edx
  mov    %eax,0xc(%rsp)
  call   <aaip_encode_acl_text@plt>
  movslq %eax,%r12
  cmp    $0xfffffffe,%r12d
  jl     L72
  test   %r12d,%r12d
  js     L61
  and    $0x1,%r14d
  je     L67
  mov    %r12,(%rbx)
  mov    %r14d,%r12d
L72:
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L61:
  sub    $0x1,%r12d
  jmp    L72
  xchg   %ax,%ax
L67:
  lea    0x1(%r12),%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,0x0(%rbp)
  test   %rax,%rax
  je     L106
  movslq %r12d,%rdx
  mov    0xc(%rsp),%r8d
  mov    %r13d,%esi
  mov    %r15,%rdi
  movb   $0x0,(%rax,%rdx,1)
  mov    0x0(%rbp),%rcx
  mov    %rdx,(%rbx)
  call   <aaip_encode_acl_text@plt>
  movslq %eax,%r12
  cmp    $0xfffffffe,%r12d
  jl     L72
  test   %r12d,%r12d
  js     L61
  cmp    (%rbx),%r12
  jne    L86
  mov    $0x1,%r12d
  jmp    L72
L86:
  movq   $0x0,(%rbx)
  mov    $0xfffffffe,%r12d
  jmp    L72
L106:
  mov    $0xffffffff,%r12d
  jmp    L72
```

assistant: ```c
int aaip_encode_acl(char *acl_text, mode_t st_mode, size_t *result_len, unsigned char **result, int flag) {
  ssize_t bytes;
  *result = NULL;
  *result_len = 0;
  bytes = aaip_encode_acl_text(acl_text, st_mode, (size_t)0, NULL, 1 | (flag & (2 | 4 | 8)));
  if (bytes < -2)
    return (bytes);
  if (bytes < 0)
    return ((int)bytes - 1);
  if (flag & 1) {
    *result_len = bytes;
    return (1);
  }
  *result = calloc(bytes + 1, 1);
  if (*result == NULL)
    return (-1);
  (*result)[bytes] = 0;
  *result_len = bytes;
  bytes = aaip_encode_acl_text(acl_text, st_mode, *result_len, *result, (flag & (2 | 4 | 8)));
  if (bytes < -2)
    return (bytes);
  if (bytes < 0)
    return ((int)bytes - 1);
  if ((size_t)bytes != *result_len) {
    *result_len = 0;
    return (-2);
  }
  return (1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<aaip_encode_acl>:
  endbr64
  push   %r15
  mov    %r8d,%eax
  mov    %rdi,%r15
  push   %r14
  and    $0xe,%eax
  mov    %r8d,%r14d
  push   %r13
  mov    %eax,%r8d
  mov    %esi,%r13d
  push   %r12
  or     $0x1,%r8d
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x18,%rsp
  movq   $0x0,(%rcx)
  xor    %ecx,%ecx
  movq   $0x0,(%rdx)
  xor    %edx,%edx
  mov    %eax,0xc(%rsp)
  call   <aaip_encode_acl_text@plt>
  movslq %eax,%r12
  cmp    $0xfffffffe,%r12d
  jl     L46
  test   %r12d,%r12d
  js     L99
  and    $0x1,%r14d
  je     L90
  mov    %r12,(%rbx)
  mov    %r14d,%r12d
L46:
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L99:
  sub    $0x1,%r12d
  jmp    L46
  xchg   %ax,%ax
L90:
  lea    0x1(%r12),%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,0x0(%rbp)
  test   %rax,%rax
  je     L49
  movslq %r12d,%rdx
  mov    0xc(%rsp),%r8d
  mov    %r13d,%esi
  mov    %r15,%rdi
  movb   $0x0,(%rax,%rdx,1)
  mov    0x0(%rbp),%rcx
  mov    %rdx,(%rbx)
  call   <aaip_encode_acl_text@plt>
  movslq %eax,%r12
  cmp    $0xfffffffe,%r12d
  jl     L46
  test   %r12d,%r12d
  js     L99
  cmp    (%rbx),%r12
  jne    L55
  mov    $0x1,%r12d
  jmp    L46
L55:
  movq   $0x0,(%rbx)
  mov    $0xfffffffe,%r12d
  jmp    L46
L49:
  mov    $0xffffffff,%r12d
  jmp    L46
```

assistant: ```c
int aaip_encode_acl(char *acl_text, mode_t st_mode, size_t *result_len, unsigned char **result, int flag) {
  ssize_t bytes;
  *result = NULL;
  *result_len = 0;
  bytes = aaip_encode_acl_text(acl_text, st_mode, (size_t)0, NULL, 1 | (flag & (2 | 4 | 8)));
  if (bytes < -2)
    return (bytes);
  if (bytes < 0)
    return ((int)bytes - 1);
  if (flag & 1) {
    *result_len = bytes;
    return (1);
  }
  *result = calloc(bytes + 1, 1);
  if (*result == NULL)
    return (-1);
  (*result)[bytes] = 0;
  *result_len = bytes;
  bytes = aaip_encode_acl_text(acl_text, st_mode, *result_len, *result, (flag & (2 | 4 | 8)));
  if (bytes < -2)
    return (bytes);
  if (bytes < 0)
    return ((int)bytes - 1);
  if ((size_t)bytes != *result_len) {
    *result_len = 0;
    return (-2);
  }
  return (1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<powerof2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L15:
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L39
  sarl   -0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  sete   %al
  movzbl %al,%eax
  jmp    L3
L39:
  sarl   -0x14(%rbp)
  addl   $0x1,-0x4(%rbp)
L11:
  cmpl   $0x1f,-0x4(%rbp)
  jle    L15
  mov    $0x0,%eax
L3:
  pop    %rbp
  ret
```

assistant: ```c
int powerof2(int x) {
  int i;
  for (i = 0; i < 32; i++) {
    if (x & 1) {
      x >>= 1;
      return (x == 0);
    }
    x >>= 1;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<powerof2>:
  endbr64
  test   $0x1,%dil
  jne    L15
  sar    %edi
  mov    $0x1f,%eax
L30:
  mov    %edi,%edx
  and    $0x1,%edx
  jne    L15
  sar    %edi
  sub    $0x1,%eax
  jne    L30
  jmp    L2
L15:
  shr    %edi
  sete   %dl
  movzbl %dl,%edx
L2:
  mov    %edx,%eax
  ret
```

assistant: ```c
int powerof2(int x) {
  int i;
  for (i = 0; i < 32; i++) {
    if (x & 1) {
      x >>= 1;
      return (x == 0);
    }
    x >>= 1;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<powerof2>:
  endbr64
  mov    $0x20,%eax
  jmp    L19
  nopl   0x0(%rax,%rax,1)
L25:
  sub    $0x1,%eax
  je     L28
L19:
  mov    %edi,%r8d
  sar    %edi
  and    $0x1,%r8d
  test   %r8d,%r8d
  je     L25
  xor    %r8d,%r8d
  test   %edi,%edi
  sete   %r8b
L28:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int powerof2(int x) {
  int i;
  for (i = 0; i < 32; i++) {
    if (x & 1) {
      x >>= 1;
      return (x == 0);
    }
    x >>= 1;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<powerof2>:
  endbr64
  mov    $0x20,%eax
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L4:
  sub    $0x1,%eax
  je     L14
L18:
  mov    %edi,%r8d
  sar    %edi
  and    $0x1,%r8d
  test   %r8d,%r8d
  je     L4
  xor    %r8d,%r8d
  test   %edi,%edi
  sete   %r8b
L14:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int powerof2(int x) {
  int i;
  for (i = 0; i < 32; i++) {
    if (x & 1) {
      x >>= 1;
      return (x == 0);
    }
    x >>= 1;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<substr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L1
L9:
  mov    -0x4(%rbp),%edx
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
L1:
  mov    -0x4(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jge    L17
  mov    -0x24(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L9
L17:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
substr(char dest[], char src[], int offset, int len) {
  int i;
  for (i = 0; i < len && src[offset + i] != '\0'; i++)
    dest[i] = src[i + offset];
  dest[i] = '\0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<substr>:
  endbr64
  mov    %rdi,%r8
  test   %ecx,%ecx
  jle    L12
  mov    %ecx,%ecx
  mov    $0x0,%eax
  movslq %edx,%rdx
  add    %rdx,%rsi
L4:
  mov    %eax,%edi
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  je     L14
  mov    %dl,(%r8,%rax,1)
  lea    0x1(%rax),%edi
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L4
L14:
  movslq %edi,%rdi
  movb   $0x0,(%r8,%rdi,1)
  ret
L12:
  mov    $0x0,%edi
  jmp    L14
```

assistant: ```c
substr(char dest[], char src[], int offset, int len) {
  int i;
  for (i = 0; i < len && src[offset + i] != '\0'; i++)
    dest[i] = src[i + offset];
  dest[i] = '\0';
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<substr>:
  endbr64
  mov    %rdi,%r9
  test   %ecx,%ecx
  jle    L7
  movslq %edx,%rdx
  lea    0x1(%rcx),%r8d
  mov    $0x1,%eax
  add    %rdx,%rsi
  jmp    L42
  nopl   0x0(%rax)
L20:
  mov    %dl,(%rdi)
  movslq %eax,%rdx
  add    $0x1,%rax
  lea    0x1(%rcx),%rdi
  cmp    %rax,%r8
  je     L4
L42:
  movzbl -0x1(%rsi,%rax,1),%edx
  mov    %rdi,%rcx
  test   %dl,%dl
  jne    L20
L8:
  movb   $0x0,(%rcx)
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  lea    (%r9,%rdx,1),%rcx
  movb   $0x0,(%rcx)
  ret
L7:
  mov    %rdi,%rcx
  jmp    L8
```

assistant: ```c
substr(char dest[], char src[], int offset, int len) {
  int i;
  for (i = 0; i < len && src[offset + i] != '\0'; i++)
    dest[i] = src[i + offset];
  dest[i] = '\0';
}
```