ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<any_directional_flips>:
  endbr64
  movslq %esi,%rsi
  xor    %r8d,%r8d
  lea    0x0(,%rsi,4),%rax
  add    %rax,%rdi
  cmp    %ecx,(%rdi)
  je     L41
  mov    %r8d,%eax
  ret
  nopl   (%rax)
L41:
  add    %rax,%rdi
  mov    (%rdi),%r8d
  cmp    %ecx,%r8d
  je     L12
L67:
  cmp    %r8d,%edx
  sete   %r8b
  movzbl %r8b,%r8d
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  add    %rax,%rdi
  mov    (%rdi),%r8d
  cmp    %ecx,%r8d
  jne    L67
  add    %rax,%rdi
  mov    (%rdi),%r8d
  cmp    %ecx,%r8d
  jne    L67
  add    %rax,%rdi
  mov    (%rdi),%r8d
  cmp    %ecx,%r8d
  jne    L67
  add    %rax,%rdi
  mov    (%rdi),%r8d
  cmp    %ecx,%r8d
  jne    L67
  mov    (%rdi,%rsi,4),%r8d
  jmp    L67
```

assistant: ```c
int any_directional_flips(uint8_t *sq, int inc, int color, int oppcol) {
  uint8_t *pt = sq + inc;
  if (*pt == oppcol) {
    pt += inc;
    if (*pt == oppcol) {
      pt += inc;
      if (*pt == oppcol) {
        pt += inc;
        if (*pt == oppcol) {
          pt += inc;
          if (*pt == oppcol) {
            pt += inc;
            if (*pt == oppcol) {
              pt += inc;
            }
          }
        }
      }
    }
    if (*pt == color)
      return 1;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<any_directional_flips>:
  endbr64
  movslq %esi,%rsi
  xor    %r8d,%r8d
  lea    0x0(,%rsi,4),%rax
  add    %rax,%rdi
  cmp    %ecx,(%rdi)
  je     L23
  mov    %r8d,%eax
  ret
  nopl   (%rax)
L23:
  add    %rax,%rdi
  mov    (%rdi),%r8d
  cmp    %ecx,%r8d
  je     L39
L35:
  cmp    %r8d,%edx
  sete   %r8b
  movzbl %r8b,%r8d
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L39:
  add    %rax,%rdi
  mov    (%rdi),%r8d
  cmp    %ecx,%r8d
  jne    L35
  add    %rax,%rdi
  mov    (%rdi),%r8d
  cmp    %ecx,%r8d
  jne    L35
  add    %rax,%rdi
  mov    (%rdi),%r8d
  cmp    %ecx,%r8d
  jne    L35
  add    %rax,%rdi
  mov    (%rdi),%r8d
  cmp    %ecx,%r8d
  jne    L35
  mov    (%rdi,%rsi,4),%r8d
  jmp    L35
```

assistant: ```c
int any_directional_flips(uint8_t *sq, int inc, int color, int oppcol) {
  uint8_t *pt = sq + inc;
  if (*pt == oppcol) {
    pt += inc;
    if (*pt == oppcol) {
      pt += inc;
      if (*pt == oppcol) {
        pt += inc;
        if (*pt == oppcol) {
          pt += inc;
          if (*pt == oppcol) {
            pt += inc;
            if (*pt == oppcol) {
              pt += inc;
            }
          }
        }
      }
    }
    if (*pt == color)
      return 1;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movl   $0x0,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  mov    $0x3e8,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  lea    D66(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D55(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L70
L67:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L32
  addl   $0x1,-0x18(%rbp)
  jmp    L71
L32:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  jne    L29
  addl   $0x1,-0x10(%rbp)
  jmp    L71
L29:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L15
  addl   $0x1,-0x14(%rbp)
  jmp    L71
L15:
  addl   $0x1,-0xc(%rbp)
L71:
  addq   $0x1,-0x8(%rbp)
L70:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L67
  lea    D61(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D62(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D60(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D66
data_type: string
name: unpack_data
arguments: data_tag: D55
data_type: string
name: unpack_data
arguments: data_tag: D61
data_type: string
name: unpack_data
arguments: data_tag: D16
data_type: string
name: unpack_data
arguments: data_tag: D62
data_type: string
name: unpack_data
arguments: data_tag: D31
data_type: string
name: unpack_data
arguments: data_tag: D60
data_type: string

tool: D66: Enter any string: 

tool: D55: %[^'!']

tool: D61: The Result: 

tool: D16: 	Blank:   %d


tool: D62: 	Tab:     %d


tool: D31: 	Newline: %d


tool: D60: 	Unknow:  %d


assistant: ```c
int main(void) {
  int cb, cn, ct, unknown;
  cb = cn = ct = unknown = 0;
  char *str = malloc(1000);
  printf("Enter any string: ");
  scanf("%[^'!']", str);
  while (*str) {
    if (*str == ' ') {
      cb++;
    } else if (*str == '\t') {
      ct++;
    } else if (*str == '\n') {
      cn++;
    } else
      unknown++;
    str++;
  }
  printf("The Result: \n");
  printf("\tBlank:   %d\n", cb);
  printf("\tTab:     %d\n", ct);
  printf("\tNewline: %d\n", cn);
  printf("\tUnknow:  %d\n", unknown);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x3e8,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  lea    D47(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D15(%rip),%rdi
  call   <scanf@plt>
  movzbl (%rbx),%eax
  test   %al,%al
  je     L75
  mov    $0x0,%r12d
  mov    $0x0,%r13d
  mov    $0x0,%r14d
  mov    $0x0,%ebp
  jmp    L34
L71:
  add    $0x1,%ebp
L19:
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  test   %al,%al
  je     L45
L34:
  cmp    $0x20,%al
  je     L71
  cmp    $0x9,%al
  je     L53
  cmp    $0xa,%al
  je     L83
  add    $0x1,%r12d
  jmp    L19
L53:
  add    $0x1,%r13d
  jmp    L19
L83:
  add    $0x1,%r14d
  jmp    L19
L75:
  mov    $0x0,%r12d
  mov    $0x0,%r13d
  mov    $0x0,%r14d
  mov    $0x0,%ebp
L45:
  lea    D19(%rip),%rdi
  call   <puts@plt>
  mov    %ebp,%esi
  lea    D53(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r13d,%esi
  lea    D67(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r14d,%esi
  lea    D69(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r12d,%esi
  lea    D65(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D47
type: string
name: parse_data
arguments: data_tag: D15
type: string
name: parse_data
arguments: data_tag: D19
type: string
name: parse_data
arguments: data_tag: D53
type: string
name: parse_data
arguments: data_tag: D67
type: string
name: parse_data
arguments: data_tag: D69
type: string
name: parse_data
arguments: data_tag: D65
type: string

tool: D47: Enter any string: 

tool: D15: %[^'!']

tool: D19: The Result: 

tool: D53: 	Blank:   %d


tool: D67: 	Tab:     %d


tool: D69: 	Newline: %d


tool: D65: 	Unknow:  %d


assistant: ```c
int main(void) {
  int cb, cn, ct, unknown;
  cb = cn = ct = unknown = 0;
  char *str = malloc(1000);
  printf("Enter any string: ");
  scanf("%[^'!']", str);
  while (*str) {
    if (*str == ' ') {
      cb++;
    } else if (*str == '\t') {
      ct++;
    } else if (*str == '\n') {
      cn++;
    } else
      unknown++;
    str++;
  }
  printf("The Result: \n");
  printf("\tBlank:   %d\n", cb);
  printf("\tTab:     %d\n", ct);
  printf("\tNewline: %d\n", cn);
  printf("\tUnknow:  %d\n", unknown);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x3e8,%edi
  xor    %r14d,%r14d
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  call   <malloc@plt>
  lea    D35(%rip),%rdi
  mov    %rax,%rbx
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D34(%rip),%rdi
  call   <scanf@plt>
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L29
  jmp    L73
  nopw   0x0(%rax,%rax,1)
L58:
  cmp    $0x9,%al
  je     L44
  cmp    $0xa,%al
  je     L50
  add    $0x1,%r12d
L14:
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  test   %al,%al
  je     L73
L29:
  cmp    $0x20,%al
  jne    L58
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  add    $0x1,%ebp
  test   %al,%al
  jne    L29
L73:
  lea    D6(%rip),%rdi
  call   <puts@plt>
  mov    %ebp,%esi
  lea    D65(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13d,%esi
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r14d,%esi
  lea    D46(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12d,%esi
  lea    D4(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L44:
  add    $0x1,%r13d
  jmp    L14
  nopl   0x0(%rax)
L50:
  add    $0x1,%r14d
  jmp    L14
```

tool_calls: name: disassemble_data
arguments: data_label: D35
type_guessed: string
name: disassemble_data
arguments: data_label: D34
type_guessed: string
name: disassemble_data
arguments: data_label: D6
type_guessed: string
name: disassemble_data
arguments: data_label: D65
type_guessed: string
name: disassemble_data
arguments: data_label: D14
type_guessed: string
name: disassemble_data
arguments: data_label: D46
type_guessed: string
name: disassemble_data
arguments: data_label: D4
type_guessed: string

tool: D35: Enter any string: 

tool: D34: %[^'!']

tool: D6: The Result: 

tool: D65: 	Blank:   %d


tool: D14: 	Tab:     %d


tool: D46: 	Newline: %d


tool: D4: 	Unknow:  %d


assistant: ```c
int main(void) {
  int cb, cn, ct, unknown;
  cb = cn = ct = unknown = 0;
  char *str = malloc(1000);
  printf("Enter any string: ");
  scanf("%[^'!']", str);
  while (*str) {
    if (*str == ' ') {
      cb++;
    } else if (*str == '\t') {
      ct++;
    } else if (*str == '\n') {
      cn++;
    } else
      unknown++;
    str++;
  }
  printf("The Result: \n");
  printf("\tBlank:   %d\n", cb);
  printf("\tTab:     %d\n", ct);
  printf("\tNewline: %d\n", cn);
  printf("\tUnknow:  %d\n", unknown);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x3e8,%edi
  xor    %r14d,%r14d
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  call   <malloc@plt>
  lea    D61(%rip),%rdi
  mov    %rax,%rbx
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L65
  jmp    L76
  nopw   0x0(%rax,%rax,1)
L9:
  cmp    $0x9,%al
  je     L79
  cmp    $0xa,%al
  je     L71
  add    $0x1,%r12d
L84:
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  test   %al,%al
  je     L76
L65:
  cmp    $0x20,%al
  jne    L9
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  add    $0x1,%ebp
  test   %al,%al
  jne    L65
L76:
  lea    D13(%rip),%rdi
  call   <puts@plt>
  mov    %ebp,%esi
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13d,%esi
  lea    D1(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r14d,%esi
  lea    D66(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12d,%esi
  lea    D47(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L79:
  add    $0x1,%r13d
  jmp    L84
  nopl   0x0(%rax)
L71:
  add    $0x1,%r14d
  jmp    L84
```

tool_calls: name: parse_data
arguments: label: D61
type_guessed: string
name: parse_data
arguments: label: D24
type_guessed: string
name: parse_data
arguments: label: D13
type_guessed: string
name: parse_data
arguments: label: D38
type_guessed: string
name: parse_data
arguments: label: D1
type_guessed: string
name: parse_data
arguments: label: D66
type_guessed: string
name: parse_data
arguments: label: D47
type_guessed: string

tool: D61: Enter any string: 

tool: D24: %[^'!']

tool: D13: The Result: 

tool: D38: 	Blank:   %d


tool: D1: 	Tab:     %d


tool: D66: 	Newline: %d


tool: D47: 	Unknow:  %d


assistant: ```c
int main(void) {
  int cb, cn, ct, unknown;
  cb = cn = ct = unknown = 0;
  char *str = malloc(1000);
  printf("Enter any string: ");
  scanf("%[^'!']", str);
  while (*str) {
    if (*str == ' ') {
      cb++;
    } else if (*str == '\t') {
      ct++;
    } else if (*str == '\n') {
      cn++;
    } else
      unknown++;
    str++;
  }
  printf("The Result: \n");
  printf("\tBlank:   %d\n", cb);
  printf("\tTab:     %d\n", ct);
  printf("\tNewline: %d\n", cn);
  printf("\tUnknow:  %d\n", unknown);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<soft_left_degrees>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    $0x0,%eax
  call   <soft_left@plt>
  shll   -0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <angle_rotate@plt>
  nop
  leave
  ret
```

assistant: ```c
void soft_left_degrees(unsigned int Degrees) {
  soft_left();
  Degrees = Degrees * 2;
  angle_rotate(Degrees);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<soft_left_degrees>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0x0,%eax
  call   <soft_left@plt>
  lea    (%rbx,%rbx,1),%edi
  call   <angle_rotate@plt>
  pop    %rbx
  ret
```

assistant: ```c
void soft_left_degrees(unsigned int Degrees) {
  soft_left();
  Degrees = Degrees * 2;
  angle_rotate(Degrees);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<soft_left_degrees>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  mov    %edi,%ebx
  call   <soft_left@plt>
  lea    (%rbx,%rbx,1),%edi
  pop    %rbx
  jmp    1070 <angle_rotate@plt>
```

assistant: ```c
void soft_left_degrees(unsigned int Degrees) {
  soft_left();
  Degrees = Degrees * 2;
  angle_rotate(Degrees);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<soft_left_degrees>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  mov    %edi,%ebx
  call   <soft_left@plt>
  lea    (%rbx,%rbx,1),%edi
  pop    %rbx
  jmp    1070 <angle_rotate@plt>
```

assistant: ```c
void soft_left_degrees(unsigned int Degrees) {
  soft_left();
  Degrees = Degrees * 2;
  angle_rotate(Degrees);
}
```
user: Translate the below assembly code into C programming language:

```asm
<LimitSym>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L9
  mov    -0x8(%rbp),%eax
  jmp    L27
L9:
  mov    -0x8(%rbp),%eax
  neg    %eax
  cmp    %eax,-0x4(%rbp)
  jge    L18
  mov    -0x8(%rbp),%eax
  neg    %eax
  jmp    L27
L18:
  mov    -0x4(%rbp),%eax
L27:
  pop    %rbp
  ret
```

assistant: ```c
int LimitSym(int in, int max) {
  if (in > max)
    return (max);
  if (in < -max)
    return (-max);
  return (in);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<LimitSym>:
  endbr64
  mov    %esi,%eax
  neg    %eax
  cmp    %edi,%eax
  cmovl  %edi,%eax
  cmp    %esi,%edi
  cmovg  %esi,%eax
  ret
```

assistant: ```c
int LimitSym(int in, int max) {
  if (in > max)
    return (max);
  if (in < -max)
    return (-max);
  return (in);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<LimitSym>:
  endbr64
  mov    %esi,%eax
  neg    %eax
  cmp    %edi,%eax
  cmovl  %edi,%eax
  cmp    %esi,%edi
  cmovg  %esi,%eax
  ret
```

assistant: ```c
int LimitSym(int in, int max) {
  if (in > max)
    return (max);
  if (in < -max)
    return (-max);
  return (in);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<LimitSym>:
  endbr64
  mov    %esi,%eax
  neg    %eax
  cmp    %edi,%eax
  cmovl  %edi,%eax
  cmp    %esi,%edi
  cmovg  %esi,%eax
  ret
```

assistant: ```c
int LimitSym(int in, int max) {
  if (in > max)
    return (max);
  if (in < -max)
    return (-max);
  return (in);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<DES_crypt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  lea    D8(%rip),%rdx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <DES_fcrypt@plt>
  leave
  ret
```

assistant: ```c
char *DES_crypt(const char *buf, const char *salt) {
  static char buff[14];
  return DES_fcrypt(buf, salt, buff);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<DES_crypt>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdx
  call   <DES_fcrypt@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
char *DES_crypt(const char *buf, const char *salt) {
  static char buff[14];
  return DES_fcrypt(buf, salt, buff);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<DES_crypt>:
  endbr64
  lea    D4(%rip),%rdx
  jmp    1050 <DES_fcrypt@plt>
```

assistant: ```c
char *DES_crypt(const char *buf, const char *salt) {
  static char buff[14];
  return DES_fcrypt(buf, salt, buff);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<DES_crypt>:
  endbr64
  lea    D0(%rip),%rdx
  jmp    1050 <DES_fcrypt@plt>
```

assistant: ```c
char *DES_crypt(const char *buf, const char *salt) {
  static char buff[14];
  return DES_fcrypt(buf, salt, buff);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  imul   $0xffffff9c,%eax,%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  imul   $0xffffff9c,%eax,%edx
  mov    -0x18(%rbp),%eax
  add    %eax,%edx
  mov    -0x10(%rbp),%eax
  imul   $0xfffffff6,%eax,%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D24
type_guessed: string
name: unpack_data
arguments: data_label: D3
type_guessed: string
name: unpack_data
arguments: data_label: D27
type_guessed: string

tool: D24: What's the 3 digit number you wish reversed? 

tool: D3: %3d

tool: D27: Your number reversed is %1d%1d%1d.


assistant: ```c
int main(void) {
  int input_number, first_digit, second_digit, third_digit;
  printf("What's the 3 digit number you wish reversed? ");
  scanf("%3d", &input_number);
  first_digit = input_number / 100;
  second_digit = (input_number - first_digit * 100) / 10;
  third_digit = input_number - first_digit * 100 - second_digit * 10;
  printf("Your number reversed is %1d%1d%1d.\n", third_digit, second_digit, first_digit);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D29(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  movslq %edx,%rcx
  imul   $0x51eb851f,%rcx,%rcx
  sar    $0x25,%rcx
  mov    %edx,%eax
  sar    $0x1f,%eax
  sub    %eax,%ecx
  imul   $0xffffff9c,%ecx,%eax
  add    %edx,%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%esi
  sar    $0x1f,%esi
  sub    %esi,%edx
  imul   $0xfffffff6,%edx,%esi
  add    %eax,%esi
  lea    D10(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D0
data_type: string
name: retrieve_data
arguments: label: D29
data_type: string
name: retrieve_data
arguments: label: D10
data_type: string

tool: D0: What's the 3 digit number you wish reversed? 

tool: D29: %3d

tool: D10: Your number reversed is %1d%1d%1d.


assistant: ```c
int main(void) {
  int input_number, first_digit, second_digit, third_digit;
  printf("What's the 3 digit number you wish reversed? ");
  scanf("%3d", &input_number);
  first_digit = input_number / 100;
  second_digit = (input_number - first_digit * 100) / 10;
  third_digit = input_number - first_digit * 100 - second_digit * 10;
  printf("Your number reversed is %1d%1d%1d.\n", third_digit, second_digit, first_digit);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D25(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D28(%rip),%rdi
  call   <scanf@plt>
  movslq 0x4(%rsp),%rcx
  lea    D0(%rip),%rdi
  mov    %rcx,%rdx
  imul   $0x51eb851f,%rcx,%rcx
  mov    %edx,%eax
  sar    $0x1f,%eax
  sar    $0x25,%rcx
  sub    %eax,%ecx
  imul   $0xffffff9c,%ecx,%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    %eax,%esi
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%esi
  sar    $0x22,%rdx
  sub    %esi,%edx
  imul   $0xfffffff6,%edx,%esi
  add    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D25
type: string
name: retrieve_data
arguments: tag: D28
type: string
name: retrieve_data
arguments: tag: D0
type: string

tool: D25: What's the 3 digit number you wish reversed? 

tool: D28: %3d

tool: D0: Your number reversed is %1d%1d%1d.


assistant: ```c
int main(void) {
  int input_number, first_digit, second_digit, third_digit;
  printf("What's the 3 digit number you wish reversed? ");
  scanf("%3d", &input_number);
  first_digit = input_number / 100;
  second_digit = (input_number - first_digit * 100) / 10;
  third_digit = input_number - first_digit * 100 - second_digit * 10;
  printf("Your number reversed is %1d%1d%1d.\n", third_digit, second_digit, first_digit);
  return 0;
}
```