ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<add_option_request_list>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    $0x6,%ecx
  lea    D8(%rip),%rdx
  mov    $0x37,%esi
  mov    %rax,%rdi
  call   <add_option_string@plt>
  leave
  ret
```

assistant: ```c
size_t add_option_request_list(struct dhcpmsg *packet) {
  static const char reqdata[] = {
      0x01, 0x03, 0x06, 0x0c, 0x0f, 0x1c,
  };
  return add_option_string(packet, 0x37, reqdata, sizeof reqdata);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<add_option_request_list>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x6,%ecx
  lea    D0(%rip),%rdx
  mov    $0x37,%esi
  call   <add_option_string@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
size_t add_option_request_list(struct dhcpmsg *packet) {
  static const char reqdata[] = {
      0x01, 0x03, 0x06, 0x0c, 0x0f, 0x1c,
  };
  return add_option_string(packet, 0x37, reqdata, sizeof reqdata);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<add_option_request_list>:
  endbr64
  mov    $0x6,%ecx
  lea    D0(%rip),%rdx
  mov    $0x37,%esi
  jmp    1050 <add_option_string@plt>
```

assistant: ```c
size_t add_option_request_list(struct dhcpmsg *packet) {
  static const char reqdata[] = {
      0x01, 0x03, 0x06, 0x0c, 0x0f, 0x1c,
  };
  return add_option_string(packet, 0x37, reqdata, sizeof reqdata);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<add_option_request_list>:
  endbr64
  mov    $0x6,%ecx
  lea    D3(%rip),%rdx
  mov    $0x37,%esi
  jmp    1050 <add_option_string@plt>
```

assistant: ```c
size_t add_option_request_list(struct dhcpmsg *packet) {
  static const char reqdata[] = {
      0x01, 0x03, 0x06, 0x0c, 0x0f, 0x1c,
  };
  return add_option_string(packet, 0x37, reqdata, sizeof reqdata);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <stringLength@plt>
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <stringLength@plt>
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  call   <stringLength@plt>
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D40
datatype: string
name: extract_symbolic_data
arguments: data_label: D23
datatype: string
name: extract_symbolic_data
arguments: data_label: D10
datatype: string
name: extract_symbolic_data
arguments: data_label: D51
datatype: string
name: extract_symbolic_data
arguments: data_label: D3
datatype: string

tool: D40: stringLength test

tool: D23: %i 

tool: 

tool: D51: complete

tool: D3: %i


assistant: ```c
int main(void) {
  int stringLength(const char *string);
  printf("%i ", stringLength("stringLength test"));
  printf("%i ", stringLength(""));
  printf("%i\n", stringLength("complete"));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D24(%rip),%rdi
  call   <stringLength@plt>
  mov    %eax,%esi
  lea    D2(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  lea    D39(%rip),%rdi
  call   <stringLength@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  lea    D28(%rip),%rdi
  call   <stringLength@plt>
  mov    %eax,%esi
  lea    D23(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D24
type_guessed: string
name: retrieve_data
arguments: data_label: D2
type_guessed: string
name: retrieve_data
arguments: data_label: D39
type_guessed: string
name: retrieve_data
arguments: data_label: D28
type_guessed: string
name: retrieve_data
arguments: data_label: D23
type_guessed: string

tool: D24: stringLength test

tool: D2: %i 

tool: 

tool: D28: complete

tool: D23: %i


assistant: ```c
int main(void) {
  int stringLength(const char *string);
  printf("%i ", stringLength("stringLength test"));
  printf("%i ", stringLength(""));
  printf("%i\n", stringLength("complete"));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D11(%rip),%rdi
  lea    D30(%rip),%rbp
  call   <stringLength@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  lea    D39(%rip),%rdi
  call   <stringLength@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  lea    D4(%rip),%rdi
  call   <stringLength@plt>
  lea    D23(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D11
datatype: string
name: unpack_data
arguments: data_tag: D30
datatype: string
name: unpack_data
arguments: data_tag: D39
datatype: string
name: unpack_data
arguments: data_tag: D4
datatype: string
name: unpack_data
arguments: data_tag: D23
datatype: string

tool: D11: stringLength test

tool: D30: %i 

tool: 

tool: D4: complete

tool: D23: %i


assistant: ```c
int main(void) {
  int stringLength(const char *string);
  printf("%i ", stringLength("stringLength test"));
  printf("%i ", stringLength(""));
  printf("%i\n", stringLength("complete"));
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D12(%rip),%rdi
  lea    D32(%rip),%rbp
  call   <stringLength@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  lea    D45(%rip),%rdi
  call   <stringLength@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  lea    D33(%rip),%rdi
  call   <stringLength@plt>
  lea    D47(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D12
type_infered: string
name: parse_data
arguments: data_label: D32
type_infered: string
name: parse_data
arguments: data_label: D45
type_infered: string
name: parse_data
arguments: data_label: D33
type_infered: string
name: parse_data
arguments: data_label: D47
type_infered: string

tool: D12: stringLength test

tool: D32: %i 

tool: 

tool: D33: complete

tool: D47: %i


assistant: ```c
int main(void) {
  int stringLength(const char *string);
  printf("%i ", stringLength("stringLength test"));
  printf("%i ", stringLength(""));
  printf("%i\n", stringLength("complete"));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<m_max>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movss  %xmm0,-0x4(%rbp)
  movss  %xmm1,-0x8(%rbp)
  movss  -0x4(%rbp),%xmm0
  comiss -0x8(%rbp),%xmm0
  jbe    L8
  movss  -0x4(%rbp),%xmm0
  jmp    L19
L8:
  movss  -0x8(%rbp),%xmm0
L19:
  pop    %rbp
  ret
```

assistant: ```c
float m_max(float value_a, float value_b) {
  if (value_a > value_b)
    return value_a;
  return value_b;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<m_max>:
  endbr64
  maxss  %xmm1,%xmm0
  ret
```

assistant: ```c
float m_max(float value_a, float value_b) {
  if (value_a > value_b)
    return value_a;
  return value_b;
}
```
user: Translate the below assembly code into C programming language:

```asm
<m_max>:
  endbr64
  maxss  %xmm1,%xmm0
  ret
```

assistant: ```c
float m_max(float value_a, float value_b) {
  if (value_a > value_b)
    return value_a;
  return value_b;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<m_max>:
  endbr64
  maxss  %xmm1,%xmm0
  ret
```

assistant: ```c
float m_max(float value_a, float value_b) {
  if (value_a > value_b)
    return value_a;
  return value_b;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strtonum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L10
  mov    $0x0,%eax
  jmp    L20
L10:
  movq   $0x0,-0x10(%rbp)
  jmp    L30
L8:
  mov    -0x8(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  cltq
  add    %rcx,%rax
  mov    %rax,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L30:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %rax,-0x10(%rbp)
  jb     L8
  mov    -0x8(%rbp),%rax
L20:
  leave
  ret
```

assistant: ```c
size_t strtonum(char *palab) {
  size_t i;
  size_t numero = 0;
  if (!palab)
    return 0;
  for (i = 0; i < strlen(palab); i++) {
    numero = numero * 10 + (palab[i] - '0');
  }
  return numero;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strtonum>:
  endbr64
  test   %rdi,%rdi
  je     L24
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %rbx,%rdx
  lea    (%rbx,%rax,1),%rdi
  mov    $0x0,%eax
  jmp    L3
L7:
  lea    (%rax,%rax,4),%rcx
  movsbl (%rdx),%eax
  sub    $0x30,%eax
  cltq
  lea    (%rax,%rcx,2),%rax
  add    $0x1,%rdx
L3:
  cmp    %rdi,%rdx
  jne    L7
  pop    %rbx
  ret
L24:
  mov    $0x0,%eax
  ret
```

assistant: ```c
size_t strtonum(char *palab) {
  size_t i;
  size_t numero = 0;
  if (!palab)
    return 0;
  for (i = 0; i < strlen(palab); i++) {
    numero = numero * 10 + (palab[i] - '0');
  }
  return numero;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<strtonum>:
  endbr64
  test   %rdi,%rdi
  je     L17
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %rbx,%rdi
  lea    (%rbx,%rax,1),%rsi
  xor    %eax,%eax
  jmp    L29
  nopl   (%rax)
L26:
  movsbl (%rdi),%edx
  lea    (%rax,%rax,4),%rcx
  add    $0x1,%rdi
  sub    $0x30,%edx
  movslq %edx,%rdx
  lea    (%rdx,%rcx,2),%rax
L29:
  cmp    %rsi,%rdi
  jne    L26
  pop    %rbx
  ret
  nopl   0x0(%rax)
L17:
  xor    %eax,%eax
  ret
```

assistant: ```c
size_t strtonum(char *palab) {
  size_t i;
  size_t numero = 0;
  if (!palab)
    return 0;
  for (i = 0; i < strlen(palab); i++) {
    numero = numero * 10 + (palab[i] - '0');
  }
  return numero;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strtonum>:
  endbr64
  test   %rdi,%rdi
  je     L25
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %rax,%rax
  je     L23
  lea    (%rbx,%rax,1),%rcx
  mov    %rbx,%rdi
  xor    %eax,%eax
L3:
  lea    (%rax,%rax,4),%rdx
  movsbl (%rdi),%eax
  add    $0x1,%rdi
  sub    $0x30,%eax
  cltq
  lea    (%rax,%rdx,2),%rax
  cmp    %rdi,%rcx
  jne    L3
L23:
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L25:
  xor    %eax,%eax
  ret
```

assistant: ```c
size_t strtonum(char *palab) {
  size_t i;
  size_t numero = 0;
  if (!palab)
    return 0;
  for (i = 0; i < strlen(palab); i++) {
    numero = numero * 10 + (palab[i] - '0');
  }
  return numero;
}
```
user: Translate the below assembly code into C programming language:

```asm
<shift>:
  endbr64
  cmp    $0xb,%sil
  ja     L7
  movzbl %sil,%esi
  lea    D7(%rip),%rcx
  movslq (%rcx,%rsi,4),%rax
  add    %rcx,%rax
  notrack jmp *%rax
  lea    0x1(%rdi),%eax
  ret
  mov    $0x67,%ecx
  mov    %ecx,%eax
  imul   %dil
  sar    $0x9,%ax
  mov    %edi,%ecx
  sar    $0x7,%cl
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    0x6(%rax,%rdi,1),%eax
  ret
  mov    $0x67,%ecx
  mov    %ecx,%eax
  imul   %dil
  sar    $0x9,%ax
  mov    %edi,%ecx
  sar    $0x7,%cl
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    0x5(%rax,%rdi,1),%eax
  ret
  lea    0xa(%rdi),%eax
  ret
  mov    $0x67,%ecx
  mov    %ecx,%eax
  imul   %dil
  sar    $0x9,%ax
  mov    %edi,%ecx
  sar    $0x7,%cl
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    0x4(%rax,%rdi,1),%eax
  ret
  mov    $0x67,%ecx
  mov    %ecx,%eax
  imul   %dil
  sar    $0x9,%ax
  mov    %edi,%ecx
  sar    $0x7,%cl
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    0x3(%rax,%rdi,1),%eax
  ret
  lea    -0x1(%rdi),%eax
  ret
  mov    $0x67,%ecx
  mov    %ecx,%eax
  imul   %dil
  sar    $0x9,%ax
  mov    %edi,%ecx
  sar    $0x7,%cl
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    -0x7(%rax,%rdi,1),%eax
  ret
  mov    $0x67,%ecx
  mov    %ecx,%eax
  imul   %dil
  sar    $0x9,%ax
  mov    %edi,%ecx
  sar    $0x7,%cl
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    -0x6(%rax,%rdi,1),%eax
  ret
  lea    -0xa(%rdi),%eax
  ret
  mov    $0x67,%ecx
  mov    %ecx,%eax
  imul   %dil
  sar    $0x9,%ax
  mov    %edi,%ecx
  sar    $0x7,%cl
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    -0x5(%rax,%rdi,1),%eax
  ret
  mov    $0x67,%ecx
  mov    %ecx,%eax
  imul   %dil
  sar    $0x9,%ax
  mov    %edi,%ecx
  sar    $0x7,%cl
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    -0x4(%rax,%rdi,1),%eax
  ret
L7:
  mov    %edi,%eax
  ret
```

assistant: ```c
char shift(char cell, char dir) {
  switch (dir) {
  case 0:
    return cell + 1;
  case 1:
    if ((cell / 5) % 2)
      return cell + 7;
    else
      return cell + 6;
  case 2:
    if ((cell / 5) % 2)
      return cell + 6;
    else
      return cell + 5;
  case 3:
    return cell + 10;
  case 4:
    if ((cell / 5) % 2)
      return cell + 5;
    else
      return cell + 4;
  case 5:
    if ((cell / 5) % 2)
      return cell + 4;
    else
      return cell + 3;
  case 6:
    return cell - 1;
  case 7:
    if ((cell / 5) % 2)
      return cell - 6;
    else
      return cell - 7;
  case 8:
    if ((cell / 5) % 2)
      return cell - 5;
    else
      return cell - 6;
  case 9:
    return cell - 10;
  case 10:
    if ((cell / 5) % 2)
      return cell - 4;
    else
      return cell - 5;
  case 11:
    if ((cell / 5) % 2)
      return cell - 3;
    else
      return cell - 4;
  default:
    return cell;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<shift>:
  endbr64
  cmp    $0xb,%sil
  ja     1040 <shift.cold>
  lea    D9(%rip),%rcx
  movzbl %sil,%esi
  movslq (%rcx,%rsi,4),%rax
  add    %rcx,%rax
  notrack jmp *%rax
  nopl   0x0(%rax,%rax,1)
  mov    $0x67,%eax
  mov    %edi,%ecx
  imul   %dil
  sar    $0x7,%cl
  sar    $0x9,%ax
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    -0x5(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  mov    $0x67,%eax
  mov    %edi,%ecx
  imul   %dil
  sar    $0x7,%cl
  sar    $0x9,%ax
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    -0x4(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0x1(%rdi),%eax
  ret
  nopl   0x0(%rax)
  mov    $0x67,%eax
  mov    %edi,%ecx
  imul   %dil
  sar    $0x7,%cl
  sar    $0x9,%ax
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    0x6(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  mov    $0x67,%eax
  mov    %edi,%ecx
  imul   %dil
  sar    $0x7,%cl
  sar    $0x9,%ax
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    0x5(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0xa(%rdi),%eax
  ret
  nopl   0x0(%rax)
  mov    $0x67,%eax
  mov    %edi,%ecx
  imul   %dil
  sar    $0x7,%cl
  sar    $0x9,%ax
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    0x4(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  mov    $0x67,%eax
  mov    %edi,%ecx
  imul   %dil
  sar    $0x7,%cl
  sar    $0x9,%ax
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    0x3(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    -0x1(%rdi),%eax
  ret
  nopl   0x0(%rax)
  mov    $0x67,%eax
  mov    %edi,%ecx
  imul   %dil
  sar    $0x7,%cl
  sar    $0x9,%ax
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    -0x7(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  mov    $0x67,%eax
  mov    %edi,%ecx
  imul   %dil
  sar    $0x7,%cl
  sar    $0x9,%ax
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    -0x6(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    -0xa(%rdi),%eax
  ret
```

assistant: ```c
char shift(char cell, char dir) {
  switch (dir) {
  case 0:
    return cell + 1;
  case 1:
    if ((cell / 5) % 2)
      return cell + 7;
    else
      return cell + 6;
  case 2:
    if ((cell / 5) % 2)
      return cell + 6;
    else
      return cell + 5;
  case 3:
    return cell + 10;
  case 4:
    if ((cell / 5) % 2)
      return cell + 5;
    else
      return cell + 4;
  case 5:
    if ((cell / 5) % 2)
      return cell + 4;
    else
      return cell + 3;
  case 6:
    return cell - 1;
  case 7:
    if ((cell / 5) % 2)
      return cell - 6;
    else
      return cell - 7;
  case 8:
    if ((cell / 5) % 2)
      return cell - 5;
    else
      return cell - 6;
  case 9:
    return cell - 10;
  case 10:
    if ((cell / 5) % 2)
      return cell - 4;
    else
      return cell - 5;
  case 11:
    if ((cell / 5) % 2)
      return cell - 3;
    else
      return cell - 4;
  default:
    return cell;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<shift>:
  endbr64
  cmp    $0xb,%sil
  ja     1040 <shift.cold>
  lea    D8(%rip),%rcx
  movzbl %sil,%esi
  movslq (%rcx,%rsi,4),%rax
  add    %rcx,%rax
  notrack jmp *%rax
  nopl   0x0(%rax,%rax,1)
  mov    $0x67,%eax
  mov    %edi,%ecx
  imul   %dil
  sar    $0x7,%cl
  sar    $0x9,%ax
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    -0x5(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  mov    $0x67,%eax
  mov    %edi,%ecx
  imul   %dil
  sar    $0x7,%cl
  sar    $0x9,%ax
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    -0x4(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0x1(%rdi),%eax
  ret
  nopl   0x0(%rax)
  mov    $0x67,%eax
  mov    %edi,%ecx
  imul   %dil
  sar    $0x7,%cl
  sar    $0x9,%ax
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    0x6(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  mov    $0x67,%eax
  mov    %edi,%ecx
  imul   %dil
  sar    $0x7,%cl
  sar    $0x9,%ax
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    0x5(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0xa(%rdi),%eax
  ret
  nopl   0x0(%rax)
  mov    $0x67,%eax
  mov    %edi,%ecx
  imul   %dil
  sar    $0x7,%cl
  sar    $0x9,%ax
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    0x4(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  mov    $0x67,%eax
  mov    %edi,%ecx
  imul   %dil
  sar    $0x7,%cl
  sar    $0x9,%ax
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    0x3(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    -0x1(%rdi),%eax
  ret
  nopl   0x0(%rax)
  mov    $0x67,%eax
  mov    %edi,%ecx
  imul   %dil
  sar    $0x7,%cl
  sar    $0x9,%ax
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    -0x7(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  mov    $0x67,%eax
  mov    %edi,%ecx
  imul   %dil
  sar    $0x7,%cl
  sar    $0x9,%ax
  sub    %ecx,%eax
  and    $0x1,%eax
  lea    -0x6(%rax,%rdi,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    -0xa(%rdi),%eax
  ret
```

assistant: ```c
char shift(char cell, char dir) {
  switch (dir) {
  case 0:
    return cell + 1;
  case 1:
    if ((cell / 5) % 2)
      return cell + 7;
    else
      return cell + 6;
  case 2:
    if ((cell / 5) % 2)
      return cell + 6;
    else
      return cell + 5;
  case 3:
    return cell + 10;
  case 4:
    if ((cell / 5) % 2)
      return cell + 5;
    else
      return cell + 4;
  case 5:
    if ((cell / 5) % 2)
      return cell + 4;
    else
      return cell + 3;
  case 6:
    return cell - 1;
  case 7:
    if ((cell / 5) % 2)
      return cell - 6;
    else
      return cell - 7;
  case 8:
    if ((cell / 5) % 2)
      return cell - 5;
    else
      return cell - 6;
  case 9:
    return cell - 10;
  case 10:
    if ((cell / 5) % 2)
      return cell - 4;
    else
      return cell - 5;
  case 11:
    if ((cell / 5) % 2)
      return cell - 3;
    else
      return cell - 4;
  default:
    return cell;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<skip_to_newline>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  jmp    L6
L11:
  addl   $0x1,-0xc(%rbp)
L6:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L11
  addl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int skip_to_newline(char *input, int offset) {
  while (input[offset] != 10) {
    offset++;
  }
  return ++offset;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<skip_to_newline>:
  endbr64
  movslq %esi,%rax
  cmpb   $0xa,(%rdi,%rax,1)
  je     L6
  add    $0x1,%esi
  movslq %esi,%rdx
L1:
  mov    %edx,%esi
  add    $0x1,%rdx
  cmpb   $0xa,-0x1(%rdi,%rdx,1)
  jne    L1
L6:
  lea    0x1(%rsi),%eax
  ret
```

assistant: ```c
int skip_to_newline(char *input, int offset) {
  while (input[offset] != 10) {
    offset++;
  }
  return ++offset;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<skip_to_newline>:
  endbr64
  movslq %esi,%rax
  cmpb   $0xa,(%rdi,%rax,1)
  je     L0
  add    $0x1,%esi
  movslq %esi,%rdx
  nopl   0x0(%rax,%rax,1)
L8:
  mov    %edx,%esi
  add    $0x1,%rdx
  cmpb   $0xa,-0x1(%rdi,%rdx,1)
  jne    L8
L0:
  lea    0x1(%rsi),%eax
  ret
```

assistant: ```c
int skip_to_newline(char *input, int offset) {
  while (input[offset] != 10) {
    offset++;
  }
  return ++offset;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<skip_to_newline>:
  endbr64
  movslq %esi,%rax
  cmpb   $0xa,(%rdi,%rax,1)
  je     L14
  add    $0x1,%esi
  movslq %esi,%rdx
  nopl   0x0(%rax,%rax,1)
L1:
  mov    %edx,%esi
  add    $0x1,%rdx
  cmpb   $0xa,-0x1(%rdi,%rdx,1)
  jne    L1
L14:
  lea    0x1(%rsi),%eax
  ret
```

assistant: ```c
int skip_to_newline(char *input, int offset) {
  while (input[offset] != 10) {
    offset++;
  }
  return ++offset;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  sub    $0x3f60a4d,%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L20
L17:
  movq   $0x0,-0x20(%rbp)
  jmp    L48
L3:
  lea    -0x10(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  movzwl (%rax),%eax
  mov    %ax,-0x22(%rbp)
  lea    -0x10(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  add    $0x2,%rax
  lea    -0x10(%rbp),%rcx
  mov    -0x20(%rbp),%rdx
  shl    $0x3,%rdx
  add    %rcx,%rdx
  movzwl (%rax),%eax
  mov    %ax,(%rdx)
  lea    -0x10(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  lea    0x2(%rax),%rdx
  movzwl -0x22(%rbp),%eax
  mov    %ax,(%rdx)
  addq   $0x1,-0x20(%rbp)
L48:
  cmpq   $0x0,-0x20(%rbp)
  je     L3
  addq   $0x1,-0x18(%rbp)
L20:
  cmpq   $0x0,-0x18(%rbp)
  je     L17
  mov    -0x10(%rbp),%rax
  lea    -0x4331cbd1(%rax),%rdx
  mov    -0x40(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local2;
  unsigned long local1;
  unsigned short copy11;
  {
    state[0UL] = (input[0UL] + 914778474UL) - 981234615UL;
    local1 = 0UL;
    while (local1 < 1UL) {
      local2 = 0UL;
      while (local2 < 1UL) {
        copy11 = *((unsigned short *)(&state[local2]) + 0);
        *((unsigned short *)(&state[local2]) + 0) = *((unsigned short *)(&state[local2]) + 1);
        *((unsigned short *)(&state[local2]) + 1) = copy11;
        local2++;
      }
      local1++;
    }
    output[0UL] = (state[0UL] + 690137161UL) - 1817474074UL;
  }
}
```