ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<meta_db_string_hash>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    $0x8,%edx
  mov    %rdi,%rsi
  mov    %rbx,%rdi
  call   <strncpy@plt>
  mov    %rbx,%rbp
  lea    0x8(%rsp),%r12
  mov    %rbx,%rax
  jmp    L30
L34:
  add    $0x1,%rax
  cmp    %r12,%rax
  je     L67
L30:
  movzbl (%rax),%edx
  lea    -0x61(%rdx),%ecx
  cmp    $0x19,%cl
  ja     L34
  sub    $0x20,%edx
  mov    %dl,(%rax)
  jmp    L34
L67:
  mov    %rsp,%rdi
  lea    D4(%rip),%rsi
  call   <strcmp@plt>
  mov    %rax,%rdx
  mov    $0x38,%ecx
  mov    $0x0,%eax
  test   %rdx,%rdx
  je     L76
L29:
  movsbl 0x0(%rbp),%edx
  shl    %cl,%edx
  add    %edx,%eax
  add    $0x1,%rbp
  sub    $0x8,%ecx
  cmp    %r12,%rbp
  jne    L29
L71:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L76:
  lea    0x4(%r13),%rdi
  call   <meta_db_string_hash>
  jmp    L71
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D4
type: string

tool: D4: THE 

assistant: ```c
uint64_t meta_db_string_hash(char *input) {
  uint64_t hash = 0;
  char temp[8];
  int i;
  strncpy(temp, input, 8);
  for (i = 0; i < 8; i++) {
    if ((temp[i] >= 'a') && (temp[i] <= 'z')) {
      temp[i] -= ('a' - 'A');
    }
  }
  if (strcmp(temp, "THE ") == 0) {
    return meta_db_string_hash(&input[4]);
  }
  for (i = 0; i < 8; i++) {
    hash += (temp[i] << (8 * (7 - i)));
  }
  return hash;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<meta_db_string_hash>:
  endbr64
  push   %r13
  mov    %rdi,%rsi
  mov    $0x8,%edx
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  lea    0x8(%rsp),%r12
  mov    %r13,%rdi
  mov    %r13,%rbx
  call   <strncpy@plt>
  mov    %r13,%rax
  nop
L33:
  movzbl (%rax),%edx
  lea    -0x61(%rdx),%ecx
  cmp    $0x19,%cl
  ja     L58
  sub    $0x20,%edx
  mov    %dl,(%rax)
L58:
  add    $0x1,%rax
  cmp    %r12,%rax
  jne    L33
  lea    D8(%rip),%rsi
  mov    %r13,%rdi
  call   <strcmp@plt>
  mov    $0x38,%ecx
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  je     L14
  nopw   0x0(%rax,%rax,1)
L20:
  movsbl (%rbx),%edx
  add    $0x1,%rbx
  shl    %cl,%edx
  sub    $0x8,%ecx
  add    %edx,%eax
  cmp    %r12,%rbx
  jne    L20
L13:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L51
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L14:
  lea    0x4(%rbp),%rdi
  call   <meta_db_string_hash>
  jmp    L13
L51:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D8
type: string

tool: D8: THE 

assistant: ```c
uint64_t meta_db_string_hash(char *input) {
  uint64_t hash = 0;
  char temp[8];
  int i;
  strncpy(temp, input, 8);
  for (i = 0; i < 8; i++) {
    if ((temp[i] >= 'a') && (temp[i] <= 'z')) {
      temp[i] -= ('a' - 'A');
    }
  }
  if (strcmp(temp, "THE ") == 0) {
    return meta_db_string_hash(&input[4]);
  }
  for (i = 0; i < 8; i++) {
    hash += (temp[i] << (8 * (7 - i)));
  }
  return hash;
}
```
user: What is the c source code of the assembly code below?

```asm
<meta_db_string_hash>:
  endbr64
  push   %rbp
  mov    $0x8,%edx
  mov    %rdi,%rsi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <strncpy@plt>
  movzbl (%rsp),%eax
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L30
  sub    $0x20,%eax
  mov    %al,(%rsp)
L30:
  movzbl 0x1(%rsp),%eax
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L64
  sub    $0x20,%eax
  mov    %al,0x1(%rsp)
L64:
  movzbl 0x2(%rsp),%eax
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L29
  sub    $0x20,%eax
  mov    %al,0x2(%rsp)
L29:
  movzbl 0x3(%rsp),%eax
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L100
  sub    $0x20,%eax
  mov    %al,0x3(%rsp)
L100:
  movzbl 0x4(%rsp),%eax
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L80
  sub    $0x20,%eax
  mov    %al,0x4(%rsp)
L80:
  movzbl 0x5(%rsp),%eax
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L11
  sub    $0x20,%eax
  mov    %al,0x5(%rsp)
L11:
  movzbl 0x6(%rsp),%eax
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L0
  sub    $0x20,%eax
  mov    %al,0x6(%rsp)
L0:
  movzbl 0x7(%rsp),%eax
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L94
  sub    $0x20,%eax
  mov    %al,0x7(%rsp)
L94:
  lea    D8(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L104
  movsbl 0x1(%rsp),%eax
  mov    $0x30,%ecx
  movsbl (%rsp),%edx
  shl    %cl,%eax
  mov    $0x38,%ecx
  shl    %cl,%edx
  mov    $0x28,%ecx
  add    %edx,%eax
  movsbl 0x2(%rsp),%edx
  shl    %cl,%edx
  mov    $0x20,%ecx
  add    %edx,%eax
  movsbl 0x3(%rsp),%edx
  shl    %cl,%edx
  add    %edx,%eax
  add    0x7(%rsp),%al
L53:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L104:
  lea    0x4(%rbx),%rdi
  call   <meta_db_string_hash>
  jmp    L53
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D8
type: string

tool: D8: THE 

assistant: ```c
uint64_t meta_db_string_hash(char *input) {
  uint64_t hash = 0;
  char temp[8];
  int i;
  strncpy(temp, input, 8);
  for (i = 0; i < 8; i++) {
    if ((temp[i] >= 'a') && (temp[i] <= 'z')) {
      temp[i] -= ('a' - 'A');
    }
  }
  if (strcmp(temp, "THE ") == 0) {
    return meta_db_string_hash(&input[4]);
  }
  for (i = 0; i < 8; i++) {
    hash += (temp[i] << (8 * (7 - i)));
  }
  return hash;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<read_ushort>:
  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),%rdx
  lea    -0xa(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x1,%edx
  mov    $0x2,%esi
  mov    %rax,%rdi
  call   <fread@plt>
  cmp    $0x1,%eax
  je     L4
  mov    $0x0,%eax
  jmp    L24
L4:
  movzbl -0xa(%rbp),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x9(%rbp),%eax
  movzbl %al,%eax
  add    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %dx,(%rax)
  mov    $0x1,%eax
L24:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
int read_ushort(FILE *file, unsigned short *shortp) {
  unsigned char file_short[2];
  if (fread((char *)file_short, (int)sizeof(file_short), 1, file) != 1)
    return 0;
  *shortp = file_short[0] * 256 + file_short[1];
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<read_ushort>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rcx
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rdi
  mov    $0x1,%edx
  mov    $0x2,%esi
  call   <fread@plt>
  cmp    $0x1,%eax
  jne    L15
  movzbl 0x6(%rsp),%edx
  shl    $0x8,%edx
  movzbl 0x7(%rsp),%ecx
  add    %ecx,%edx
  mov    %dx,(%rbx)
L24:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x10,%rsp
  pop    %rbx
  ret
L15:
  mov    $0x0,%eax
  jmp    L24
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_ushort(FILE *file, unsigned short *shortp) {
  unsigned char file_short[2];
  if (fread((char *)file_short, (int)sizeof(file_short), 1, file) != 1)
    return 0;
  *shortp = file_short[0] * 256 + file_short[1];
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<read_ushort>:
  endbr64
  push   %rbx
  mov    %rdi,%rcx
  mov    %rsi,%rbx
  mov    $0x1,%edx
  mov    $0x2,%esi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rdi
  call   <fread@plt>
  cmp    $0x1,%eax
  jne    L28
  movzbl 0x6(%rsp),%edx
  movzbl 0x7(%rsp),%ecx
  shl    $0x8,%edx
  add    %ecx,%edx
  mov    %dx,(%rbx)
L10:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L24
  add    $0x10,%rsp
  pop    %rbx
  ret
L28:
  xor    %eax,%eax
  jmp    L10
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_ushort(FILE *file, unsigned short *shortp) {
  unsigned char file_short[2];
  if (fread((char *)file_short, (int)sizeof(file_short), 1, file) != 1)
    return 0;
  *shortp = file_short[0] * 256 + file_short[1];
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<read_ushort>:
  endbr64
  push   %rbx
  mov    %rdi,%rcx
  mov    %rsi,%rbx
  mov    $0x1,%edx
  mov    $0x2,%esi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rdi
  call   <fread@plt>
  cmp    $0x1,%eax
  jne    L7
  movzbl 0x6(%rsp),%edx
  movzbl 0x7(%rsp),%ecx
  shl    $0x8,%edx
  add    %ecx,%edx
  mov    %dx,(%rbx)
L10:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x10,%rsp
  pop    %rbx
  ret
L7:
  xor    %eax,%eax
  jmp    L10
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_ushort(FILE *file, unsigned short *shortp) {
  unsigned char file_short[2];
  if (fread((char *)file_short, (int)sizeof(file_short), 1, file) != 1)
    return 0;
  *shortp = file_short[0] * 256 + file_short[1];
  return 1;
}
```
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
  movl   $0x3,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,-0x4(%rbp)
  notl   -0x4(%rbp)
  negl   -0x4(%rbp)
  cmpl   $0x2,-0x4(%rbp)
  je     L4
  mov    $0x1,%eax
  jmp    L19
L4:
  mov    $0x0,%eax
L19:
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int x;
  x = 3;
  x = !x;
  x = !x;
  x = ~x;
  x = -x;
  if (x != 2)
    return 1;
  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
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  int x;
  x = 3;
  x = !x;
  x = !x;
  x = ~x;
  x = -x;
  if (x != 2)
    return 1;
  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
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int x;
  x = 3;
  x = !x;
  x = !x;
  x = ~x;
  x = -x;
  if (x != 2)
    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
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int x;
  x = 3;
  x = !x;
  x = !x;
  x = ~x;
  x = -x;
  if (x != 2)
    return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<subsys_network_initialize>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <network_init@plt>
  test   %rax,%rax
  jns    L14
  mov    $0xffffffff,%eax
  jmp    L4
L14:
  mov    $0x0,%eax
L4:
  pop    %rbp
  ret
```

assistant: ```c
int subsys_network_initialize(void) {
  if (network_init() < 0)
    return -1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<subsys_network_initialize>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <network_init@plt>
  sar    $0x3f,%rax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int subsys_network_initialize(void) {
  if (network_init() < 0)
    return -1;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<subsys_network_initialize>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <network_init@plt>
  add    $0x8,%rsp
  sar    $0x3f,%rax
  ret
```

assistant: ```c
int subsys_network_initialize(void) {
  if (network_init() < 0)
    return -1;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<subsys_network_initialize>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <network_init@plt>
  add    $0x8,%rsp
  sar    $0x3f,%rax
  ret
```

assistant: ```c
int subsys_network_initialize(void) {
  if (network_init() < 0)
    return -1;
  return 0;
}
```
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
  movw   $0x4240,-0x2(%rbp)
  movswl -0x2(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D9
datatype: string

tool: D9: %d


assistant: ```c
int main(void) {
  short a = 1000 * 1000;
  printf("%d\n", a);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4240,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
type_infered: string

tool: D5: %d


assistant: ```c
int main(void) {
  short a = 1000 * 1000;
  printf("%d\n", a);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4240,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D9
datatype: string

tool: D9: %d


assistant: ```c
int main(void) {
  short a = 1000 * 1000;
  printf("%d\n", a);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4240,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D3
data_type: string

tool: D3: %d


assistant: ```c
int main(void) {
  short a = 1000 * 1000;
  printf("%d\n", a);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<framecpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x14(%rbp),%eax
  leave
  ret
```

assistant: ```c
unsigned framecpy(unsigned char *addr, unsigned char *data, unsigned len) {
  memcpy(addr, data, len);
  return len;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<framecpy>:
  endbr64
  push   %rbx
  mov    %edx,%ebx
  call   <memcpy@plt>
  mov    %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
unsigned framecpy(unsigned char *addr, unsigned char *data, unsigned len) {
  memcpy(addr, data, len);
  return len;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<framecpy>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  call   <memcpy@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
unsigned framecpy(unsigned char *addr, unsigned char *data, unsigned len) {
  memcpy(addr, data, len);
  return len;
}
```
user: What is the c source code of the assembly code below:

```asm
<framecpy>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  call   <memcpy@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
unsigned framecpy(unsigned char *addr, unsigned char *data, unsigned len) {
  memcpy(addr, data, len);
  return len;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rotr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rotr(int *a, int *b, int *c) {
  int dnum = *c;
  *c = *b;
  *b = *a;
  *a = dnum;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<rotr>:
  endbr64
  mov    (%rdx),%eax
  mov    (%rsi),%ecx
  mov    %ecx,(%rdx)
  mov    (%rdi),%edx
  mov    %edx,(%rsi)
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void rotr(int *a, int *b, int *c) {
  int dnum = *c;
  *c = *b;
  *b = *a;
  *a = dnum;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<rotr>:
  endbr64
  mov    (%rsi),%ecx
  mov    (%rdx),%eax
  mov    %ecx,(%rdx)
  mov    (%rdi),%edx
  mov    %edx,(%rsi)
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void rotr(int *a, int *b, int *c) {
  int dnum = *c;
  *c = *b;
  *b = *a;
  *a = dnum;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rotr>:
  endbr64
  mov    (%rsi),%ecx
  mov    (%rdx),%eax
  mov    %ecx,(%rdx)
  mov    (%rdi),%edx
  mov    %edx,(%rsi)
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void rotr(int *a, int *b, int *c) {
  int dnum = *c;
  *c = *b;
  *b = *a;
  *a = dnum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L9
L1:
  mov    -0x8(%rbp),%eax
  add    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L9:
  cmpl   $0x14,-0x8(%rbp)
  jle    L1
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D3
type_guessed: string

tool: D3: Suma liczb naturalnych od 1 do 20 to %d


assistant: ```c
int main() {
  int i, sum = 0;
  for (i = 1; i <= 20; i++) {
    sum = sum + i;
  }
  printf("Suma liczb naturalnych od 1 do 20 to %d\n", sum);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x14,%eax
L7:
  sub    $0x1,%eax
  jne    L7
  mov    $0xd2,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D8
type_infered: string

tool: D8: Suma liczb naturalnych od 1 do 20 to %d


assistant: ```c
int main() {
  int i, sum = 0;
  for (i = 1; i <= 20; i++) {
    sum = sum + i;
  }
  printf("Suma liczb naturalnych od 1 do 20 to %d\n", sum);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xd2,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D7
type_guessed: string

tool: D7: Suma liczb naturalnych od 1 do 20 to %d


assistant: ```c
int main() {
  int i, sum = 0;
  for (i = 1; i <= 20; i++) {
    sum = sum + i;
  }
  printf("Suma liczb naturalnych od 1 do 20 to %d\n", sum);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xd2,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D3
type: string

tool: D3: Suma liczb naturalnych od 1 do 20 to %d


assistant: ```c
int main() {
  int i, sum = 0;
  for (i = 1; i <= 20; i++) {
    sum = sum + i;
  }
  printf("Suma liczb naturalnych od 1 do 20 to %d\n", sum);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getinput>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  lea    D9(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x8(%rbp)
  lea    D9(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D9
type_guessed: string

tool: D9: =

assistant: ```c
char *getinput(char *input) {
  char *next;
  next = strtok(input, "=");
  next = strtok(NULL, "=");
  return next;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<getinput>:
  endbr64
  push   %rbx
  lea    D6(%rip),%rbx
  mov    %rbx,%rsi
  call   <strtok@plt>
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D6
type_guessed: string

tool: D6: =

assistant: ```c
char *getinput(char *input) {
  char *next;
  next = strtok(input, "=");
  next = strtok(NULL, "=");
  return next;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<getinput>:
  endbr64
  push   %rbp
  lea    D3(%rip),%rbp
  mov    %rbp,%rsi
  call   <strtok@plt>
  mov    %rbp,%rsi
  xor    %edi,%edi
  pop    %rbp
  jmp    1050 <strtok@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D3
type_infered: string

tool: D3: =

assistant: ```c
char *getinput(char *input) {
  char *next;
  next = strtok(input, "=");
  next = strtok(NULL, "=");
  return next;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<getinput>:
  endbr64
  push   %rbp
  lea    D8(%rip),%rbp
  mov    %rbp,%rsi
  call   <strtok@plt>
  mov    %rbp,%rsi
  xor    %edi,%edi
  pop    %rbp
  jmp    1050 <strtok@plt>
```

tool_calls: name: unpack_data
arguments: label: D8
type_infered: string

tool: D8: =

assistant: ```c
char *getinput(char *input) {
  char *next;
  next = strtok(input, "=");
  next = strtok(NULL, "=");
  return next;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<isbadanswer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D35(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L17
  mov    -0x8(%rbp),%rax
  lea    D37(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L17
  mov    -0x8(%rbp),%rax
  lea    D30(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L17
  mov    -0x8(%rbp),%rax
  lea    D36(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L17
  mov    $0x1,%eax
  jmp    L29
L17:
  mov    $0x0,%eax
L29:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D35
type_guessed: string
name: unpack_data
arguments: data_tag: D37
type_guessed: string
name: unpack_data
arguments: data_tag: D30
type_guessed: string
name: unpack_data
arguments: data_tag: D36
type_guessed: string

tool: D35: yes

tool: D37: Yes

tool: D30: no

tool: D36: No

assistant: ```c
int isbadanswer(char *answer) {
  if (!(!strcmp(answer, "yes") || !strcmp(answer, "Yes") || !strcmp(answer, "no") || !strcmp(answer, "No")))
    return 1;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<isbadanswer>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  lea    D21(%rip),%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L10
  lea    D35(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L10
  lea    D12(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L10
  lea    D4(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
L10:
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D21
data_type: string
name: parse_data
arguments: data_tag: D35
data_type: string
name: parse_data
arguments: data_tag: D12
data_type: string
name: parse_data
arguments: data_tag: D4
data_type: string

tool: D21: yes

tool: D35: Yes

tool: D12: no

tool: D4: No

assistant: ```c
int isbadanswer(char *answer) {
  if (!(!strcmp(answer, "yes") || !strcmp(answer, "Yes") || !strcmp(answer, "no") || !strcmp(answer, "No")))
    return 1;
  return 0;
}
```