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

```asm
<mem_vla_range_store_idx>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x3(%rdi),%eax
  cmp    $0x2,%eax
  mov    $0x3,%eax
  cmovae %eax,%edi
  movslq %edi,%rdi
  lea    0x10(%rdi),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L33:
  cmp    %rdx,%rsp
  je     L28
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L33
L28:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L17
  orq    $0x0,-0x8(%rsp,%rax,1)
L17:
  movb   $0x0,(%rsp)
  movb   $0x0,(%rsp,%rdi,1)
  movb   $0x1,0x1(%rsp)
  movb   $0x2,0x2(%rsp)
  lea    0x2(%rsp),%rsi
  mov    %rsp,%rdi
  mov    $0x0,%eax
  call   <sink@plt>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  leave
  ret
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void mem_vla_range_store_idx(int n) {
  if (n < 3 || 4 < n)
    n = 3;
  struct {
    char a[n], b;
  } s;
  char *p = s.a;
  s.a[0] = 0;
  s.b = 0;
  *++p = 1;
  *++p = 2;
  sink(&s, p);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<mem_vla_range_store_idx>:
  endbr64
  push   %rbp
  movslq %edi,%rcx
  lea    -0x3(%rdi),%edx
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    0x1(%rcx),%rax
  cmp    $0x1,%edx
  jbe    L7
  mov    $0x4,%eax
  mov    $0x3,%ecx
L7:
  add    $0xf,%rax
  mov    %rsp,%rsi
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rdx
  cmp    %rsi,%rsp
  je     L34
L2:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L2
L34:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L18
L35:
  mov    $0x100,%eax
  movb   $0x0,(%rsp,%rcx,1)
  lea    0x2(%rsp),%rsi
  mov    %rsp,%rdi
  mov    %ax,(%rsp)
  xor    %eax,%eax
  movb   $0x2,0x2(%rsp)
  call   <sink@plt>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  leave
  ret
  nopl   (%rax)
L18:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L35
L48:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void mem_vla_range_store_idx(int n) {
  if (n < 3 || 4 < n)
    n = 3;
  struct {
    char a[n], b;
  } s;
  char *p = s.a;
  s.a[0] = 0;
  s.b = 0;
  *++p = 1;
  *++p = 2;
  sink(&s, p);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<mem_vla_range_store_idx>:
  endbr64
  push   %rbp
  movslq %edi,%rcx
  lea    -0x3(%rdi),%edx
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    0x1(%rcx),%rax
  cmp    $0x1,%edx
  jbe    L20
  mov    $0x4,%eax
  mov    $0x3,%ecx
L20:
  add    $0xf,%rax
  mov    %rsp,%rsi
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rdx
  cmp    %rsi,%rsp
  je     L2
L4:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L4
L2:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L24
L1:
  mov    $0x100,%eax
  movb   $0x0,(%rsp,%rcx,1)
  lea    0x2(%rsp),%rsi
  mov    %rsp,%rdi
  mov    %ax,(%rsp)
  xor    %eax,%eax
  movb   $0x2,0x2(%rsp)
  call   <sink@plt>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L52
  leave
  ret
  nopl   (%rax)
L24:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L1
L52:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void mem_vla_range_store_idx(int n) {
  if (n < 3 || 4 < n)
    n = 3;
  struct {
    char a[n], b;
  } s;
  char *p = s.a;
  s.a[0] = 0;
  s.b = 0;
  *++p = 1;
  *++p = 2;
  sink(&s, p);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<count_number_of_threads>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D1(%rip),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x98(%rbp)
  jmp    L11
L29:
  mov    -0x98(%rbp),%rdx
  lea    -0x90(%rbp),%rax
  mov    $0x80,%esi
  mov    %rax,%rdi
  call   <fgets@plt>
  lea    -0x9c(%rbp),%rdx
  lea    -0x90(%rbp),%rax
  lea    D31(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sscanf@plt>
  cmp    $0x1,%eax
  je     L54
  jmp    L11
L54:
  mov    -0x98(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    -0x9c(%rbp),%eax
  jmp    L51
L11:
  mov    -0x98(%rbp),%rax
  mov    %rax,%rdi
  call   <feof@plt>
  test   %eax,%eax
  je     L29
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <abort@plt>
L51:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L31
  call   <__stack_chk_fail@plt>
L31:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D1
datatype: string
name: parse_data
arguments: data_label: D14
datatype: string
name: parse_data
arguments: data_label: D31
datatype: string
name: parse_data
arguments: data_label: D26
datatype: string

tool: D1: r

tool: D14: /proc/self/status

tool: D31: Threads: %d


tool: D26: Internal error, please send email to gback@cs.vt.edu


assistant: ```c
int count_number_of_threads(void) {
  FILE *p = fopen("/proc/self/status", "r");
  while (!feof(p)) {
    int threadsleft;
    char buf[128];
    fgets(buf, sizeof buf, p);
    if (sscanf(buf, "Threads: %d\n", &threadsleft) != 1)
      continue;
    fclose(p);
    return threadsleft;
  }
  printf("Internal error, please send email to gback@cs.vt.edu\n");
  abort();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<count_number_of_threads>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    D1(%rip),%rsi
  lea    D0(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbx
  lea    D21(%rip),%r12
L6:
  mov    %rbx,%rdi
  call   <feof@plt>
  test   %eax,%eax
  jne    L30
  lea    0x10(%rsp),%rbp
  mov    %rbx,%rdx
  mov    $0x80,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  lea    0xc(%rsp),%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <sscanf@plt>
  cmp    $0x1,%eax
  jne    L6
  mov    %rbx,%rdi
  call   <fclose@plt>
  mov    0xc(%rsp),%eax
L19:
  mov    0x98(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0xa0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L30:
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L19
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D1
data_type: string
name: disassemble_data
arguments: tag: D0
data_type: string
name: disassemble_data
arguments: tag: D21
data_type: string
name: disassemble_data
arguments: tag: D2
data_type: string

tool: D1: r

tool: D0: /proc/self/status

tool: D21: Threads: %d


tool: D2: Internal error, please send email to gback@cs.vt.edu


assistant: ```c
int count_number_of_threads(void) {
  FILE *p = fopen("/proc/self/status", "r");
  while (!feof(p)) {
    int threadsleft;
    char buf[128];
    fgets(buf, sizeof buf, p);
    if (sscanf(buf, "Threads: %d\n", &threadsleft) != 1)
      continue;
    fclose(p);
    return threadsleft;
  }
  printf("Internal error, please send email to gback@cs.vt.edu\n");
  abort();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<count_number_of_threads>:
  endbr64
  push   %r12
  lea    D19(%rip),%rsi
  lea    D10(%rip),%rdi
  push   %rbp
  push   %rbx
  lea    D22(%rip),%rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  call   <fopen@plt>
  mov    %rax,%rbp
L27:
  mov    %rbp,%rdi
  call   <feof@plt>
  test   %eax,%eax
  jne    L11
  lea    0x10(%rsp),%r12
  mov    %rbp,%rdx
  mov    $0x80,%esi
  mov    %r12,%rdi
  call   <fgets@plt>
  lea    0xc(%rsp),%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <sscanf@plt>
  cmp    $0x1,%eax
  jne    L27
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    0xc(%rsp),%eax
  mov    0x98(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L38
  add    $0xa0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  lea    D27(%rip),%rdi
  call   <printf@plt>
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  add    $0xa0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    10c0 <abort@plt>
L38:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D19
datatype: string
name: extract_symbolic_data
arguments: data_tag: D10
datatype: string
name: extract_symbolic_data
arguments: data_tag: D22
datatype: string
name: extract_symbolic_data
arguments: data_tag: D27
datatype: string

tool: D19: r

tool: D10: /proc/self/status

tool: D22: Threads: %d


tool: D27: Internal error, please send email to gback@cs.vt.edu


assistant: ```c
int count_number_of_threads(void) {
  FILE *p = fopen("/proc/self/status", "r");
  while (!feof(p)) {
    int threadsleft;
    char buf[128];
    fgets(buf, sizeof buf, p);
    if (sscanf(buf, "Threads: %d\n", &threadsleft) != 1)
      continue;
    fclose(p);
    return threadsleft;
  }
  printf("Internal error, please send email to gback@cs.vt.edu\n");
  abort();
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<count_number_of_threads>:
  endbr64
  push   %r12
  lea    D35(%rip),%rsi
  lea    D0(%rip),%rdi
  push   %rbp
  push   %rbx
  lea    D11(%rip),%rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  call   <fopen@plt>
  mov    %rax,%rbp
L9:
  mov    %rbp,%rdi
  call   <feof@plt>
  test   %eax,%eax
  jne    L20
  lea    0x10(%rsp),%r12
  mov    %rbp,%rdx
  mov    $0x80,%esi
  mov    %r12,%rdi
  call   <fgets@plt>
  lea    0xc(%rsp),%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <sscanf@plt>
  cmp    $0x1,%eax
  jne    L9
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    0xc(%rsp),%eax
  mov    0x98(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L39
  add    $0xa0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L20:
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  add    $0xa0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    10c0 <abort@plt>
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D35
datatype: string
name: parse_data
arguments: label: D0
datatype: string
name: parse_data
arguments: label: D11
datatype: string
name: parse_data
arguments: label: D14
datatype: string

tool: D35: r

tool: D0: /proc/self/status

tool: D11: Threads: %d


tool: D14: Internal error, please send email to gback@cs.vt.edu


assistant: ```c
int count_number_of_threads(void) {
  FILE *p = fopen("/proc/self/status", "r");
  while (!feof(p)) {
    int threadsleft;
    char buf[128];
    fgets(buf, sizeof buf, p);
    if (sscanf(buf, "Threads: %d\n", &threadsleft) != 1)
      continue;
    fclose(p);
    return threadsleft;
  }
  printf("Internal error, please send email to gback@cs.vt.edu\n");
  abort();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<keyword>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x230,%rsp
  mov    %rdi,-0x228(%rbp)
  mov    %rsi,-0x230(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x214(%rbp)
  jmp    L26
L39:
  mov    -0x214(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x228(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x214(%rbp),%eax
  cltq
  mov    %dl,-0x210(%rbp,%rax,1)
  addl   $0x1,-0x214(%rbp)
L26:
  mov    -0x214(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x228(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %eax,%eax
  jne    L2
  mov    -0x214(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x228(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L39
L2:
  mov    -0x214(%rbp),%eax
  cltq
  movb   $0x0,-0x210(%rbp,%rax,1)
  mov    -0x230(%rbp),%rdx
  lea    -0x210(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  sete   %al
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
char keyword(char *line, char *code) {
  int i;
  char key[512];
  for (i = 0; !isspace(line[i]) && line[i] != '\0'; i++)
    key[i] = line[i];
  key[i] = '\0';
  return (strcmp(key, code) == 0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<keyword>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x218,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    $0x0,%ebx
  jmp    L38
L6:
  mov    %al,(%rsp,%rbx,1)
  add    $0x1,%rbx
L38:
  mov    %ebx,%r12d
  movsbl 0x0(%rbp,%rbx,1),%edi
  call   <isspace@plt>
  test   %eax,%eax
  jne    L8
  movzbl 0x0(%rbp,%rbx,1),%eax
  test   %al,%al
  jne    L6
L8:
  movslq %r12d,%r12
  movb   $0x0,(%rsp,%r12,1)
  mov    %rsp,%rdi
  mov    %r13,%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  sete   %al
  mov    0x208(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L28
  add    $0x218,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char keyword(char *line, char *code) {
  int i;
  char key[512];
  for (i = 0; !isspace(line[i]) && line[i] != '\0'; i++)
    key[i] = line[i];
  key[i] = '\0';
  return (strcmp(key, code) == 0);
}
```
user: What is the c source code of the assembly code below:

```asm
<keyword>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  jmp    L23
  nopl   0x0(%rax,%rax,1)
L15:
  movzbl 0x0(%rbp,%rbx,1),%eax
  test   %al,%al
  je     L21
  mov    %al,0x0(%r13,%rbx,1)
  add    $0x1,%rbx
L23:
  movsbl 0x0(%rbp,%rbx,1),%edi
  movslq %ebx,%r12
  call   <isspace@plt>
  test   %eax,%eax
  je     L15
L21:
  mov    %r14,%rsi
  mov    %r13,%rdi
  movb   $0x0,(%rsp,%r12,1)
  call   <strcmp@plt>
  test   %rax,%rax
  sete   %al
  mov    0x208(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L14
  add    $0x210,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char keyword(char *line, char *code) {
  int i;
  char key[512];
  for (i = 0; !isspace(line[i]) && line[i] != '\0'; i++)
    key[i] = line[i];
  key[i] = '\0';
  return (strcmp(key, code) == 0);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<keyword>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  jmp    L2
  nopl   0x0(%rax,%rax,1)
L4:
  movzbl 0x0(%rbp,%rbx,1),%eax
  test   %al,%al
  je     L22
  mov    %al,0x0(%r13,%rbx,1)
  add    $0x1,%rbx
L2:
  movsbl 0x0(%rbp,%rbx,1),%edi
  movslq %ebx,%r12
  call   <isspace@plt>
  test   %eax,%eax
  je     L4
L22:
  mov    %r14,%rsi
  mov    %r13,%rdi
  movb   $0x0,(%rsp,%r12,1)
  call   <strcmp@plt>
  test   %rax,%rax
  sete   %al
  mov    0x208(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L23
  add    $0x210,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char keyword(char *line, char *code) {
  int i;
  char key[512];
  for (i = 0; !isspace(line[i]) && line[i] != '\0'; i++)
    key[i] = line[i];
  key[i] = '\0';
  return (strcmp(key, code) == 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rules>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D70(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D68(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x9(%rbp),%rax
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movzbl -0x9(%rbp),%eax
  cmp    $0x79,%al
  je     L25
  movzbl -0x9(%rbp),%eax
  cmp    $0x59,%al
  jne    L20
L25:
  lea    D62(%rip),%rax
  mov    %rax,%rdi
  call   <system@plt>
L20:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D27
type_guessed: string
name: disassemble_data
arguments: data_tag: D70
type_guessed: string
name: disassemble_data
arguments: data_tag: D22
type_guessed: string
name: disassemble_data
arguments: data_tag: D59
type_guessed: string
name: disassemble_data
arguments: data_tag: D42
type_guessed: string
name: disassemble_data
arguments: data_tag: D68
type_guessed: string
name: disassemble_data
arguments: data_tag: D12
type_guessed: string
name: disassemble_data
arguments: data_tag: D49
type_guessed: string
name: disassemble_data
arguments: data_tag: D62
type_guessed: string

tool: D27: 	Tic-Tac-Toe



tool: D70: Welcome to the most played 2D game and a sort of fun using X and O



tool: D22: Rules:-


tool: D59: 
1:Each player will be entering the number to put respective X or O in the desired position

tool: D42: 
2:Player who gets a combination of 3 same characters either diagonal or horizontally or 
  vertically will be declared as the winner

tool: D68: 

Enjoy the game! Be a Winner!



tool: D12: For more clarifications press Y else type any other character:- 

tool: D49: %c

tool: D62: start http://www.wikihow.com/Play-Tic-Tac-Toe

assistant: ```c
void rules() {
  char link;
  printf("\tTic-Tac-Toe\n\n");
  printf("Welcome to the most played 2D game and a sort of fun using X and O\n\n");
  printf("Rules:-\n");
  printf("\n1:Each player will be entering the number to put respective X or O in the desired position");
  printf("\n2:Player who gets a combination of 3 same characters either diagonal or horizontally or \n  vertically will be declared as the winner");
  printf("\n\nEnjoy the game! Be a Winner!\n\n");
  printf("For more clarifications press Y else type any other character:- ");
  scanf("%c", &link);
  if (link == 'y' || link == 'Y') {
    system("start http://www.wikihow.com/Play-Tic-Tac-Toe");
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rules>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D11(%rip),%rdi
  call   <printf@plt>
  lea    D60(%rip),%rdi
  call   <printf@plt>
  lea    D85(%rip),%rdi
  call   <printf@plt>
  lea    D18(%rip),%rdi
  call   <printf@plt>
  lea    D68(%rip),%rdi
  call   <printf@plt>
  lea    D45(%rip),%rdi
  call   <printf@plt>
  lea    D36(%rip),%rdi
  call   <printf@plt>
  lea    0x7(%rsp),%rsi
  lea    D71(%rip),%rdi
  call   <scanf@plt>
  movzbl 0x7(%rsp),%eax
  and    $0xffffffdf,%eax
  cmp    $0x59,%al
  je     L11
L23:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x18,%rsp
  ret
L11:
  lea    D70(%rip),%rdi
  call   <system@plt>
  jmp    L23
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D11
type_infered: string
name: unpack_data
arguments: data_tag: D60
type_infered: string
name: unpack_data
arguments: data_tag: D85
type_infered: string
name: unpack_data
arguments: data_tag: D18
type_infered: string
name: unpack_data
arguments: data_tag: D68
type_infered: string
name: unpack_data
arguments: data_tag: D45
type_infered: string
name: unpack_data
arguments: data_tag: D36
type_infered: string
name: unpack_data
arguments: data_tag: D71
type_infered: string
name: unpack_data
arguments: data_tag: D70
type_infered: string

tool: D11: 	Tic-Tac-Toe



tool: D60: Welcome to the most played 2D game and a sort of fun using X and O



tool: D85: Rules:-


tool: D18: 
1:Each player will be entering the number to put respective X or O in the desired position

tool: D68: 
2:Player who gets a combination of 3 same characters either diagonal or horizontally or 
  vertically will be declared as the winner

tool: D45: 

Enjoy the game! Be a Winner!



tool: D36: For more clarifications press Y else type any other character:- 

tool: D71: %c

tool: D70: start http://www.wikihow.com/Play-Tic-Tac-Toe

assistant: ```c
void rules() {
  char link;
  printf("\tTic-Tac-Toe\n\n");
  printf("Welcome to the most played 2D game and a sort of fun using X and O\n\n");
  printf("Rules:-\n");
  printf("\n1:Each player will be entering the number to put respective X or O in the desired position");
  printf("\n2:Player who gets a combination of 3 same characters either diagonal or horizontally or \n  vertically will be declared as the winner");
  printf("\n\nEnjoy the game! Be a Winner!\n\n");
  printf("For more clarifications press Y else type any other character:- ");
  scanf("%c", &link);
  if (link == 'y' || link == 'Y') {
    system("start http://www.wikihow.com/Play-Tic-Tac-Toe");
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rules>:
  endbr64
  sub    $0x18,%rsp
  lea    D72(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D37(%rip),%rdi
  call   <printf@plt>
  lea    D58(%rip),%rdi
  call   <printf@plt>
  lea    D5(%rip),%rdi
  call   <printf@plt>
  lea    D73(%rip),%rdi
  call   <printf@plt>
  lea    D2(%rip),%rdi
  call   <printf@plt>
  lea    D34(%rip),%rdi
  call   <printf@plt>
  lea    0x7(%rsp),%rsi
  lea    D39(%rip),%rdi
  call   <scanf@plt>
  movzbl 0x7(%rsp),%eax
  and    $0xffffffdf,%eax
  cmp    $0x59,%al
  je     L19
L10:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x18,%rsp
  ret
  xchg   %ax,%ax
L19:
  lea    D42(%rip),%rdi
  call   <system@plt>
  jmp    L10
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D72
data_type: string
name: parse_data
arguments: tag: D37
data_type: string
name: parse_data
arguments: tag: D58
data_type: string
name: parse_data
arguments: tag: D5
data_type: string
name: parse_data
arguments: tag: D73
data_type: string
name: parse_data
arguments: tag: D2
data_type: string
name: parse_data
arguments: tag: D34
data_type: string
name: parse_data
arguments: tag: D39
data_type: string
name: parse_data
arguments: tag: D42
data_type: string

tool: D72: 	Tic-Tac-Toe



tool: D37: Welcome to the most played 2D game and a sort of fun using X and O



tool: D58: Rules:-


tool: D5: 
1:Each player will be entering the number to put respective X or O in the desired position

tool: D73: 
2:Player who gets a combination of 3 same characters either diagonal or horizontally or 
  vertically will be declared as the winner

tool: D2: 

Enjoy the game! Be a Winner!



tool: D34: For more clarifications press Y else type any other character:- 

tool: D39: %c

tool: D42: start http://www.wikihow.com/Play-Tic-Tac-Toe

assistant: ```c
void rules() {
  char link;
  printf("\tTic-Tac-Toe\n\n");
  printf("Welcome to the most played 2D game and a sort of fun using X and O\n\n");
  printf("Rules:-\n");
  printf("\n1:Each player will be entering the number to put respective X or O in the desired position");
  printf("\n2:Player who gets a combination of 3 same characters either diagonal or horizontally or \n  vertically will be declared as the winner");
  printf("\n\nEnjoy the game! Be a Winner!\n\n");
  printf("For more clarifications press Y else type any other character:- ");
  scanf("%c", &link);
  if (link == 'y' || link == 'Y') {
    system("start http://www.wikihow.com/Play-Tic-Tac-Toe");
  }
}
```