ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<retab>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x3c(%rbp),%eax
  sar    $0x11,%eax
  and    $0x1,%eax
  mov    %eax,-0x2c(%rbp)
  mov    -0x3c(%rbp),%eax
  sar    $0x10,%eax
  and    $0x1,%eax
  mov    %eax,-0x28(%rbp)
  mov    -0x3c(%rbp),%eax
  movzwl %ax,%eax
  mov    %eax,-0x24(%rbp)
  movq   $0x0,-0x10(%rbp)
  mov    -0x38(%rbp),%rax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <msg@plt>
  mov    -0x24(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <msg@plt>
  cmpl   $0x0,-0x2c(%rbp)
  je     L20
  mov    $0x59,%eax
  jmp    L7
L20:
  mov    $0x4e,%eax
L7:
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <msg@plt>
  cmpl   $0x0,-0x28(%rbp)
  je     L19
  mov    $0x59,%eax
  jmp    L1
L19:
  mov    $0x4e,%eax
L1:
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <msg@plt>
L90:
  mov    -0x38(%rbp),%rdx
  lea    -0x18(%rbp),%rcx
  lea    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <getline@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jle    L6
  cmpl   $0x0,-0x28(%rbp)
  je     L51
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <trim_back@plt>
L51:
  cmpl   $0x0,-0x2c(%rbp)
  je     L73
  mov    -0x20(%rbp),%rax
  mov    -0x24(%rbp),%edx
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <indent_tabs@plt>
  jmp    L61
L73:
  mov    -0x20(%rbp),%rax
  mov    -0x24(%rbp),%edx
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <indent_spaces@plt>
L61:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L90
L6:
  nop
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L41
  call   <__stack_chk_fail@plt>
L41:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D36
type_infered: string
name: read_data
arguments: data_label: D24
type_infered: string
name: read_data
arguments: data_label: D22
type_infered: string
name: read_data
arguments: data_label: D18
type_infered: string

tool: D36: fp         %08X

tool: D24: indent     %d

tool: D22: use tabs   %c

tool: D18: trim right %c

assistant: ```c
void retab(FILE *fp, int flags) {
  int use_tabs = ((1 << 17) & flags) ? 1 : 0;
  int trim_right = ((1 << 16) & flags) ? 1 : 0;
  int indent_size = 0xFFFF & flags;
  ssize_t n = 0;
  char *line;
  size_t len;
  msg("fp         %08X", (void *)fp);
  msg("indent     %d", indent_size);
  msg("use tabs   %c", use_tabs ? 'Y' : 'N');
  msg("trim right %c", trim_right ? 'Y' : 'N');
  for (;;) {
    n = getline(&line, &len, fp);
    if (n <= 0)
      break;
    if (trim_right)
      trim_back(line);
    if (use_tabs)
      indent_tabs(line, indent_size);
    else
      indent_spaces(line, indent_size);
    puts(line);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<retab>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %esi,%r13d
  sar    $0x11,%r13d
  and    $0x1,%r13d
  mov    %esi,%r12d
  sar    $0x10,%r12d
  and    $0x1,%r12d
  movzwl %si,%ebx
  movsbl %dil,%esi
  lea    D24(%rip),%rdi
  call   <msg@plt>
  movsbl %bl,%esi
  lea    D9(%rip),%rdi
  call   <msg@plt>
  cmp    $0x1,%r13d
  sbb    %esi,%esi
  and    $0xfffffff5,%esi
  add    $0x59,%esi
  lea    D33(%rip),%rdi
  call   <msg@plt>
  cmp    $0x1,%r12d
  sbb    %esi,%esi
  and    $0xfffffff5,%esi
  add    $0x59,%esi
  lea    D20(%rip),%rdi
  call   <msg@plt>
  lea    0x10(%rsp),%r14
  jmp    L45
L56:
  mov    0x8(%rsp),%rdi
  call   <trim_back@plt>
  jmp    L24
L5:
  mov    %ebx,%esi
  mov    0x8(%rsp),%rdi
  call   <indent_spaces@plt>
L62:
  mov    0x8(%rsp),%rdi
  call   <puts@plt>
L45:
  lea    0x8(%rsp),%rdi
  mov    %rbp,%rdx
  mov    %r14,%rsi
  call   <getline@plt>
  test   %rax,%rax
  jle    L4
  test   %r12d,%r12d
  jne    L56
L24:
  test   %r13d,%r13d
  je     L5
  mov    %ebx,%esi
  mov    0x8(%rsp),%rdi
  call   <indent_tabs@plt>
  jmp    L62
L4:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D24
datatype: string
name: disassemble_data
arguments: tag: D9
datatype: string
name: disassemble_data
arguments: tag: D33
datatype: string
name: disassemble_data
arguments: tag: D20
datatype: string

tool: D24: fp         %08X

tool: D9: indent     %d

tool: D33: use tabs   %c

tool: D20: trim right %c

assistant: ```c
void retab(FILE *fp, int flags) {
  int use_tabs = ((1 << 17) & flags) ? 1 : 0;
  int trim_right = ((1 << 16) & flags) ? 1 : 0;
  int indent_size = 0xFFFF & flags;
  ssize_t n = 0;
  char *line;
  size_t len;
  msg("fp         %08X", (void *)fp);
  msg("indent     %d", indent_size);
  msg("use tabs   %c", use_tabs ? 'Y' : 'N');
  msg("trim right %c", trim_right ? 'Y' : 'N');
  for (;;) {
    n = getline(&line, &len, fp);
    if (n <= 0)
      break;
    if (trim_right)
      trim_back(line);
    if (use_tabs)
      indent_tabs(line, indent_size);
    else
      indent_spaces(line, indent_size);
    puts(line);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<retab>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  sar    $0x11,%r13d
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  sar    $0x10,%r12d
  and    $0x1,%r13d
  push   %rbx
  movzwl %si,%ebx
  movsbl %dil,%esi
  lea    D16(%rip),%rdi
  and    $0x1,%r12d
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r15
  lea    0x8(%rsp),%r14
  call   <msg@plt>
  movsbl %bl,%esi
  lea    D30(%rip),%rdi
  call   <msg@plt>
  cmp    $0x1,%r13d
  lea    D15(%rip),%rdi
  sbb    %esi,%esi
  and    $0xfffffff5,%esi
  add    $0x59,%esi
  call   <msg@plt>
  cmp    $0x1,%r12d
  lea    D10(%rip),%rdi
  sbb    %esi,%esi
  and    $0xfffffff5,%esi
  add    $0x59,%esi
  call   <msg@plt>
  jmp    L33
  nopl   0x0(%rax)
L1:
  call   <indent_tabs@plt>
L29:
  mov    0x8(%rsp),%rdi
  call   <puts@plt>
L33:
  mov    %rbp,%rdx
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <getline@plt>
  test   %rax,%rax
  jle    L22
  test   %r12d,%r12d
  jne    L61
L31:
  mov    0x8(%rsp),%rdi
  mov    %ebx,%esi
  test   %r13d,%r13d
  jne    L1
  call   <indent_spaces@plt>
  jmp    L29
  nopw   0x0(%rax,%rax,1)
L61:
  mov    0x8(%rsp),%rdi
  call   <trim_back@plt>
  jmp    L31
  nopl   0x0(%rax)
L22:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L62
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L62:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D16
type_guessed: string
name: retrieve_data
arguments: tag: D30
type_guessed: string
name: retrieve_data
arguments: tag: D15
type_guessed: string
name: retrieve_data
arguments: tag: D10
type_guessed: string

tool: D16: fp         %08X

tool: D30: indent     %d

tool: D15: use tabs   %c

tool: D10: trim right %c

assistant: ```c
void retab(FILE *fp, int flags) {
  int use_tabs = ((1 << 17) & flags) ? 1 : 0;
  int trim_right = ((1 << 16) & flags) ? 1 : 0;
  int indent_size = 0xFFFF & flags;
  ssize_t n = 0;
  char *line;
  size_t len;
  msg("fp         %08X", (void *)fp);
  msg("indent     %d", indent_size);
  msg("use tabs   %c", use_tabs ? 'Y' : 'N');
  msg("trim right %c", trim_right ? 'Y' : 'N');
  for (;;) {
    n = getline(&line, &len, fp);
    if (n <= 0)
      break;
    if (trim_right)
      trim_back(line);
    if (use_tabs)
      indent_tabs(line, indent_size);
    else
      indent_spaces(line, indent_size);
    puts(line);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<DoAllMarker>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <DoMarker@plt>
  test   %rax,%rax
  jns    L2
  mov    $0xffffffff,%eax
  jmp    L22
L2:
  nop
L8:
  mov    $0x0,%eax
  call   <ScreenMarker@plt>
  test   %rax,%rax
  jns    L8
  mov    $0x0,%eax
L22:
  pop    %rbp
  ret
```

assistant: ```c
int DoAllMarker() {
  static char RoutineName[] = "DoAllMarker";
  if (DoMarker() < 0) {
    return (-1);
  }
  while (ScreenMarker() >= 0)
    ;
  return (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<DoAllMarker>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <DoMarker@plt>
  test   %rax,%rax
  js     L5
L19:
  mov    $0x0,%eax
  call   <ScreenMarker@plt>
  test   %rax,%rax
  jns    L19
  mov    $0x0,%eax
L12:
  add    $0x8,%rsp
  ret
L5:
  mov    $0xffffffff,%eax
  jmp    L12
```

assistant: ```c
int DoAllMarker() {
  static char RoutineName[] = "DoAllMarker";
  if (DoMarker() < 0) {
    return (-1);
  }
  while (ScreenMarker() >= 0)
    ;
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<DoAllMarker>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <DoMarker@plt>
  test   %rax,%rax
  js     L7
  nopl   0x0(%rax)
L17:
  xor    %eax,%eax
  call   <ScreenMarker@plt>
  test   %rax,%rax
  jns    L17
  xor    %eax,%eax
L13:
  add    $0x8,%rsp
  ret
L7:
  mov    $0xffffffff,%eax
  jmp    L13
```

assistant: ```c
int DoAllMarker() {
  static char RoutineName[] = "DoAllMarker";
  if (DoMarker() < 0) {
    return (-1);
  }
  while (ScreenMarker() >= 0)
    ;
  return (0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<DoAllMarker>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <DoMarker@plt>
  test   %rax,%rax
  js     L29
  nopl   0x0(%rax)
L19:
  xor    %eax,%eax
  call   <ScreenMarker@plt>
  test   %rax,%rax
  jns    L19
  xor    %eax,%eax
L18:
  add    $0x8,%rsp
  ret
L29:
  mov    $0xffffffff,%eax
  jmp    L18
```

assistant: ```c
int DoAllMarker() {
  static char RoutineName[] = "DoAllMarker";
  if (DoMarker() < 0) {
    return (-1);
  }
  while (ScreenMarker() >= 0)
    ;
  return (0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<tick>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  lea    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <clock_gettime@plt>
  mov    -0x20(%rbp),%eax
  imul   $0x3b9aca00,%eax,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
uint64_t tick() {
  struct timespec tp;
  clock_gettime(CLOCK_MONOTONIC, &tp);
  return tp.tv_sec * 1000000000 + tp.tv_nsec;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<tick>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    D7(%rip),%rax
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  imul   $0x3b9aca00,(%rsp),%eax
  cltq
  add    0x8(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x28,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t tick() {
  struct timespec tp;
  clock_gettime(CLOCK_MONOTONIC, &tp);
  return tp.tv_sec * 1000000000 + tp.tv_nsec;
}
```
user: Translate the below assembly code into C programming language:

```asm
<tick>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D1(%rip),%rax
  mov    %rsp,%rsi
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  imul   $0x3b9aca00,(%rsp),%eax
  cltq
  add    0x8(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x28,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t tick() {
  struct timespec tp;
  clock_gettime(CLOCK_MONOTONIC, &tp);
  return tp.tv_sec * 1000000000 + tp.tv_nsec;
}
```
user: Translate the below assembly code into C programming language:

```asm
<tick>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D0(%rip),%rax
  mov    %rsp,%rsi
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  imul   $0x3b9aca00,(%rsp),%eax
  cltq
  add    0x8(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x28,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t tick() {
  struct timespec tp;
  clock_gettime(CLOCK_MONOTONIC, &tp);
  return tp.tv_sec * 1000000000 + tp.tv_nsec;
}
```
user: Translate the below assembly code into C programming language:

```asm
<call_to_mayinit_and_return_good>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x60(%rbp),%rax
  mov    %rax,%rdi
  call   <mayinit_vpp@plt>
  mov    -0x60(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
void *call_to_mayinit_and_return_good() {
  void *obj[10];
  mayinit_vpp((const void **)obj);
  return obj[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
<call_to_mayinit_and_return_good>:
  endbr64
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <mayinit_vpp@plt>
  mov    (%rsp),%rax
  mov    0x58(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x68,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *call_to_mayinit_and_return_good() {
  void *obj[10];
  mayinit_vpp((const void **)obj);
  return obj[0];
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<call_to_mayinit_and_return_good>:
  endbr64
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <mayinit_vpp@plt>
  mov    (%rsp),%rax
  mov    0x58(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x68,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *call_to_mayinit_and_return_good() {
  void *obj[10];
  mayinit_vpp((const void **)obj);
  return obj[0];
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<call_to_mayinit_and_return_good>:
  endbr64
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <mayinit_vpp@plt>
  mov    (%rsp),%rax
  mov    0x58(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x68,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *call_to_mayinit_and_return_good() {
  void *obj[10];
  mayinit_vpp((const void **)obj);
  return obj[0];
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<mul>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jne    L113
  mov    $0x0,%eax
  jmp    L111
L113:
  cmpl   $0x0,-0x4(%rbp)
  jne    L92
  mov    $0x0,%eax
  jmp    L111
L92:
  cmpl   $0x0,-0x8(%rbp)
  jle    L14
  cmpl   $0x0,-0x4(%rbp)
  jle    L14
  mov    -0x8(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <mul>
  mov    -0x4(%rbp),%edx
  add    %edx,%eax
  jmp    L111
L14:
  cmpl   $0x0,-0x4(%rbp)
  jns    L1
  cmpl   $0x0,-0x8(%rbp)
  jns    L1
  mov    -0x8(%rbp),%eax
  not    %eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  neg    %eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <mul>
  sub    -0x4(%rbp),%eax
  jmp    L111
L1:
  cmpl   $0x0,-0x8(%rbp)
  jle    L39
  cmpl   $0x0,-0x4(%rbp)
  jns    L39
  mov    -0x8(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <mul>
  mov    -0x4(%rbp),%edx
  add    %edx,%eax
  jmp    L111
L39:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  mov    %edx,%edi
  call   <mul>
  mov    -0x8(%rbp),%edx
  add    %edx,%eax
L111:
  leave
  ret
```

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

```asm
<mul>:
  endbr64
  test   %edi,%edi
  je     L41
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  test   %esi,%esi
  mov    $0x0,%eax
  je     L54
  setg   %al
  test   %edi,%edi
  jle    L16
  test   %al,%al
  jne    L7
L16:
  mov    %ebx,%edx
  shr    $0x1f,%edx
  test   %ebx,%ebp
  js     L89
  test   %al,%al
  je     L66
  test   %dl,%dl
  je     L66
  lea    -0x1(%rbp),%esi
  mov    %ebx,%edi
  call   <mul>
  add    %ebx,%eax
L54:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L7:
  lea    -0x1(%rsi),%esi
  call   <mul>
  add    %ebx,%eax
  jmp    L54
L89:
  mov    %ebp,%esi
  not    %esi
  mov    %ebx,%edi
  neg    %edi
  call   <mul>
  sub    %ebx,%eax
  jmp    L54
L66:
  lea    -0x1(%rbx),%edi
  mov    %ebp,%esi
  call   <mul>
  add    %ebp,%eax
  jmp    L54
L41:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int mul(int x, int y) {
  if (y == 0)
    return 0;
  else if (x == 0)
    return 0;
  else if (y > 0 && x > 0)
    return (x + mul(x, y - 1));
  else if (x < 0 && y < 0)
    return (-x + mul(-x, -y - 1));
  else if (y > 0 && x < 0)
    return x + mul(x, y - 1);
  else
    return y + mul(x - 1, y);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mul>:
  endbr64
  xor    %eax,%eax
  test   %esi,%esi
  jne    L92
  jmp    L4
  nopl   0x0(%rax)
L32:
  sub    $0x1,%esi
  add    %edi,%eax
L79:
  test   %esi,%esi
  je     L4
L92:
  test   %edi,%edi
  je     L4
  test   %esi,%esi
  setg   %dl
  test   %edi,%edi
  jle    L45
  test   %dl,%dl
  jne    L32
L45:
  mov    %edi,%ecx
  shr    $0x1f,%ecx
  test   %edi,%esi
  jns    L86
  not    %esi
  sub    %edi,%eax
  neg    %edi
  test   %esi,%esi
  jne    L92
L4:
  ret
  xchg   %ax,%ax
L86:
  test   %dl,%dl
  je     L82
  test   %cl,%cl
  jne    L32
L82:
  sub    $0x1,%edi
  add    %esi,%eax
  jmp    L79
```

assistant: ```c
int mul(int x, int y) {
  if (y == 0)
    return 0;
  else if (x == 0)
    return 0;
  else if (y > 0 && x > 0)
    return (x + mul(x, y - 1));
  else if (x < 0 && y < 0)
    return (-x + mul(-x, -y - 1));
  else if (y > 0 && x < 0)
    return x + mul(x, y - 1);
  else
    return y + mul(x - 1, y);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mul>:
  endbr64
  xor    %eax,%eax
  test   %esi,%esi
  je     L25
  test   %edi,%edi
  jne    L39
  jmp    L25
L77:
  sub    $0x1,%esi
  add    %edi,%eax
L11:
  test   %edi,%edi
  je     L25
L112:
  test   %esi,%esi
  je     L119
L39:
  test   %esi,%esi
  setg   %dl
  test   %edi,%edi
  jle    L17
  test   %dl,%dl
  jne    L77
L17:
  mov    %edi,%ecx
  shr    $0x1f,%ecx
  test   %esi,%edi
  jns    L3
  sub    %edi,%eax
  neg    %edi
  not    %esi
  test   %edi,%edi
  jne    L112
L25:
  ret
  xchg   %ax,%ax
L3:
  test   %dl,%dl
  je     L33
  test   %cl,%cl
  jne    L77
L33:
  sub    $0x1,%edi
  add    %esi,%eax
  jmp    L11
  nop
L119:
  ret
```

assistant: ```c
int mul(int x, int y) {
  if (y == 0)
    return 0;
  else if (x == 0)
    return 0;
  else if (y > 0 && x > 0)
    return (x + mul(x, y - 1));
  else if (x < 0 && y < 0)
    return (-x + mul(-x, -y - 1));
  else if (y > 0 && x < 0)
    return x + mul(x, y - 1);
  else
    return y + mul(x - 1, y);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  sub    $0x4d,%eax
  mov    %al,-0x9(%rbp)
  movb   $0x0,-0xa(%rbp)
  nop
  movb   $0x0,-0xa(%rbp)
  nop
  movzbl -0x9(%rbp),%eax
  lea    -0x12(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = (input[0UL] + 914778474UL) - (unsigned char)183;
    local1 = 0UL;
    while (local1 < (unsigned char)0) {
      state[0UL] += state[0UL];
      state[local1] = state[0UL] + state[0UL];
      local1++;
    }
    local1 = 0UL;
    while (local1 < (unsigned char)0) {
      state[local1] += state[local1];
      state[local1] = state[0UL] + state[local1];
      local1++;
    }
    output[0UL] = state[0UL] - 681723154UL;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  sub    $0x5f,%eax
  mov    %al,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = (input[0UL] + 914778474UL) - (unsigned char)183;
    local1 = 0UL;
    while (local1 < (unsigned char)0) {
      state[0UL] += state[0UL];
      state[local1] = state[0UL] + state[0UL];
      local1++;
    }
    local1 = 0UL;
    while (local1 < (unsigned char)0) {
      state[local1] += state[local1];
      state[local1] = state[0UL] + state[local1];
      local1++;
    }
    output[0UL] = state[0UL] - 681723154UL;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  sub    $0x5f,%eax
  mov    %al,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = (input[0UL] + 914778474UL) - (unsigned char)183;
    local1 = 0UL;
    while (local1 < (unsigned char)0) {
      state[0UL] += state[0UL];
      state[local1] = state[0UL] + state[0UL];
      local1++;
    }
    local1 = 0UL;
    while (local1 < (unsigned char)0) {
      state[local1] += state[local1];
      state[local1] = state[0UL] + state[local1];
      local1++;
    }
    output[0UL] = state[0UL] - 681723154UL;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  sub    $0x5f,%eax
  mov    %al,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = (input[0UL] + 914778474UL) - (unsigned char)183;
    local1 = 0UL;
    while (local1 < (unsigned char)0) {
      state[0UL] += state[0UL];
      state[local1] = state[0UL] + state[0UL];
      local1++;
    }
    local1 = 0UL;
    while (local1 < (unsigned char)0) {
      state[local1] += state[local1];
      state[local1] = state[0UL] + state[local1];
      local1++;
    }
    output[0UL] = state[0UL] - 681723154UL;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<display_score>:
  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    $0x28,%ecx
  mov    $0xf0,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <my_gfx_cleararea@plt>
  mov    $0x5c,%edx
  mov    $0x38,%esi
  mov    $0x94,%edi
  call   <gfx_color@plt>
  mov    -0x24(%rbp),%edx
  lea    -0x20(%rbp),%rax
  lea    D5(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    -0x20(%rbp),%rdx
  mov    D11(%rip),%rax
  mov    %rdx,%rsi
  movq   %rax,%xmm0
  mov    $0x28,%edi
  call   <gfx_text@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D5
datatype: string
name: unpack_data
arguments: data_tag: D11
datatype: byte[8]

tool: D5: %d

tool: {"D11": [0x00,0x00,0x00,0x00,0x00,0x00,0x3e,0x40]}

assistant: ```c
void display_score(int score) {
  char score_string[20];
  my_gfx_cleararea(0, 0, 12 * 20, 2 * 20);
  gfx_color(148, 56, 92);
  sprintf(score_string, "%d", score);
  gfx_text(2 * 20, 1.5 * 20, score_string);
}
```