ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  lea    D20(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  movdqa D6(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    D28(%rip),%rax
  mov    %rsp,%r14
  lea    0x28(%rsp),%r12
  movaps %xmm0,(%rsp)
  movdqa D11(%rip),%xmm0
  mov    %r14,%rbx
  mov    %r14,%rbp
  mov    %rax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  nopw   0x0(%rax,%rax,1)
L20:
  mov    0x0(%rbp),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %rbp,%r12
  jne    L20
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x9,%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  call   <msort@plt>
  nop
L2:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L2
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D20
type_guessed: string
name: retrieve_data
arguments: data_tag: D6
type_guessed: byte[16]
name: retrieve_data
arguments: data_tag: D28
type_guessed: byte[8]
name: retrieve_data
arguments: data_tag: D11
type_guessed: byte[16]

tool: D20: %d 

tool: {"D6": [0x09,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0xee,0xff,0xff,0xff]}

tool: {"D28": [0x00,0x00,0x00,0x00,0xfe,0xff,0xff,0xff]}

tool: {"D11": [0x03,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x02,0x00,0x00,0x00]}

assistant: ```c
int main(void) {
  int n = 10;
  int test[n];
  test[0] = 9;
  test[1] = 4;
  test[2] = 1;
  test[3] = -18;
  test[4] = 3;
  test[5] = 6;
  test[6] = 7;
  test[7] = 2;
  test[8] = 0;
  test[9] = -2;
  for (int i = 0; i < n; i++) {
    printf("%d ", test[i]);
  }
  printf("\n");
  msort(test, 0, n - 1);
  for (int i = 0; i < n; i++) {
    printf("%d ", test[i]);
  }
  printf("\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<compare_fn_default>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jne    L30
  mov    $0x0,%eax
  jmp    L24
L30:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jae    L19
  mov    $0xffffffff,%eax
  jmp    L24
L19:
  mov    $0x1,%eax
L24:
  pop    %rbp
  ret
```

assistant: ```c
int compare_fn_default(const void *obj1, const void *obj2) {
  if (obj1 == obj2)
    return 0;
  return (obj1 < obj2 ? -1 : 1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<compare_fn_default>:
  endbr64
  cmp    %rsi,%rdi
  je     L1
  sbb    %eax,%eax
  or     $0x1,%eax
  ret
L1:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int compare_fn_default(const void *obj1, const void *obj2) {
  if (obj1 == obj2)
    return 0;
  return (obj1 < obj2 ? -1 : 1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<compare_fn_default>:
  endbr64
  cmp    %rsi,%rdi
  je     L0
  sbb    %eax,%eax
  or     $0x1,%eax
  ret
  nop
L0:
  xor    %eax,%eax
  ret
```

assistant: ```c
int compare_fn_default(const void *obj1, const void *obj2) {
  if (obj1 == obj2)
    return 0;
  return (obj1 < obj2 ? -1 : 1);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<compare_fn_default>:
  endbr64
  cmp    %rsi,%rdi
  je     L4
  sbb    %eax,%eax
  or     $0x1,%eax
  ret
  nop
L4:
  xor    %eax,%eax
  ret
```

assistant: ```c
int compare_fn_default(const void *obj1, const void *obj2) {
  if (obj1 == obj2)
    return 0;
  return (obj1 < obj2 ? -1 : 1);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<vec2_subtract>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  movss  (%rax),%xmm0
  mov    -0x10(%rbp),%rax
  movss  (%rax),%xmm1
  subss  %xmm1,%xmm0
  mov    -0x8(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm1
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  subss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void vec2_subtract(float *dst, float *b) {
  dst[0] = dst[0] - b[0];
  dst[1] = dst[1] - b[1];
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<vec2_subtract>:
  endbr64
  movss  (%rdi),%xmm0
  subss  (%rsi),%xmm0
  movss  %xmm0,(%rdi)
  movss  0x4(%rdi),%xmm0
  subss  0x4(%rsi),%xmm0
  movss  %xmm0,0x4(%rdi)
  ret
```

assistant: ```c
void vec2_subtract(float *dst, float *b) {
  dst[0] = dst[0] - b[0];
  dst[1] = dst[1] - b[1];
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<vec2_subtract>:
  endbr64
  movss  (%rdi),%xmm0
  subss  (%rsi),%xmm0
  movss  %xmm0,(%rdi)
  movss  0x4(%rdi),%xmm0
  subss  0x4(%rsi),%xmm0
  movss  %xmm0,0x4(%rdi)
  ret
```

assistant: ```c
void vec2_subtract(float *dst, float *b) {
  dst[0] = dst[0] - b[0];
  dst[1] = dst[1] - b[1];
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<vec2_subtract>:
  endbr64
  movss  (%rdi),%xmm0
  subss  (%rsi),%xmm0
  movss  %xmm0,(%rdi)
  movss  0x4(%rdi),%xmm0
  subss  0x4(%rsi),%xmm0
  movss  %xmm0,0x4(%rdi)
  ret
```

assistant: ```c
void vec2_subtract(float *dst, float *b) {
  dst[0] = dst[0] - b[0];
  dst[1] = dst[1] - b[1];
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<http_mth>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D19(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcasecmp@plt>
  test   %rax,%rax
  jne    L18
  mov    $0x1,%eax
  jmp    L28
L18:
  mov    -0x8(%rbp),%rax
  lea    D5(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcasecmp@plt>
  test   %rax,%rax
  jne    L48
  mov    $0x2,%eax
  jmp    L28
L48:
  mov    -0x8(%rbp),%rax
  lea    D10(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcasecmp@plt>
  test   %rax,%rax
  jne    L49
  mov    $0x4,%eax
  jmp    L28
L49:
  mov    $0x8,%eax
L28:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D19
datatype: string
name: read_data
arguments: tag: D5
datatype: string
name: read_data
arguments: tag: D10
datatype: string

tool: D19: GET

tool: D5: POST

tool: D10: HEAD

assistant: ```c
int http_mth(char *s) {
  if (strcasecmp(s, "GET") == 0) {
    return 1;
  }
  if (strcasecmp(s, "POST") == 0) {
    return 2;
  }
  if (strcasecmp(s, "HEAD") == 0) {
    return 4;
  }
  return 8;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<http_mth>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  lea    D22(%rip),%rsi
  call   <strcasecmp@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L1
L14:
  pop    %rbx
  ret
L1:
  lea    D14(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcasecmp@plt>
  mov    %rax,%rdx
  mov    $0x2,%eax
  test   %rdx,%rdx
  je     L14
  lea    D28(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcasecmp@plt>
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0xfffffffc,%eax
  add    $0x8,%eax
  jmp    L14
```

tool_calls: name: disassemble_data
arguments: tag: D22
type_infered: string
name: disassemble_data
arguments: tag: D14
type_infered: string
name: disassemble_data
arguments: tag: D28
type_infered: string

tool: D22: GET

tool: D14: POST

tool: D28: HEAD

assistant: ```c
int http_mth(char *s) {
  if (strcasecmp(s, "GET") == 0) {
    return 1;
  }
  if (strcasecmp(s, "POST") == 0) {
    return 2;
  }
  if (strcasecmp(s, "HEAD") == 0) {
    return 4;
  }
  return 8;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<http_mth>:
  endbr64
  push   %rbp
  lea    D16(%rip),%rsi
  mov    %rdi,%rbp
  call   <strcasecmp@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  jne    L9
L13:
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  lea    D9(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcasecmp@plt>
  mov    %rax,%r8
  mov    $0x2,%eax
  test   %r8,%r8
  je     L13
  mov    %rbp,%rdi
  lea    D3(%rip),%rsi
  call   <strcasecmp@plt>
  pop    %rbp
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0xfffffffc,%eax
  add    $0x8,%eax
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D16
type: string
name: disassemble_data
arguments: data_tag: D9
type: string
name: disassemble_data
arguments: data_tag: D3
type: string

tool: D16: GET

tool: D9: POST

tool: D3: HEAD

assistant: ```c
int http_mth(char *s) {
  if (strcasecmp(s, "GET") == 0) {
    return 1;
  }
  if (strcasecmp(s, "POST") == 0) {
    return 2;
  }
  if (strcasecmp(s, "HEAD") == 0) {
    return 4;
  }
  return 8;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<http_mth>:
  endbr64
  push   %rbp
  lea    D15(%rip),%rsi
  mov    %rdi,%rbp
  call   <strcasecmp@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  jne    L9
L6:
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  lea    D28(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcasecmp@plt>
  mov    %rax,%r8
  mov    $0x2,%eax
  test   %r8,%r8
  je     L6
  mov    %rbp,%rdi
  lea    D1(%rip),%rsi
  call   <strcasecmp@plt>
  pop    %rbp
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0xfffffffc,%eax
  add    $0x8,%eax
  ret
```

tool_calls: name: unpack_data
arguments: tag: D15
data_type: string
name: unpack_data
arguments: tag: D28
data_type: string
name: unpack_data
arguments: tag: D1
data_type: string

tool: D15: GET

tool: D28: POST

tool: D1: HEAD

assistant: ```c
int http_mth(char *s) {
  if (strcasecmp(s, "GET") == 0) {
    return 1;
  }
  if (strcasecmp(s, "POST") == 0) {
    return 2;
  }
  if (strcasecmp(s, "HEAD") == 0) {
    return 4;
  }
  return 8;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_region_name_GQ>:
  endbr64
  sub    $0x3,%edi
  cmp    $0x6,%edi
  ja     L7
  mov    %edi,%edi
  lea    D10(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D65(%rip),%rax
  ret
  lea    D42(%rip),%rax
  ret
  lea    D41(%rip),%rax
  ret
  lea    D2(%rip),%rax
  ret
  lea    D30(%rip),%rax
  ret
  lea    D62(%rip),%rax
  ret
L7:
  mov    $0x0,%eax
  ret
  lea    D24(%rip),%rax
  ret
```

tool_calls: name: read_data
arguments: data_tag: D65
type_guessed: string
name: read_data
arguments: data_tag: D42
type_guessed: string
name: read_data
arguments: data_tag: D41
type_guessed: string
name: read_data
arguments: data_tag: D2
type_guessed: string
name: read_data
arguments: data_tag: D30
type_guessed: string
name: read_data
arguments: data_tag: D62
type_guessed: string
name: read_data
arguments: data_tag: D24
type_guessed: string

tool: D65: Bioko Norte

tool: D42: Bioko Sur

tool: D41: Centro Sur

tool: D2: Kie-Ntem

tool: D30: Litoral

tool: D62: Wele-Nzas

tool: D24: Annobon

assistant: ```c
const char *get_region_name_GQ(int region_code) {
  switch (region_code) {
  case 3:
    return "Annobon";
  case 4:
    return "Bioko Norte";
  case 5:
    return "Bioko Sur";
  case 6:
    return "Centro Sur";
  case 7:
    return "Kie-Ntem";
  case 8:
    return "Litoral";
  case 9:
    return "Wele-Nzas";
  default:
    return NULL;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_region_name_GQ>:
  endbr64
  sub    $0x3,%edi
  cmp    $0x6,%edi
  ja     L6
  lea    D49(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nopl   (%rax)
  lea    D75(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D76(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D37(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D36(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D20(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D18(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D11(%rip),%rax
  ret
L6:
  xor    %eax,%eax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D75
type_guessed: string
name: extract_symbolic_data
arguments: tag: D76
type_guessed: string
name: extract_symbolic_data
arguments: tag: D37
type_guessed: string
name: extract_symbolic_data
arguments: tag: D36
type_guessed: string
name: extract_symbolic_data
arguments: tag: D20
type_guessed: string
name: extract_symbolic_data
arguments: tag: D18
type_guessed: string
name: extract_symbolic_data
arguments: tag: D11
type_guessed: string

tool: D75: Bioko Norte

tool: D76: Annobon

tool: D37: Bioko Sur

tool: D36: Centro Sur

tool: D20: Wele-Nzas

tool: D18: Kie-Ntem

tool: D11: Litoral

assistant: ```c
const char *get_region_name_GQ(int region_code) {
  switch (region_code) {
  case 3:
    return "Annobon";
  case 4:
    return "Bioko Norte";
  case 5:
    return "Bioko Sur";
  case 6:
    return "Centro Sur";
  case 7:
    return "Kie-Ntem";
  case 8:
    return "Litoral";
  case 9:
    return "Wele-Nzas";
  default:
    return NULL;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_region_name_GQ>:
  endbr64
  sub    $0x3,%edi
  cmp    $0x6,%edi
  ja     L8
  lea    D36(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nopl   (%rax)
  lea    D30(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D26(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D20(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D78(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D11(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D51(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D67(%rip),%rax
  ret
L8:
  xor    %eax,%eax
  ret
```

tool_calls: name: unpack_data
arguments: tag: D30
type_infered: string
name: unpack_data
arguments: tag: D26
type_infered: string
name: unpack_data
arguments: tag: D20
type_infered: string
name: unpack_data
arguments: tag: D78
type_infered: string
name: unpack_data
arguments: tag: D11
type_infered: string
name: unpack_data
arguments: tag: D51
type_infered: string
name: unpack_data
arguments: tag: D67
type_infered: string

tool: D30: Bioko Norte

tool: D26: Annobon

tool: D20: Bioko Sur

tool: D78: Centro Sur

tool: D11: Wele-Nzas

tool: D51: Kie-Ntem

tool: D67: Litoral

assistant: ```c
const char *get_region_name_GQ(int region_code) {
  switch (region_code) {
  case 3:
    return "Annobon";
  case 4:
    return "Bioko Norte";
  case 5:
    return "Bioko Sur";
  case 6:
    return "Centro Sur";
  case 7:
    return "Kie-Ntem";
  case 8:
    return "Litoral";
  case 9:
    return "Wele-Nzas";
  default:
    return NULL;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<print_vfpu_reg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %ecx,%eax
  mov    %r8,-0x28(%rbp)
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0x20,%eax
  test   %eax,%eax
  je     L13
  mov    -0x14(%rbp),%eax
  and    $0x3,%eax
  mov    %eax,%edi
  mov    -0x14(%rbp),%eax
  sar    $0x2,%eax
  and    $0x7,%eax
  mov    %eax,%ecx
  movsbl -0x20(%rbp),%edx
  mov    -0x18(%rbp),%esi
  mov    -0x28(%rbp),%rax
  mov    %edi,%r9d
  mov    %esi,%r8d
  lea    D10(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L9
L13:
  mov    -0x14(%rbp),%eax
  and    $0x3,%eax
  mov    %eax,%edi
  mov    -0x14(%rbp),%eax
  sar    $0x2,%eax
  and    $0x7,%eax
  mov    %eax,%ecx
  movsbl -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%esi
  mov    -0x28(%rbp),%rax
  mov    %esi,%r9d
  mov    %edi,%r8d
  lea    D10(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    %eax,-0x4(%rbp)
L9:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D10
datatype: string

tool: D10: %c%d%d%d

assistant: ```c
char *print_vfpu_reg(int reg, int offset, char one, char two, char *output) {
  int len;
  if ((reg >> 5) & 1) {
    len = sprintf(output, "%c%d%d%d", two, (reg >> 2) & 7, offset, reg & 3);
  } else {
    len = sprintf(output, "%c%d%d%d", one, (reg >> 2) & 7, reg & 3, offset);
  }
  return output + len;
}
```
user: Translate the below assembly code into C programming language:

```asm
<print_vfpu_reg>:
  endbr64
  push   %rbx
  mov    %r8,%rbx
  test   $0x20,%dil
  je     L15
  mov    %edi,%eax
  sar    $0x2,%eax
  and    $0x7,%eax
  movsbl %cl,%edx
  and    $0x3,%edi
  mov    %edi,%r9d
  mov    %esi,%r8d
  mov    %eax,%ecx
  lea    D5(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
L8:
  cltq
  add    %rbx,%rax
  pop    %rbx
  ret
L15:
  mov    %edi,%ecx
  sar    $0x2,%ecx
  and    $0x7,%ecx
  movsbl %dl,%edx
  mov    %esi,%r9d
  and    $0x3,%edi
  mov    %edi,%r8d
  lea    D5(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  jmp    L8
```

tool_calls: name: unpack_data
arguments: tag: D5
type: string

tool: D5: %c%d%d%d

assistant: ```c
char *print_vfpu_reg(int reg, int offset, char one, char two, char *output) {
  int len;
  if ((reg >> 5) & 1) {
    len = sprintf(output, "%c%d%d%d", two, (reg >> 2) & 7, offset, reg & 3);
  } else {
    len = sprintf(output, "%c%d%d%d", one, (reg >> 2) & 7, reg & 3, offset);
  }
  return output + len;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print_vfpu_reg>:
  endbr64
  mov    %edi,%r10d
  push   %rbx
  mov    %r8,%rbx
  mov    %edi,%r8d
  sar    $0x2,%r10d
  and    $0x3,%r8d
  and    $0x7,%r10d
  and    $0x20,%edi
  je     L3
  movsbl %cl,%edx
  mov    %r8d,%r9d
  mov    %rbx,%rdi
  mov    %esi,%r8d
  mov    %r10d,%ecx
  lea    D11(%rip),%rsi
  call   <sprintf@plt>
  cltq
  add    %rbx,%rax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L3:
  mov    %esi,%r9d
  mov    %rbx,%rdi
  movsbl %dl,%edx
  mov    %r10d,%ecx
  lea    D11(%rip),%rsi
  call   <sprintf@plt>
  cltq
  add    %rbx,%rax
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: tag: D11
type_guessed: string

tool: D11: %c%d%d%d

assistant: ```c
char *print_vfpu_reg(int reg, int offset, char one, char two, char *output) {
  int len;
  if ((reg >> 5) & 1) {
    len = sprintf(output, "%c%d%d%d", two, (reg >> 2) & 7, offset, reg & 3);
  } else {
    len = sprintf(output, "%c%d%d%d", one, (reg >> 2) & 7, reg & 3, offset);
  }
  return output + len;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<print_vfpu_reg>:
  endbr64
  mov    %edi,%r10d
  push   %rbx
  mov    %r8,%rbx
  mov    %edi,%r8d
  sar    $0x2,%r10d
  and    $0x3,%r8d
  and    $0x7,%r10d
  and    $0x20,%edi
  je     L0
  movsbl %cl,%edx
  mov    %r8d,%r9d
  mov    %rbx,%rdi
  mov    %esi,%r8d
  mov    %r10d,%ecx
  lea    D16(%rip),%rsi
  call   <sprintf@plt>
  cltq
  add    %rbx,%rax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L0:
  mov    %esi,%r9d
  mov    %rbx,%rdi
  movsbl %dl,%edx
  mov    %r10d,%ecx
  lea    D16(%rip),%rsi
  call   <sprintf@plt>
  cltq
  add    %rbx,%rax
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D16
type: string

tool: D16: %c%d%d%d

assistant: ```c
char *print_vfpu_reg(int reg, int offset, char one, char two, char *output) {
  int len;
  if ((reg >> 5) & 1) {
    len = sprintf(output, "%c%d%d%d", two, (reg >> 2) & 7, offset, reg & 3);
  } else {
    len = sprintf(output, "%c%d%d%d", one, (reg >> 2) & 7, reg & 3, offset);
  }
  return output + len;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  lea    D18(%rip),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <isSubstr@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D18
type_infered: string
name: retrieve_data
arguments: label: D13
type_infered: string

tool: D18: RuchikD

tool: D13: RuchikRuchikARuchikSRuchikDRuchikM

assistant: ```c
int main(int argc, const char *argv[]) {
  isSubstr("RuchikRuchikARuchikSRuchikDRuchikM", "RuchikD");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D10(%rip),%rsi
  lea    D8(%rip),%rdi
  call   <isSubstr@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D10
data_type: string
name: parse_data
arguments: data_tag: D8
data_type: string

tool: D10: RuchikD

tool: D8: RuchikRuchikARuchikSRuchikDRuchikM

assistant: ```c
int main(int argc, const char *argv[]) {
  isSubstr("RuchikRuchikARuchikSRuchikDRuchikM", "RuchikD");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D10(%rip),%rsi
  lea    D7(%rip),%rdi
  call   <isSubstr@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D10
type_infered: string
name: disassemble_data
arguments: label: D7
type_infered: string

tool: D10: RuchikD

tool: D7: RuchikRuchikARuchikSRuchikDRuchikM

assistant: ```c
int main(int argc, const char *argv[]) {
  isSubstr("RuchikRuchikARuchikSRuchikDRuchikM", "RuchikD");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D18(%rip),%rsi
  lea    D17(%rip),%rdi
  call   <isSubstr@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D18
type: string
name: parse_data
arguments: data_tag: D17
type: string

tool: D18: RuchikD

tool: D17: RuchikRuchikARuchikSRuchikDRuchikM

assistant: ```c
int main(int argc, const char *argv[]) {
  isSubstr("RuchikRuchikARuchikSRuchikDRuchikM", "RuchikD");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x4,-0x10(%rbp)
  movl   $0x2,-0xc(%rbp)
  movl   $0x3,-0x8(%rbp)
  movl   $0x6,-0x4(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  imul   -0x4(%rbp),%eax
  sub    -0x10(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  sub    -0x8(%rbp),%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%eax
  imul   %eax,%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x4(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  add    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int a = 4;
  int b = 2;
  int c = 3;
  int d = 6;
  a = (b + c) * d - a;
  b = a - c + a * a + d / b;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  int a = 4;
  int b = 2;
  int c = 3;
  int d = 6;
  a = (b + c) * d - a;
  b = a - c + a * a + d / b;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int a = 4;
  int b = 2;
  int c = 3;
  int d = 6;
  a = (b + c) * d - a;
  b = a - c + a * a + d / b;
}
```