ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<coffset>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L25
  jmp    L17
  cs nopw 0x0(%rax,%rax,1)
L28:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  je     L46
L25:
  cmp    $0x2e,%al
  jne    L28
L17:
  test   %al,%al
  je     L46
  add    $0x1,%rdi
  call   <atoi@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <atoi@plt>
  add    $0x8,%rsp
  mov    %eax,%r8d
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  sub    %r8d,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L46:
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1050 <atoi@plt>
```

assistant: ```c
int coffset(char *str) {
  register char *p;
  for (p = str; *p && *p != '.'; ++p)
    ;
  return (*p ? atoi(p + 1) - atoi(str) : atoi(str));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<coffset>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L13
  jmp    L21
  cs nopw 0x0(%rax,%rax,1)
L6:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  je     L10
L13:
  cmp    $0x2e,%al
  jne    L6
L21:
  test   %al,%al
  je     L10
  add    $0x1,%rdi
  call   <atoi@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <atoi@plt>
  add    $0x8,%rsp
  mov    %eax,%r8d
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  sub    %r8d,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L10:
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1050 <atoi@plt>
```

assistant: ```c
int coffset(char *str) {
  register char *p;
  for (p = str; *p && *p != '.'; ++p)
    ;
  return (*p ? atoi(p + 1) - atoi(str) : atoi(str));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ParseHex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L39
L46:
  shll   $0x4,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L30
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jg     L30
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  add    %eax,-0xc(%rbp)
  jmp    L91
L30:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x60,%al
  jle    L87
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x66,%al
  jg     L87
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x57,%eax
  add    %eax,-0xc(%rbp)
  jmp    L91
L87:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jle    L21
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x46,%al
  jg     L21
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x37,%eax
  add    %eax,-0xc(%rbp)
  jmp    L91
L21:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <Error@plt>
L91:
  addq   $0x1,-0x8(%rbp)
L39:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L46
  mov    -0xc(%rbp),%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D8
type_guessed: string

tool: D8: Bad hex number: %s

assistant: ```c
int ParseHex(char *hex) {
  char *str;
  int num;
  num = 0;
  str = hex;
  while (*str) {
    num <<= 4;
    if (*str >= '0' && *str <= '9')
      num += *str - '0';
    else if (*str >= 'a' && *str <= 'f')
      num += 10 + *str - 'a';
    else if (*str >= 'A' && *str <= 'F')
      num += 10 + *str - 'A';
    else
      Error("Bad hex number: %s", hex);
    str++;
  }
  return num;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ParseHex>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%eax
  test   %al,%al
  je     L61
  mov    %rdi,%r12
  mov    %rdi,%rbp
  mov    $0x0,%ebx
  lea    D6(%rip),%r13
  jmp    L42
L14:
  lea    -0x61(%rax),%edx
  cmp    $0x5,%dl
  ja     L77
  movsbl %al,%eax
  lea    -0x57(%rbx,%rax,1),%ebx
L64:
  add    $0x1,%rbp
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L25
L42:
  shl    $0x4,%ebx
  mov    %eax,%ecx
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  ja     L14
  movsbl %al,%eax
  lea    -0x30(%rbx,%rax,1),%ebx
  jmp    L64
L77:
  sub    $0x41,%ecx
  cmp    $0x5,%cl
  ja     L74
  movsbl %al,%eax
  lea    -0x37(%rbx,%rax,1),%ebx
  jmp    L64
L74:
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <Error@plt>
  jmp    L64
L61:
  mov    $0x0,%ebx
L25:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
type: string

tool: D6: Bad hex number: %s

assistant: ```c
int ParseHex(char *hex) {
  char *str;
  int num;
  num = 0;
  str = hex;
  while (*str) {
    num <<= 4;
    if (*str >= '0' && *str <= '9')
      num += *str - '0';
    else if (*str >= 'a' && *str <= 'f')
      num += 10 + *str - 'a';
    else if (*str >= 'A' && *str <= 'F')
      num += 10 + *str - 'A';
    else
      Error("Bad hex number: %s", hex);
    str++;
  }
  return num;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ParseHex>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%eax
  test   %al,%al
  je     L24
  mov    %rdi,%rbp
  mov    %rdi,%rbx
  lea    D8(%rip),%r13
  xor    %r12d,%r12d
  jmp    L68
  nopl   0x0(%rax,%rax,1)
L58:
  movsbl %dl,%edx
  add    %edx,%r12d
L55:
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  test   %al,%al
  je     L80
L68:
  lea    -0x30(%rax),%edx
  shl    $0x4,%r12d
  cmp    $0x9,%dl
  jbe    L58
  lea    -0x61(%rax),%edx
  cmp    $0x5,%dl
  ja     L74
  sub    $0x57,%eax
  add    $0x1,%rbx
  movsbl %al,%eax
  add    %eax,%r12d
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L68
L80:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L74:
  lea    -0x41(%rax),%edx
  cmp    $0x5,%dl
  ja     L17
  sub    $0x37,%eax
  movsbl %al,%eax
  add    %eax,%r12d
  jmp    L55
  nopl   0x0(%rax,%rax,1)
L17:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <Error@plt>
  jmp    L55
  nopl   (%rax)
L24:
  add    $0x8,%rsp
  xor    %r12d,%r12d
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
datatype: string

tool: D8: Bad hex number: %s

assistant: ```c
int ParseHex(char *hex) {
  char *str;
  int num;
  num = 0;
  str = hex;
  while (*str) {
    num <<= 4;
    if (*str >= '0' && *str <= '9')
      num += *str - '0';
    else if (*str >= 'a' && *str <= 'f')
      num += 10 + *str - 'a';
    else if (*str >= 'A' && *str <= 'F')
      num += 10 + *str - 'A';
    else
      Error("Bad hex number: %s", hex);
    str++;
  }
  return num;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ParseHex>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%eax
  test   %al,%al
  je     L50
  mov    %rdi,%rbp
  mov    %rdi,%rbx
  lea    D1(%rip),%r13
  xor    %r12d,%r12d
  jmp    L80
  nopl   0x0(%rax,%rax,1)
L14:
  movsbl %dl,%edx
  add    %edx,%r12d
L44:
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  test   %al,%al
  je     L43
L80:
  lea    -0x30(%rax),%edx
  shl    $0x4,%r12d
  cmp    $0x9,%dl
  jbe    L14
  lea    -0x61(%rax),%edx
  cmp    $0x5,%dl
  ja     L89
  sub    $0x57,%eax
  add    $0x1,%rbx
  movsbl %al,%eax
  add    %eax,%r12d
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L80
L43:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L89:
  lea    -0x41(%rax),%edx
  cmp    $0x5,%dl
  ja     L52
  sub    $0x37,%eax
  movsbl %al,%eax
  add    %eax,%r12d
  jmp    L44
  nopl   0x0(%rax,%rax,1)
L52:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <Error@plt>
  jmp    L44
  nopl   (%rax)
L50:
  add    $0x8,%rsp
  xor    %r12d,%r12d
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D1
data_type: string

tool: D1: Bad hex number: %s

assistant: ```c
int ParseHex(char *hex) {
  char *str;
  int num;
  num = 0;
  str = hex;
  while (*str) {
    num <<= 4;
    if (*str >= '0' && *str <= '9')
      num += *str - '0';
    else if (*str >= 'a' && *str <= 'f')
      num += 10 + *str - 'a';
    else if (*str >= 'A' && *str <= 'F')
      num += 10 + *str - 'A';
    else
      Error("Bad hex number: %s", hex);
    str++;
  }
  return num;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sxerox>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *sxerox(char *s) {
  char *rv = (char *)malloc(strlen(s) + 1);
  strcpy(rv, s);
  return rv;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sxerox>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *sxerox(char *s) {
  char *rv = (char *)malloc(strlen(s) + 1);
  strcpy(rv, s);
  return rv;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sxerox>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *sxerox(char *s) {
  char *rv = (char *)malloc(strlen(s) + 1);
  strcpy(rv, s);
  return rv;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sxerox>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *sxerox(char *s) {
  char *rv = (char *)malloc(strlen(s) + 1);
  strcpy(rv, s);
  return rv;
}
```
user: Translate the below assembly code into C programming language:

```asm
<append_crc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L26
L36:
  shll   -0x10(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x10(%rbp),%edx
  shr    $0xf,%edx
  and    $0x1,%edx
  cmp    %edx,%eax
  je     L19
  xorl   $0x4599,-0x10(%rbp)
L19:
  andl   $0x7fff,-0x10(%rbp)
  addl   $0x1,-0xc(%rbp)
L26:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L36
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0xe,-0x4(%rbp)
  jmp    L40
L20:
  mov    -0x4(%rbp),%eax
  mov    -0x10(%rbp),%edx
  mov    %eax,%ecx
  shr    %cl,%edx
  mov    %edx,%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  and    $0x1,%ecx
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x8(%rbp)
  subl   $0x1,-0x4(%rbp)
L40:
  cmpl   $0x0,-0x4(%rbp)
  jns    L20
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int append_crc(char *in, int in_len) {
  unsigned int crc = 0;
  for (int i = 0; i < in_len; i++) {
    crc <<= 1;
    if (((unsigned int)(in[i]) ^ ((crc >> 15) & 1U)) != 0U) {
      crc = crc ^ 0x4599U;
    }
    crc &= 0x7fffU;
  }
  int in_len_copy = in_len;
  for (int i = 14; i >= 0; i--) {
    in[in_len_copy] = (crc >> (unsigned int)(i)) & 1U;
    in_len_copy++;
  }
  return in_len_copy;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<append_crc>:
  endbr64
  mov    %esi,%eax
  test   %esi,%esi
  jle    L18
  mov    %rdi,%rcx
  lea    -0x1(%rsi),%edx
  lea    0x1(%rdi,%rdx,1),%r10
  mov    $0x0,%esi
L6:
  lea    (%rsi,%rsi,1),%edx
  movsbl (%rcx),%r9d
  shr    $0xe,%esi
  mov    %edx,%r8d
  xor    $0x4599,%r8d
  cmp    %esi,%r9d
  cmovne %r8d,%edx
  mov    %edx,%esi
  and    $0x7fff,%esi
  add    $0x1,%rcx
  cmp    %r10,%rcx
  jne    L6
L36:
  movslq %eax,%rdx
  add    %rdx,%rdi
  mov    $0xe,%ecx
L0:
  mov    %esi,%edx
  shr    %cl,%edx
  and    $0x1,%edx
  mov    %dl,(%rdi)
  sub    $0x1,%ecx
  add    $0x1,%rdi
  cmp    $0xffffffff,%ecx
  jne    L0
  add    $0xf,%eax
  ret
L18:
  mov    $0x0,%esi
  jmp    L36
```

assistant: ```c
int append_crc(char *in, int in_len) {
  unsigned int crc = 0;
  for (int i = 0; i < in_len; i++) {
    crc <<= 1;
    if (((unsigned int)(in[i]) ^ ((crc >> 15) & 1U)) != 0U) {
      crc = crc ^ 0x4599U;
    }
    crc &= 0x7fffU;
  }
  int in_len_copy = in_len;
  for (int i = 14; i >= 0; i--) {
    in[in_len_copy] = (crc >> (unsigned int)(i)) & 1U;
    in_len_copy++;
  }
  return in_len_copy;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<append_crc>:
  endbr64
  mov    %esi,%eax
  test   %esi,%esi
  jle    L36
  lea    -0x1(%rsi),%edx
  mov    %rdi,%rcx
  xor    %esi,%esi
  lea    0x1(%rdi,%rdx,1),%r10
  nopw   0x0(%rax,%rax,1)
L33:
  lea    (%rsi,%rsi,1),%edx
  movsbl (%rcx),%r9d
  shr    $0xe,%esi
  mov    %edx,%r8d
  xor    $0x4599,%r8d
  cmp    %esi,%r9d
  cmovne %r8d,%edx
  add    $0x1,%rcx
  mov    %edx,%esi
  and    $0x7fff,%esi
  cmp    %r10,%rcx
  jne    L33
L16:
  movslq %eax,%rdx
  mov    $0xe,%ecx
  add    %rdx,%rdi
  nopw   0x0(%rax,%rax,1)
L7:
  mov    %esi,%edx
  add    $0x1,%rdi
  shr    %cl,%edx
  sub    $0x1,%ecx
  and    $0x1,%edx
  mov    %dl,-0x1(%rdi)
  cmp    $0xffffffff,%ecx
  jne    L7
  add    $0xf,%eax
  ret
L36:
  xor    %esi,%esi
  jmp    L16
```

assistant: ```c
int append_crc(char *in, int in_len) {
  unsigned int crc = 0;
  for (int i = 0; i < in_len; i++) {
    crc <<= 1;
    if (((unsigned int)(in[i]) ^ ((crc >> 15) & 1U)) != 0U) {
      crc = crc ^ 0x4599U;
    }
    crc &= 0x7fffU;
  }
  int in_len_copy = in_len;
  for (int i = 14; i >= 0; i--) {
    in[in_len_copy] = (crc >> (unsigned int)(i)) & 1U;
    in_len_copy++;
  }
  return in_len_copy;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<append_crc>:
  endbr64
  push   %rbp
  mov    %esi,%eax
  push   %rbx
  test   %esi,%esi
  jle    L0
  lea    -0x1(%rax),%edx
  mov    %rdi,%rsi
  lea    0x1(%rdi,%rdx,1),%r10
  xor    %edx,%edx
  nopl   (%rax)
L23:
  lea    (%rdx,%rdx,1),%ecx
  movsbl (%rsi),%r9d
  shr    $0xe,%edx
  mov    %ecx,%r8d
  xor    $0x4599,%r8d
  cmp    %r9d,%edx
  cmovne %r8d,%ecx
  add    $0x1,%rsi
  mov    %ecx,%edx
  and    $0x7fff,%edx
  cmp    %r10,%rsi
  jne    L23
L8:
  mov    %edx,%ecx
  mov    %edx,%ebp
  mov    %edx,%ebx
  mov    %edx,%r11d
  shr    $0x7,%ecx
  shr    $0x8,%ebp
  mov    %edx,%r10d
  mov    %edx,%r9d
  movzbl %cl,%ecx
  movzbl %bpl,%ebp
  shr    $0x9,%ebx
  mov    %edx,%r8d
  shl    $0x8,%rcx
  movzbl %bl,%ebx
  shr    $0xa,%r11d
  mov    %edx,%esi
  shr    $0xb,%r10d
  or     %rbp,%rcx
  movzbl %r11b,%r11d
  shr    $0xc,%r9d
  shl    $0x8,%rcx
  movzbl %r10b,%r10d
  movzbl %r9b,%r9d
  shr    $0xd,%r8d
  or     %rbx,%rcx
  movzbl %r8b,%r8d
  shr    $0xe,%esi
  pop    %rbx
  shl    $0x8,%rcx
  movzbl %sil,%esi
  pop    %rbp
  movq   D6(%rip),%xmm0
  or     %r11,%rcx
  shl    $0x8,%rcx
  or     %r10,%rcx
  shl    $0x8,%rcx
  or     %r9,%rcx
  shl    $0x8,%rcx
  or     %r8,%rcx
  shl    $0x8,%rcx
  or     %rsi,%rcx
  mov    %edx,%esi
  movq   %rcx,%xmm1
  movslq %eax,%rcx
  shr    $0x6,%esi
  pand   %xmm1,%xmm0
  and    $0x1,%esi
  movq   %xmm0,(%rdi,%rcx,1)
  lea    0x8(%rax),%ecx
  movslq %ecx,%rcx
  mov    %sil,(%rdi,%rcx,1)
  mov    %edx,%esi
  lea    0x9(%rax),%ecx
  shr    $0x5,%esi
  movslq %ecx,%rcx
  and    $0x1,%esi
  mov    %sil,(%rdi,%rcx,1)
  mov    %edx,%esi
  lea    0xa(%rax),%ecx
  shr    $0x4,%esi
  movslq %ecx,%rcx
  and    $0x1,%esi
  mov    %sil,(%rdi,%rcx,1)
  mov    %edx,%esi
  lea    0xb(%rax),%ecx
  shr    $0x3,%esi
  movslq %ecx,%rcx
  and    $0x1,%esi
  mov    %sil,(%rdi,%rcx,1)
  mov    %edx,%esi
  lea    0xc(%rax),%ecx
  shr    $0x2,%esi
  movslq %ecx,%rcx
  and    $0x1,%esi
  mov    %sil,(%rdi,%rcx,1)
  mov    %edx,%esi
  lea    0xd(%rax),%ecx
  shr    %esi
  movslq %ecx,%rcx
  and    $0x1,%edx
  and    $0x1,%esi
  mov    %sil,(%rdi,%rcx,1)
  lea    0xe(%rax),%ecx
  add    $0xf,%eax
  movslq %ecx,%rcx
  mov    %dl,(%rdi,%rcx,1)
  ret
L0:
  xor    %edx,%edx
  jmp    L8
```

tool_calls: name: parse_data
arguments: data_tag: D6
type: byte[8]

tool: {"D6": [0x01,0x01,0x01,0x01,0x01,0x01,0x01,0x01]}

assistant: ```c
int append_crc(char *in, int in_len) {
  unsigned int crc = 0;
  for (int i = 0; i < in_len; i++) {
    crc <<= 1;
    if (((unsigned int)(in[i]) ^ ((crc >> 15) & 1U)) != 0U) {
      crc = crc ^ 0x4599U;
    }
    crc &= 0x7fffU;
  }
  int in_len_copy = in_len;
  for (int i = 14; i >= 0; i--) {
    in[in_len_copy] = (crc >> (unsigned int)(i)) & 1U;
    in_len_copy++;
  }
  return in_len_copy;
}
```
user: What is the c source code of the assembly code below?

```asm
<printArray>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0xa,%edi
  call   <putchar@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L19:
  movl   $0x0,-0x8(%rbp)
  jmp    L20
L39:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x8(%rbp)
L20:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jl     L39
  addl   $0x1,-0x4(%rbp)
L2:
  cmpl   $0x7,-0x4(%rbp)
  jle    L19
  nop
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D5
type_infered: string

tool: D5: array[%d]: %s


assistant: ```c
void printArray(char ***array, int *count) {
  int i, j;
  printf("\n");
  for (j = 0; j < 8; j++) {
    i = 0;
    while (i < count[j]) {
      printf("array[%d]: %s\n", j, array[j][i]);
      i++;
    }
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<printArray>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r15
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%ebp
  lea    D1(%rip),%r14
L2:
  mov    %r15,%r13
  mov    $0x0,%ebx
  cmpl   $0x0,(%r15)
  jle    L0
L7:
  mov    (%r12),%rax
  mov    (%rax,%rbx,8),%rdx
  mov    %ebp,%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %ebx,0x0(%r13)
  jg     L7
L0:
  add    $0x1,%ebp
  add    $0x4,%r15
  add    $0x8,%r12
  cmp    $0x8,%ebp
  jne    L2
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D1
data_type: string

tool: D1: array[%d]: %s


assistant: ```c
void printArray(char ***array, int *count) {
  int i, j;
  printf("\n");
  for (j = 0; j < 8; j++) {
    i = 0;
    while (i < count[j]) {
      printf("array[%d]: %s\n", j, array[j][i]);
      i++;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<printArray>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  mov    $0xa,%edi
  push   %r13
  lea    D9(%rip),%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  call   <putchar@plt>
  cs nopw 0x0(%rax,%rax,1)
L10:
  mov    0x0(%rbp),%eax
  xor    %ebx,%ebx
  test   %eax,%eax
  jle    L8
  nopl   0x0(%rax)
L18:
  mov    (%r14),%rax
  mov    %r12d,%esi
  mov    %r13,%rdi
  mov    (%rax,%rbx,8),%rdx
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0x0(%rbp)
  jg     L18
L8:
  add    $0x1,%r12d
  add    $0x4,%rbp
  add    $0x8,%r14
  cmp    $0x8,%r12d
  jne    L10
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D9
type_guessed: string

tool: D9: array[%d]: %s


assistant: ```c
void printArray(char ***array, int *count) {
  int i, j;
  printf("\n");
  for (j = 0; j < 8; j++) {
    i = 0;
    while (i < count[j]) {
      printf("array[%d]: %s\n", j, array[j][i]);
      i++;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<printArray>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  mov    $0xa,%edi
  push   %r13
  lea    D0(%rip),%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  call   <putchar@plt>
  cs nopw 0x0(%rax,%rax,1)
L20:
  mov    0x0(%rbp),%eax
  xor    %ebx,%ebx
  test   %eax,%eax
  jle    L15
  nopl   0x0(%rax)
L23:
  mov    (%r14),%rax
  mov    %r12d,%esi
  mov    %r13,%rdi
  mov    (%rax,%rbx,8),%rdx
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0x0(%rbp)
  jg     L23
L15:
  add    $0x1,%r12d
  add    $0x4,%rbp
  add    $0x8,%r14
  cmp    $0x8,%r12d
  jne    L20
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D0
type: string

tool: D0: array[%d]: %s


assistant: ```c
void printArray(char ***array, int *count) {
  int i, j;
  printf("\n");
  for (j = 0; j < 8; j++) {
    i = 0;
    while (i < count[j]) {
      printf("array[%d]: %s\n", j, array[j][i]);
      i++;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<divide_evenly>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L38
L21:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  cqto
  idiv   %rcx
  mov    %rdx,%rax
  test   %rax,%rax
  je     L7
  mov    $0x0,%eax
  jmp    L14
L7:
  addl   $0x1,-0x4(%rbp)
L38:
  cmpl   $0x14,-0x4(%rbp)
  jle    L21
  mov    -0x18(%rbp),%rax
L14:
  pop    %rbp
  ret
```

assistant: ```c
int divide_evenly(long dividend) {
  for (int i = 1; i <= 20; i++) {
    if (dividend % i != 0)
      return 0;
  }
  return dividend;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<divide_evenly>:
  endbr64
  mov    $0x2,%ecx
L6:
  mov    %rdi,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  jne    L10
  add    $0x1,%rcx
  cmp    $0x15,%rcx
  jne    L6
  mov    %edi,%eax
  ret
L10:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int divide_evenly(long dividend) {
  for (int i = 1; i <= 20; i++) {
    if (dividend % i != 0)
      return 0;
  }
  return dividend;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<divide_evenly>:
  endbr64
  mov    $0x2,%ecx
  jmp    L25
  nopl   0x0(%rax,%rax,1)
L10:
  cmp    $0x15,%rcx
  je     L26
L25:
  mov    %rdi,%rax
  cqto
  idiv   %rcx
  add    $0x1,%rcx
  test   %rdx,%rdx
  je     L10
  xor    %eax,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L26:
  mov    %edi,%eax
  ret
```

assistant: ```c
int divide_evenly(long dividend) {
  for (int i = 1; i <= 20; i++) {
    if (dividend % i != 0)
      return 0;
  }
  return dividend;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<divide_evenly>:
  endbr64
  mov    $0x2,%ecx
  jmp    L20
  nopl   0x0(%rax,%rax,1)
L29:
  cmp    $0x15,%rcx
  je     L5
L20:
  mov    %rdi,%rax
  cqto
  idiv   %rcx
  add    $0x1,%rcx
  test   %rdx,%rdx
  je     L29
  xor    %eax,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %edi,%eax
  ret
```

assistant: ```c
int divide_evenly(long dividend) {
  for (int i = 1; i <= 20; i++) {
    if (dividend % i != 0)
      return 0;
  }
  return dividend;
}
```