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
<jsr_inc16>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  mov    $0x5,%edi
  call   <inc@plt>
  xor    %eax,%eax
  call   <bne@plt>
  test   %rax,%rax
  je     L8
  add    $0x8,%rsp
  ret
  nopl   (%rax)
L8:
  call   <inx@plt>
  xor    %esi,%esi
  mov    $0x5,%edi
  add    $0x8,%rsp
  jmp    1090 <inc@plt>
```

assistant: ```c
void jsr_inc16() {
  inc(5, 0);
  if (bne())
    goto no_carry;
  inx();
  inc(5, 0);
no_carry:;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_calculate_k_rakp_hmac>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8,-0x30(%rbp)
  mov    %r9d,-0x28(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L83
  cmpl   $0x1,-0x14(%rbp)
  je     L83
  cmpl   $0x2,-0x14(%rbp)
  jne    L136
L83:
  cmpq   $0x0,-0x20(%rbp)
  je     L136
  cmpl   $0x0,-0x24(%rbp)
  je     L136
  cmpq   $0x0,-0x30(%rbp)
  je     L136
  cmpl   $0x0,-0x28(%rbp)
  je     L136
  cmpq   $0x0,0x10(%rbp)
  je     L136
  cmpl   $0x0,0x18(%rbp)
  je     L136
  mov    -0x24(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jb     L136
  mov    -0x28(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jb     L136
  cmpl   $0x13,0x18(%rbp)
  ja     L13
L136:
  mov    D10(%rip),%rax
  mov    (%rax),%edx
  mov    D4(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  jmp    L11
L13:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <crypt_hash_digest_len@plt>
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jns    L133
  mov    $0xffffffff,%eax
  jmp    L11
L133:
  mov    -0x8(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x8(%rbp),%ecx
  mov    -0x20(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    -0x28(%rbp),%esi
  push   %rsi
  push   -0x30(%rbp)
  mov    $0x14,%r9d
  mov    0x10(%rbp),%r8
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <crypt_hash@plt>
  add    $0x10,%rsp
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L124
  mov    $0xffffffff,%eax
  jmp    L11
L124:
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  je     L57
  mov    D10(%rip),%rax
  mov    (%rax),%edx
  mov    D4(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  jmp    L11
L57:
  mov    -0x4(%rbp),%eax
L11:
  leave
  ret
```

assistant: ```c
int _calculate_k_rakp_hmac(unsigned int hash_algorithm, unsigned int expected_digest_len, const void *sik_key, unsigned int sik_key_len, void *k, unsigned int k_len, const void *constant, unsigned int constant_len) {
  int computed_digest_len, crypt_digest_len;
  if (!(((hash_algorithm) == 0x00 || (hash_algorithm) == 0x01 || (hash_algorithm) == 0x02) ? 1 : 0) || !sik_key || !sik_key_len || !k || !k_len || !constant || !constant_len || (sik_key_len < expected_digest_len) || (k_len < expected_digest_len) || (constant_len < 20)) {
    do {
      errno = (EINVAL);
      ;
    } while (0);
    return (-1);
  }
  if ((crypt_digest_len = crypt_hash_digest_len(hash_algorithm)) < 0) {
    do {
      ;
    } while (0);
    return (-1);
  }
  assert(crypt_digest_len == expected_digest_len);
  if ((computed_digest_len = crypt_hash(hash_algorithm, 0x01, sik_key, crypt_digest_len, constant, 20, k, k_len)) < 0) {
    do {
      ;
    } while (0);
    return (-1);
  }
  if (computed_digest_len != crypt_digest_len) {
    do {
      errno = (EINVAL);
      ;
    } while (0);
    return (-1);
  }
  return (computed_digest_len);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<_calculate_k_rakp_hmac>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  ja     L8
  mov    %edi,%ebp
  mov    %esi,%r13d
  mov    %rdx,%r14
  mov    %r8,%r15
  mov    %r9d,%r12d
  test   %rdx,%rdx
  je     L8
  test   %ecx,%ecx
  je     L8
  test   %r8,%r8
  je     L8
  test   %r9d,%r9d
  je     L8
  cmpq   $0x0,0x40(%rsp)
  je     L8
  cmp    %esi,%ecx
  jb     L8
  cmpl   $0x13,0x48(%rsp)
  jbe    L8
  cmp    %esi,%r9d
  jb     L8
  call   <crypt_hash_digest_len@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  js     L117
  cmp    %r13d,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  push   %r12
  push   %r15
  mov    $0x14,%r9d
  mov    0x50(%rsp),%r8
  mov    %ebx,%ecx
  mov    %r14,%rdx
  mov    $0x1,%esi
  mov    %ebp,%edi
  call   <crypt_hash@plt>
  add    $0x10,%rsp
  test   %eax,%eax
  js     L65
  cmp    %eax,%ebx
  jne    L140
L49:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L8:
  mov    D22(%rip),%rax
  mov    (%rax),%edx
  mov    D6(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  jmp    L49
L140:
  mov    D22(%rip),%rax
  mov    (%rax),%edx
  mov    D6(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  jmp    L49
L117:
  mov    $0xffffffff,%eax
  jmp    L49
L65:
  mov    $0xffffffff,%eax
  jmp    L49
```

assistant: ```c
int _calculate_k_rakp_hmac(unsigned int hash_algorithm, unsigned int expected_digest_len, const void *sik_key, unsigned int sik_key_len, void *k, unsigned int k_len, const void *constant, unsigned int constant_len) {
  int computed_digest_len, crypt_digest_len;
  if (!(((hash_algorithm) == 0x00 || (hash_algorithm) == 0x01 || (hash_algorithm) == 0x02) ? 1 : 0) || !sik_key || !sik_key_len || !k || !k_len || !constant || !constant_len || (sik_key_len < expected_digest_len) || (k_len < expected_digest_len) || (constant_len < 20)) {
    do {
      errno = (EINVAL);
      ;
    } while (0);
    return (-1);
  }
  if ((crypt_digest_len = crypt_hash_digest_len(hash_algorithm)) < 0) {
    do {
      ;
    } while (0);
    return (-1);
  }
  assert(crypt_digest_len == expected_digest_len);
  if ((computed_digest_len = crypt_hash(hash_algorithm, 0x01, sik_key, crypt_digest_len, constant, 20, k, k_len)) < 0) {
    do {
      ;
    } while (0);
    return (-1);
  }
  if (computed_digest_len != crypt_digest_len) {
    do {
      errno = (EINVAL);
      ;
    } while (0);
    return (-1);
  }
  return (computed_digest_len);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<_calculate_k_rakp_hmac>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  ja     L83
  mov    %rdx,%r15
  test   %rdx,%rdx
  je     L83
  test   %ecx,%ecx
  je     L83
  mov    %r8,%r14
  test   %r8,%r8
  je     L83
  mov    %r9d,%r12d
  test   %r9d,%r9d
  je     L83
  cmpq   $0x0,0x40(%rsp)
  je     L83
  mov    %esi,%r13d
  cmp    %esi,%ecx
  jb     L83
  cmpl   $0x13,0x48(%rsp)
  jbe    L83
  cmp    %esi,%r9d
  jb     L83
  mov    %edi,%ebp
  call   <crypt_hash_digest_len@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  js     L84
  xor    %edi,%edi
  cmp    %r13d,%eax
  sete   %dil
  call   <assert@plt>
  push   %r12
  mov    %ebx,%ecx
  mov    %r15,%rdx
  push   %r14
  mov    0x50(%rsp),%r8
  mov    $0x14,%r9d
  mov    %ebp,%edi
  mov    $0x1,%esi
  call   <crypt_hash@plt>
  pop    %rdx
  pop    %rcx
  test   %eax,%eax
  js     L84
  cmp    %eax,%ebx
  jne    L83
L31:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L83:
  mov    D17(%rip),%rax
  mov    (%rax),%edx
  mov    D7(%rip),%rax
  mov    %edx,(%rax)
L84:
  mov    $0xffffffff,%eax
  jmp    L31
```

assistant: ```c
int _calculate_k_rakp_hmac(unsigned int hash_algorithm, unsigned int expected_digest_len, const void *sik_key, unsigned int sik_key_len, void *k, unsigned int k_len, const void *constant, unsigned int constant_len) {
  int computed_digest_len, crypt_digest_len;
  if (!(((hash_algorithm) == 0x00 || (hash_algorithm) == 0x01 || (hash_algorithm) == 0x02) ? 1 : 0) || !sik_key || !sik_key_len || !k || !k_len || !constant || !constant_len || (sik_key_len < expected_digest_len) || (k_len < expected_digest_len) || (constant_len < 20)) {
    do {
      errno = (EINVAL);
      ;
    } while (0);
    return (-1);
  }
  if ((crypt_digest_len = crypt_hash_digest_len(hash_algorithm)) < 0) {
    do {
      ;
    } while (0);
    return (-1);
  }
  assert(crypt_digest_len == expected_digest_len);
  if ((computed_digest_len = crypt_hash(hash_algorithm, 0x01, sik_key, crypt_digest_len, constant, 20, k, k_len)) < 0) {
    do {
      ;
    } while (0);
    return (-1);
  }
  if (computed_digest_len != crypt_digest_len) {
    do {
      errno = (EINVAL);
      ;
    } while (0);
    return (-1);
  }
  return (computed_digest_len);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<_calculate_k_rakp_hmac>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  ja     L98
  mov    %rdx,%r15
  test   %rdx,%rdx
  je     L98
  test   %ecx,%ecx
  je     L98
  mov    %r8,%r14
  test   %r8,%r8
  je     L98
  mov    %r9d,%r12d
  test   %r9d,%r9d
  je     L98
  cmpq   $0x0,0x40(%rsp)
  je     L98
  mov    %esi,%r13d
  cmp    %esi,%ecx
  jb     L98
  cmpl   $0x13,0x48(%rsp)
  jbe    L98
  cmp    %esi,%r9d
  jb     L98
  mov    %edi,%ebp
  call   <crypt_hash_digest_len@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  js     L7
  xor    %edi,%edi
  cmp    %r13d,%eax
  sete   %dil
  call   <assert@plt>
  push   %r12
  mov    %ebx,%ecx
  mov    %r15,%rdx
  push   %r14
  mov    0x50(%rsp),%r8
  mov    $0x14,%r9d
  mov    %ebp,%edi
  mov    $0x1,%esi
  call   <crypt_hash@plt>
  pop    %rdx
  pop    %rcx
  test   %eax,%eax
  js     L7
  cmp    %eax,%ebx
  jne    L98
L31:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L98:
  mov    D3(%rip),%rax
  mov    (%rax),%edx
  mov    D5(%rip),%rax
  mov    %edx,(%rax)
L7:
  mov    $0xffffffff,%eax
  jmp    L31
```

assistant: ```c
int _calculate_k_rakp_hmac(unsigned int hash_algorithm, unsigned int expected_digest_len, const void *sik_key, unsigned int sik_key_len, void *k, unsigned int k_len, const void *constant, unsigned int constant_len) {
  int computed_digest_len, crypt_digest_len;
  if (!(((hash_algorithm) == 0x00 || (hash_algorithm) == 0x01 || (hash_algorithm) == 0x02) ? 1 : 0) || !sik_key || !sik_key_len || !k || !k_len || !constant || !constant_len || (sik_key_len < expected_digest_len) || (k_len < expected_digest_len) || (constant_len < 20)) {
    do {
      errno = (EINVAL);
      ;
    } while (0);
    return (-1);
  }
  if ((crypt_digest_len = crypt_hash_digest_len(hash_algorithm)) < 0) {
    do {
      ;
    } while (0);
    return (-1);
  }
  assert(crypt_digest_len == expected_digest_len);
  if ((computed_digest_len = crypt_hash(hash_algorithm, 0x01, sik_key, crypt_digest_len, constant, 20, k, k_len)) < 0) {
    do {
      ;
    } while (0);
    return (-1);
  }
  if (computed_digest_len != crypt_digest_len) {
    do {
      errno = (EINVAL);
      ;
    } while (0);
    return (-1);
  }
  return (computed_digest_len);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<parse_label>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  jne    L74
  mov    -0x20(%rbp),%rax
  lea    D21(%rip),%rdx
  mov    %rdx,(%rax)
  mov    $0x0,%eax
  jmp    L23
L74:
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L10
L0:
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
  cmp    $0x3f,%rax
  jle    L85
  mov    -0x20(%rbp),%rax
  lea    D28(%rip),%rdx
  mov    %rdx,(%rax)
  mov    $0x0,%eax
  jmp    L23
L85:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  je     L50
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L21
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jle    L8
L21:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x60,%al
  jle    L44
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x7a,%al
  jle    L5
L44:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x59,%al
  jle    L159
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5a,%al
  jle    L146
L159:
  mov    -0x20(%rbp),%rax
  lea    D22(%rip),%rdx
  mov    %rdx,(%rax)
  mov    $0x0,%eax
  jmp    L23
L50:
  nop
  jmp    L143
L8:
  nop
  jmp    L143
L5:
  nop
  jmp    L143
L146:
  nop
L143:
  addq   $0x1,-0x8(%rbp)
L10:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L34
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  jne    L0
L34:
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
L23:
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D21
type: string
name: parse_data
arguments: data_label: D28
type: string
name: parse_data
arguments: data_label: D22
type: string

tool: D21: Too many '.'

tool: D28: Label is too long

tool: D22: Invalid character in host name

assistant: ```c
ssize_t parse_label(const char *buffer, const char **error) {
  char const *p;
  if (*buffer == '.') {
    *error = "Too many '.'";
    return 0;
  }
  for (p = buffer; *p && (*p != '.'); p++) {
    if ((p - buffer) > 63) {
      *error = "Label is too long";
      return 0;
    }
    if (*p == '-')
      continue;
    if ((*p >= '0') && (*p <= '9'))
      continue;
    if ((*p >= 'a') && (*p <= 'z'))
      continue;
    if ((*p >= 'Z') && (*p <= 'Z'))
      continue;
    *error = "Invalid character in host name";
    return 0;
  }
  return p - buffer;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<parse_label>:
  endbr64
  movzbl (%rdi),%edx
  cmp    $0x2e,%dl
  je     L51
  mov    %rdi,%rax
  test   %dl,%dl
  jne    L29
L63:
  sub    %edi,%eax
  ret
L51:
  lea    D12(%rip),%rax
  mov    %rax,(%rsi)
  mov    $0x0,%eax
  ret
L20:
  lea    D25(%rip),%rax
  mov    %rax,(%rsi)
  mov    $0x0,%eax
  ret
L31:
  add    $0x1,%rax
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L63
  cmp    $0x2e,%dl
  je     L63
  mov    %rax,%rcx
  sub    %rdi,%rcx
  cmp    $0x3f,%rcx
  jg     L20
L29:
  cmp    $0x2d,%dl
  je     L31
  lea    -0x30(%rdx),%ecx
  cmp    $0x9,%cl
  jbe    L31
  lea    -0x61(%rdx),%ecx
  cmp    $0x19,%cl
  jbe    L31
  cmp    $0x5a,%dl
  je     L31
  lea    D1(%rip),%rax
  mov    %rax,(%rsi)
  mov    $0x0,%eax
  ret
```

tool_calls: name: read_data
arguments: data_label: D12
data_type: string
name: read_data
arguments: data_label: D25
data_type: string
name: read_data
arguments: data_label: D1
data_type: string

tool: D12: Too many '.'

tool: D25: Label is too long

tool: D1: Invalid character in host name

assistant: ```c
ssize_t parse_label(const char *buffer, const char **error) {
  char const *p;
  if (*buffer == '.') {
    *error = "Too many '.'";
    return 0;
  }
  for (p = buffer; *p && (*p != '.'); p++) {
    if ((p - buffer) > 63) {
      *error = "Label is too long";
      return 0;
    }
    if (*p == '-')
      continue;
    if ((*p >= '0') && (*p <= '9'))
      continue;
    if ((*p >= 'a') && (*p <= 'z'))
      continue;
    if ((*p >= 'Z') && (*p <= 'Z'))
      continue;
    *error = "Invalid character in host name";
    return 0;
  }
  return p - buffer;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse_label>:
  endbr64
  movzbl (%rdi),%edx
  cmp    $0x2e,%dl
  je     L21
  mov    %rdi,%rcx
  test   %dl,%dl
  je     L49
L86:
  cmp    $0x2d,%dl
  je     L29
  lea    -0x30(%rdx),%eax
  cmp    $0x9,%al
  jbe    L29
  lea    -0x61(%rdx),%eax
  cmp    $0x19,%al
  jbe    L29
  cmp    $0x5a,%dl
  je     L29
  lea    D4(%rip),%rax
  mov    %rax,(%rsi)
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L29:
  add    $0x1,%rcx
  movzbl (%rcx),%edx
  mov    %rcx,%rax
  sub    %rdi,%rax
  test   %dl,%dl
  je     L72
  cmp    $0x2e,%dl
  je     L72
  cmp    $0x3f,%rax
  jle    L86
  lea    D21(%rip),%rax
  mov    %rax,(%rsi)
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L72:
  ret
  nopl   0x0(%rax)
L21:
  lea    D3(%rip),%rax
  mov    %rax,(%rsi)
L49:
  xor    %eax,%eax
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D4
type_guessed: string
name: disassemble_data
arguments: data_label: D21
type_guessed: string
name: disassemble_data
arguments: data_label: D3
type_guessed: string

tool: D4: Invalid character in host name

tool: D21: Label is too long

tool: D3: Too many '.'

assistant: ```c
ssize_t parse_label(const char *buffer, const char **error) {
  char const *p;
  if (*buffer == '.') {
    *error = "Too many '.'";
    return 0;
  }
  for (p = buffer; *p && (*p != '.'); p++) {
    if ((p - buffer) > 63) {
      *error = "Label is too long";
      return 0;
    }
    if (*p == '-')
      continue;
    if ((*p >= '0') && (*p <= '9'))
      continue;
    if ((*p >= 'a') && (*p <= 'z'))
      continue;
    if ((*p >= 'Z') && (*p <= 'Z'))
      continue;
    *error = "Invalid character in host name";
    return 0;
  }
  return p - buffer;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse_label>:
  endbr64
  movzbl (%rdi),%edx
  cmp    $0x2e,%dl
  je     L81
  mov    %rdi,%rcx
  test   %dl,%dl
  je     L38
L89:
  cmp    $0x2d,%dl
  je     L0
  lea    -0x30(%rdx),%eax
  cmp    $0x9,%al
  jbe    L0
  lea    -0x61(%rdx),%eax
  cmp    $0x19,%al
  jbe    L0
  cmp    $0x5a,%dl
  je     L0
  lea    D22(%rip),%rax
  mov    %rax,(%rsi)
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  add    $0x1,%rcx
  movzbl (%rcx),%edx
  mov    %rcx,%rax
  sub    %rdi,%rax
  test   %dl,%dl
  je     L75
  cmp    $0x2e,%dl
  je     L75
  cmp    $0x3f,%rax
  jle    L89
  lea    D3(%rip),%rax
  mov    %rax,(%rsi)
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L75:
  ret
  nopl   0x0(%rax)
L81:
  lea    D0(%rip),%rax
  mov    %rax,(%rsi)
L38:
  xor    %eax,%eax
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D22
type_infered: string
name: retrieve_data
arguments: tag: D3
type_infered: string
name: retrieve_data
arguments: tag: D0
type_infered: string

tool: D22: Invalid character in host name

tool: D3: Label is too long

tool: D0: Too many '.'

assistant: ```c
ssize_t parse_label(const char *buffer, const char **error) {
  char const *p;
  if (*buffer == '.') {
    *error = "Too many '.'";
    return 0;
  }
  for (p = buffer; *p && (*p != '.'); p++) {
    if ((p - buffer) > 63) {
      *error = "Label is too long";
      return 0;
    }
    if (*p == '-')
      continue;
    if ((*p >= '0') && (*p <= '9'))
      continue;
    if ((*p >= 'a') && (*p <= 'z'))
      continue;
    if ((*p >= 'Z') && (*p <= 'Z'))
      continue;
    *error = "Invalid character in host name";
    return 0;
  }
  return p - buffer;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xpmatoui>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  mov    %rdi,%rcx
  mov    %esi,-0x14(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    $0x0,%r12d
  mov    $0x0,%ebx
  jmp    L16
L47:
  movzbl (%rcx),%eax
  cmp    $0x2f,%al
  jle    L50
  movzbl (%rcx),%eax
  cmp    $0x39,%al
  jg     L50
  mov    %r12d,%eax
  shl    $0x2,%eax
  add    %r12d,%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    %rcx,%rax
  lea    0x1(%rax),%rcx
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %edx,%eax
  lea    -0x30(%rax),%r12d
  add    $0x1,%ebx
L16:
  cmp    -0x14(%rbp),%ebx
  jb     L47
L50:
  test   %ebx,%ebx
  je     L51
  cmp    -0x14(%rbp),%ebx
  jne    L51
  mov    -0x20(%rbp),%rax
  mov    %r12d,(%rax)
  mov    $0x1,%eax
  jmp    L8
L51:
  mov    $0x0,%eax
L8:
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
unsigned int xpmatoui(register char *p, unsigned int l, unsigned int *ui_return) {
  register unsigned int n, i;
  n = 0;
  for (i = 0; i < l; i++)
    if (*p >= '0' && *p <= '9')
      n = n * 10 + *p++ - '0';
    else
      break;
  if (i != 0 && i == l) {
    *ui_return = n;
    return 1;
  } else
    return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xpmatoui>:
  endbr64
  test   %esi,%esi
  je     L37
  mov    $0x0,%r8d
  mov    $0x0,%r9d
L17:
  movzbl (%rdi),%ecx
  lea    -0x30(%rcx),%r10d
  cmp    $0x9,%r10b
  ja     L42
  add    $0x1,%rdi
  lea    (%r9,%r9,4),%r9d
  movsbl %cl,%ecx
  lea    -0x30(%rcx,%r9,2),%r9d
  add    $0x1,%r8d
  cmp    %r8d,%esi
  jne    L17
L18:
  mov    %r9d,(%rdx)
  mov    $0x1,%eax
  ret
L42:
  test   %r8d,%r8d
  je     L23
  cmp    %r8d,%esi
  je     L18
L23:
  mov    $0x0,%eax
  ret
L37:
  mov    %esi,%eax
  ret
```

assistant: ```c
unsigned int xpmatoui(register char *p, unsigned int l, unsigned int *ui_return) {
  register unsigned int n, i;
  n = 0;
  for (i = 0; i < l; i++)
    if (*p >= '0' && *p <= '9')
      n = n * 10 + *p++ - '0';
    else
      break;
  if (i != 0 && i == l) {
    *ui_return = n;
    return 1;
  } else
    return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<xpmatoui>:
  endbr64
  test   %esi,%esi
  je     L37
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  jmp    L3
  nop
L7:
  lea    (%rcx,%rcx,4),%ecx
  add    $0x1,%r8d
  add    $0x1,%rdi
  lea    -0x30(%rax,%rcx,2),%ecx
  cmp    %r8d,%esi
  je     L10
L3:
  movsbl (%rdi),%eax
  lea    -0x30(%rax),%r9d
  cmp    $0x9,%r9b
  jbe    L7
L37:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L10:
  mov    %ecx,(%rdx)
  mov    $0x1,%eax
  ret
```

assistant: ```c
unsigned int xpmatoui(register char *p, unsigned int l, unsigned int *ui_return) {
  register unsigned int n, i;
  n = 0;
  for (i = 0; i < l; i++)
    if (*p >= '0' && *p <= '9')
      n = n * 10 + *p++ - '0';
    else
      break;
  if (i != 0 && i == l) {
    *ui_return = n;
    return 1;
  } else
    return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<xpmatoui>:
  endbr64
  test   %esi,%esi
  je     L12
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  jmp    L29
  nop
L33:
  lea    (%rcx,%rcx,4),%ecx
  add    $0x1,%r8d
  add    $0x1,%rdi
  lea    -0x30(%rax,%rcx,2),%ecx
  cmp    %r8d,%esi
  je     L3
L29:
  movsbl (%rdi),%eax
  lea    -0x30(%rax),%r9d
  cmp    $0x9,%r9b
  jbe    L33
L12:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L3:
  mov    %ecx,(%rdx)
  mov    $0x1,%eax
  ret
```

assistant: ```c
unsigned int xpmatoui(register char *p, unsigned int l, unsigned int *ui_return) {
  register unsigned int n, i;
  n = 0;
  for (i = 0; i < l; i++)
    if (*p >= '0' && *p <= '9')
      n = n * 10 + *p++ - '0';
    else
      break;
  if (i != 0 && i == l) {
    *ui_return = n;
    return 1;
  } else
    return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<log_meta>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %edi,-0xc4(%rbp)
  mov    %rsi,-0xd0(%rbp)
  mov    %edx,-0xc8(%rbp)
  mov    %rcx,-0xd8(%rbp)
  mov    %r8,-0xe0(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L7
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L7:
  mov    -0xe0(%rbp),%rdx
  mov    -0xb8(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb8(%rbp),%r8d
  mov    -0xe0(%rbp),%rdi
  mov    -0xd8(%rbp),%rcx
  mov    -0xc8(%rbp),%edx
  mov    -0xd0(%rbp),%rsi
  mov    -0xc4(%rbp),%eax
  mov    %r8d,%r9d
  mov    %rdi,%r8
  mov    %eax,%edi
  call   <log_metav@plt>
  mov    %eax,-0xb4(%rbp)
  mov    -0xb8(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xb4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int log_meta(int level, const char *file, int line, const char *func, const char *format, ...) {
  int r;
  va_list ap;
  va_start(ap, format);
  r = log_metav(level, file, line, func, format, ap);
  va_end(ap);
  return r;
}
```