ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<udp_open>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdx,%r13
  xor    %edx,%edx
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D30(%rip),%rax
  mov    (%rax),%esi
  mov    D17(%rip),%rax
  mov    (%rax),%edi
  call   <socket@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L10
  mov    D14(%rip),%rbx
  test   %r14,%r14
  je     L17
  mov    (%rbx),%rax
  mov    %ebp,%edi
  xor    %ebp,%ebp
  mov    %rax,0x10(%r13)
  call   <htons@plt>
  mov    %r14,%rdi
  mov    %rax,0x8(%r13)
  call   <inet_addr@plt>
  mov    %eax,0x0(%r13)
L17:
  mov    (%rbx),%rax
  mov    %ebp,%edi
  mov    %rax,0x10(%rsp)
  call   <htons@plt>
  mov    %rax,0x8(%rsp)
  mov    D37(%rip),%rax
  mov    (%rax),%edi
  call   <htonl@plt>
  mov    %rsp,%rsi
  mov    $0x18,%edx
  mov    %r12d,%edi
  mov    %eax,(%rsp)
  call   <bind@plt>
  test   %eax,%eax
  jne    L40
L10:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x20,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L40:
  mov    %r12d,%edi
  mov    $0xfffffffe,%r12d
  call   <close@plt>
  jmp    L10
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int udp_open(int port, char *addr, struct sockaddr_in *taddr2) {
  int rt;
  int fd;
  struct sockaddr_in taddr;
  fd = socket(PF_INET, SOCK_DGRAM, 0);
  if (fd == -1) {
    return -1;
  }
  if (addr) {
    taddr2->sin_family = AF_INET;
    taddr2->sin_port = htons(port);
    taddr2->sin_addr.s_addr = inet_addr(addr);
    port = 0;
  }
  taddr.sin_family = AF_INET;
  taddr.sin_port = htons(port);
  taddr.sin_addr.s_addr = htonl(INADDR_ANY);
  rt = bind(fd, (struct sockaddr *)&taddr, sizeof(taddr));
  if (rt) {
    close(fd);
    return -2;
  }
  return fd;
}
```
user: Translate the below assembly code into C programming language:

```asm
<udp_open>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdx,%r13
  xor    %edx,%edx
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D22(%rip),%rax
  mov    (%rax),%esi
  mov    D29(%rip),%rax
  mov    (%rax),%edi
  call   <socket@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L24
  mov    D26(%rip),%rbx
  test   %r14,%r14
  je     L15
  mov    (%rbx),%rax
  mov    %ebp,%edi
  xor    %ebp,%ebp
  mov    %rax,0x10(%r13)
  call   <htons@plt>
  mov    %r14,%rdi
  mov    %rax,0x8(%r13)
  call   <inet_addr@plt>
  mov    %eax,0x0(%r13)
L15:
  mov    (%rbx),%rax
  mov    %ebp,%edi
  mov    %rax,0x10(%rsp)
  call   <htons@plt>
  mov    %rax,0x8(%rsp)
  mov    D14(%rip),%rax
  mov    (%rax),%edi
  call   <htonl@plt>
  mov    %rsp,%rsi
  mov    $0x18,%edx
  mov    %r12d,%edi
  mov    %eax,(%rsp)
  call   <bind@plt>
  test   %eax,%eax
  jne    L23
L24:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x20,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L23:
  mov    %r12d,%edi
  mov    $0xfffffffe,%r12d
  call   <close@plt>
  jmp    L24
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int udp_open(int port, char *addr, struct sockaddr_in *taddr2) {
  int rt;
  int fd;
  struct sockaddr_in taddr;
  fd = socket(PF_INET, SOCK_DGRAM, 0);
  if (fd == -1) {
    return -1;
  }
  if (addr) {
    taddr2->sin_family = AF_INET;
    taddr2->sin_port = htons(port);
    taddr2->sin_addr.s_addr = inet_addr(addr);
    port = 0;
  }
  taddr.sin_family = AF_INET;
  taddr.sin_port = htons(port);
  taddr.sin_addr.s_addr = htonl(INADDR_ANY);
  rt = bind(fd, (struct sockaddr *)&taddr, sizeof(taddr));
  if (rt) {
    close(fd);
    return -2;
  }
  return fd;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x18,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x4(%rbp)
  movsbl -0xd(%rbp),%eax
  sub    $0x25,%eax
  cmp    $0xa,%eax
  ja     L22
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D3(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D3(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L21
  mov    -0x8(%rbp),%eax
  sub    -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L21
  mov    -0x8(%rbp),%eax
  imul   -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L21
  mov    -0x8(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %eax,-0xc(%rbp)
  jmp    L21
  mov    -0x8(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,-0xc(%rbp)
  jmp    L21
L22:
  movsbl -0xd(%rbp),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%eax
  jmp    L56
L21:
  movsbl -0xd(%rbp),%edx
  mov    -0xc(%rbp),%esi
  mov    -0x4(%rbp),%ecx
  mov    -0x8(%rbp),%eax
  mov    %esi,%r8d
  mov    %eax,%esi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
L56:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D4
datatype: string
name: parse_data
arguments: data_tag: D26
datatype: string

tool: D4: Unknown operation %c.


tool: D26: %d %c %d = %d


assistant: ```c
int main(int argc, char *argv[]) {
  int a = atoi(argv[1]);
  char op = argv[2][0];
  int b = atoi(argv[3]);
  int r;
  switch (op) {
  case '+':
    r = a + b;
    break;
  case '-':
    r = a - b;
    break;
  case '*':
    r = a * b;
    break;
  case '/':
    r = a / b;
    break;
  case '%':
    r = a % b;
    break;
  default:
    printf("Unknown operation %c.\n", op);
    return 1;
  }
  printf("%d %c %d = %d\n", a, op, b, r);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbp
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%r12d
  mov    0x10(%rbp),%rax
  movzbl (%rax),%ebx
  mov    0x18(%rbp),%rdi
  call   <atoi@plt>
  movsbl %bl,%esi
  sub    $0x25,%ebx
  cmp    $0xa,%bl
  ja     L2
  mov    %eax,%ecx
  movzbl %bl,%ebx
  lea    D24(%rip),%rdx
  movslq (%rdx,%rbx,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    (%r12,%rcx,1),%r8d
L21:
  mov    %esi,%edx
  mov    %r12d,%esi
  lea    D23(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  mov    %r12d,%r8d
  sub    %ecx,%r8d
  jmp    L21
  mov    %r12d,%r8d
  imul   %ecx,%r8d
  jmp    L21
  mov    %r12d,%eax
  cltd
  idiv   %ecx
  mov    %eax,%r8d
  jmp    L21
  mov    %r12d,%eax
  cltd
  idiv   %ecx
  mov    %edx,%r8d
  jmp    L21
L2:
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%eax
  jmp    L9
```

tool_calls: name: read_data
arguments: tag: D23
data_type: string
name: read_data
arguments: tag: D18
data_type: string

tool: D23: %d %c %d = %d


tool: D18: Unknown operation %c.


assistant: ```c
int main(int argc, char *argv[]) {
  int a = atoi(argv[1]);
  char op = argv[2][0];
  int b = atoi(argv[3]);
  int r;
  switch (op) {
  case '+':
    r = a + b;
    break;
  case '-':
    r = a - b;
    break;
  case '*':
    r = a * b;
    break;
  case '/':
    r = a / b;
    break;
  case '%':
    r = a % b;
    break;
  default:
    printf("Unknown operation %c.\n", op);
    return 1;
  }
  printf("%d %c %d = %d\n", a, op, b, r);
  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   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    0x18(%rbp),%rdi
  mov    %eax,%r12d
  mov    0x10(%rbp),%rax
  movzbl (%rax),%ebx
  call   <atoi@plt>
  movsbl %bl,%r9d
  sub    $0x25,%ebx
  cmp    $0xa,%bl
  ja     L2
  lea    D5(%rip),%rdx
  movzbl %bl,%ebx
  mov    %eax,%ecx
  movslq (%rdx,%rbx,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
L2:
  mov    %r9d,%esi
  lea    D8(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  mov    %r12d,%eax
  cltd
  idiv   %ecx
  mov    %edx,%r8d
L26:
  mov    %r12d,%esi
  mov    %r9d,%edx
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  mov    %r12d,%r8d
  imul   %ecx,%r8d
  jmp    L26
  lea    (%r12,%rcx,1),%r8d
  jmp    L26
  mov    %r12d,%r8d
  sub    %ecx,%r8d
  jmp    L26
  mov    %r12d,%eax
  cltd
  idiv   %ecx
  mov    %eax,%r8d
  jmp    L26
```

tool_calls: name: disassemble_data
arguments: label: D8
type_infered: string
name: disassemble_data
arguments: label: D15
type_infered: string

tool: D8: Unknown operation %c.


tool: D15: %d %c %d = %d


assistant: ```c
int main(int argc, char *argv[]) {
  int a = atoi(argv[1]);
  char op = argv[2][0];
  int b = atoi(argv[3]);
  int r;
  switch (op) {
  case '+':
    r = a + b;
    break;
  case '-':
    r = a - b;
    break;
  case '*':
    r = a * b;
    break;
  case '/':
    r = a / b;
    break;
  case '%':
    r = a % b;
    break;
  default:
    printf("Unknown operation %c.\n", op);
    return 1;
  }
  printf("%d %c %d = %d\n", a, op, b, r);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    0x18(%rbp),%rdi
  mov    %eax,%r12d
  mov    0x10(%rbp),%rax
  movzbl (%rax),%ebx
  call   <atoi@plt>
  movsbl %bl,%r9d
  sub    $0x25,%ebx
  cmp    $0xa,%bl
  ja     L35
  lea    D12(%rip),%rdx
  movzbl %bl,%ebx
  mov    %eax,%ecx
  movslq (%rdx,%rbx,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
L35:
  mov    %r9d,%esi
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  mov    %r12d,%eax
  cltd
  idiv   %ecx
  mov    %edx,%r8d
L39:
  mov    %r12d,%esi
  mov    %r9d,%edx
  lea    D20(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  mov    %r12d,%r8d
  imul   %ecx,%r8d
  jmp    L39
  lea    (%r12,%rcx,1),%r8d
  jmp    L39
  mov    %r12d,%r8d
  sub    %ecx,%r8d
  jmp    L39
  mov    %r12d,%eax
  cltd
  idiv   %ecx
  mov    %eax,%r8d
  jmp    L39
```

tool_calls: name: disassemble_data
arguments: data_tag: D14
type: string
name: disassemble_data
arguments: data_tag: D20
type: string

tool: D14: Unknown operation %c.


tool: D20: %d %c %d = %d


assistant: ```c
int main(int argc, char *argv[]) {
  int a = atoi(argv[1]);
  char op = argv[2][0];
  int b = atoi(argv[3]);
  int r;
  switch (op) {
  case '+':
    r = a + b;
    break;
  case '-':
    r = a - b;
    break;
  case '*':
    r = a * b;
    break;
  case '/':
    r = a / b;
    break;
  case '%':
    r = a % b;
    break;
  default:
    printf("Unknown operation %c.\n", op);
    return 1;
  }
  printf("%d %c %d = %d\n", a, op, b, r);
  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   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <funkcja_pi@plt>
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D79(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <funkcja_pi@plt>
  mov    %eax,%esi
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x64,%edi
  call   <funkcja_pi@plt>
  mov    %eax,%esi
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x3e8,%edi
  call   <funkcja_pi@plt>
  mov    %eax,%esi
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x2710,%edi
  call   <funkcja_pi@plt>
  mov    %eax,%esi
  lea    D89(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x186a0,%edi
  call   <funkcja_pi@plt>
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xf4240,%edi
  call   <funkcja_pi@plt>
  mov    %eax,%esi
  lea    D84(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D38
type_guessed: string
name: disassemble_data
arguments: tag: D50
type_guessed: string
name: disassemble_data
arguments: tag: D79
type_guessed: string
name: disassemble_data
arguments: tag: D37
type_guessed: string
name: disassemble_data
arguments: tag: D52
type_guessed: string
name: disassemble_data
arguments: tag: D40
type_guessed: string
name: disassemble_data
arguments: tag: D89
type_guessed: string
name: disassemble_data
arguments: tag: D17
type_guessed: string
name: disassemble_data
arguments: tag: D84
type_guessed: string

tool: D38: Podaj liczbe powtorzen: 

tool: D50: %d

tool: D79: %d: %.10Lf


tool: D37: 10: %.0Lf


tool: D52: 100: %.1Lf


tool: D40: 1000: %.2Lf


tool: D89: 10_000: %.3Lf


tool: D17: 100_000: %.4Lf


tool: D84: 1_000_000: %.5Lf


assistant: ```c
int main() {
  int liczba;
  printf("Podaj liczbe powtorzen: ");
  scanf("%d", &liczba);
  printf("%d: %.10Lf\n", liczba, funkcja_pi(liczba));
  printf("10: %.0Lf\n", funkcja_pi(10));
  printf("100: %.1Lf\n", funkcja_pi(100));
  printf("1000: %.2Lf\n", funkcja_pi(1000));
  printf("10_000: %.3Lf\n", funkcja_pi(10000));
  printf("100_000: %.4Lf\n", funkcja_pi(100000));
  printf("1_000_000: %.5Lf\n", funkcja_pi(1000000));
  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
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D73(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D31(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <funkcja_pi@plt>
  mov    %eax,%edx
  mov    0x4(%rsp),%esi
  lea    D44(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <funkcja_pi@plt>
  mov    %eax,%esi
  lea    D81(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x64,%edi
  call   <funkcja_pi@plt>
  mov    %eax,%esi
  lea    D62(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x3e8,%edi
  call   <funkcja_pi@plt>
  mov    %eax,%esi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x2710,%edi
  call   <funkcja_pi@plt>
  mov    %eax,%esi
  lea    D53(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x186a0,%edi
  call   <funkcja_pi@plt>
  mov    %eax,%esi
  lea    D20(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xf4240,%edi
  call   <funkcja_pi@plt>
  mov    %eax,%esi
  lea    D34(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D73
type_infered: string
name: parse_data
arguments: label: D31
type_infered: string
name: parse_data
arguments: label: D44
type_infered: string
name: parse_data
arguments: label: D81
type_infered: string
name: parse_data
arguments: label: D62
type_infered: string
name: parse_data
arguments: label: D16
type_infered: string
name: parse_data
arguments: label: D53
type_infered: string
name: parse_data
arguments: label: D20
type_infered: string
name: parse_data
arguments: label: D34
type_infered: string

tool: D73: Podaj liczbe powtorzen: 

tool: D31: %d

tool: D44: %d: %.10Lf


tool: D81: 10: %.0Lf


tool: D62: 100: %.1Lf


tool: D16: 1000: %.2Lf


tool: D53: 10_000: %.3Lf


tool: D20: 100_000: %.4Lf


tool: D34: 1_000_000: %.5Lf


assistant: ```c
int main() {
  int liczba;
  printf("Podaj liczbe powtorzen: ");
  scanf("%d", &liczba);
  printf("%d: %.10Lf\n", liczba, funkcja_pi(liczba));
  printf("10: %.0Lf\n", funkcja_pi(10));
  printf("100: %.1Lf\n", funkcja_pi(100));
  printf("1000: %.2Lf\n", funkcja_pi(1000));
  printf("10_000: %.3Lf\n", funkcja_pi(10000));
  printf("100_000: %.4Lf\n", funkcja_pi(100000));
  printf("1_000_000: %.5Lf\n", funkcja_pi(1000000));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D85(%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>
  mov    0x4(%rsp),%edi
  call   <funkcja_pi@plt>
  mov    0x4(%rsp),%esi
  lea    D55(%rip),%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <funkcja_pi@plt>
  lea    D83(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x64,%edi
  call   <funkcja_pi@plt>
  lea    D5(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x3e8,%edi
  call   <funkcja_pi@plt>
  lea    D34(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x2710,%edi
  call   <funkcja_pi@plt>
  lea    D78(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x186a0,%edi
  call   <funkcja_pi@plt>
  lea    D26(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xf4240,%edi
  call   <funkcja_pi@plt>
  lea    D88(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D85
type: string
name: disassemble_data
arguments: data_tag: D28
type: string
name: disassemble_data
arguments: data_tag: D55
type: string
name: disassemble_data
arguments: data_tag: D83
type: string
name: disassemble_data
arguments: data_tag: D5
type: string
name: disassemble_data
arguments: data_tag: D34
type: string
name: disassemble_data
arguments: data_tag: D78
type: string
name: disassemble_data
arguments: data_tag: D26
type: string
name: disassemble_data
arguments: data_tag: D88
type: string

tool: D85: Podaj liczbe powtorzen: 

tool: D28: %d

tool: D55: %d: %.10Lf


tool: D83: 10: %.0Lf


tool: D5: 100: %.1Lf


tool: D34: 1000: %.2Lf


tool: D78: 10_000: %.3Lf


tool: D26: 100_000: %.4Lf


tool: D88: 1_000_000: %.5Lf


assistant: ```c
int main() {
  int liczba;
  printf("Podaj liczbe powtorzen: ");
  scanf("%d", &liczba);
  printf("%d: %.10Lf\n", liczba, funkcja_pi(liczba));
  printf("10: %.0Lf\n", funkcja_pi(10));
  printf("100: %.1Lf\n", funkcja_pi(100));
  printf("1000: %.2Lf\n", funkcja_pi(1000));
  printf("10_000: %.3Lf\n", funkcja_pi(10000));
  printf("100_000: %.4Lf\n", funkcja_pi(100000));
  printf("1_000_000: %.5Lf\n", funkcja_pi(1000000));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D32(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D19(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <funkcja_pi@plt>
  mov    0x4(%rsp),%esi
  lea    D4(%rip),%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <funkcja_pi@plt>
  lea    D85(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x64,%edi
  call   <funkcja_pi@plt>
  lea    D50(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x3e8,%edi
  call   <funkcja_pi@plt>
  lea    D17(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x2710,%edi
  call   <funkcja_pi@plt>
  lea    D42(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x186a0,%edi
  call   <funkcja_pi@plt>
  lea    D10(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xf4240,%edi
  call   <funkcja_pi@plt>
  lea    D47(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D32
type_guessed: string
name: parse_data
arguments: tag: D19
type_guessed: string
name: parse_data
arguments: tag: D4
type_guessed: string
name: parse_data
arguments: tag: D85
type_guessed: string
name: parse_data
arguments: tag: D50
type_guessed: string
name: parse_data
arguments: tag: D17
type_guessed: string
name: parse_data
arguments: tag: D42
type_guessed: string
name: parse_data
arguments: tag: D10
type_guessed: string
name: parse_data
arguments: tag: D47
type_guessed: string

tool: D32: Podaj liczbe powtorzen: 

tool: D19: %d

tool: D4: %d: %.10Lf


tool: D85: 10: %.0Lf


tool: D50: 100: %.1Lf


tool: D17: 1000: %.2Lf


tool: D42: 10_000: %.3Lf


tool: D10: 100_000: %.4Lf


tool: D47: 1_000_000: %.5Lf


assistant: ```c
int main() {
  int liczba;
  printf("Podaj liczbe powtorzen: ");
  scanf("%d", &liczba);
  printf("%d: %.10Lf\n", liczba, funkcja_pi(liczba));
  printf("10: %.0Lf\n", funkcja_pi(10));
  printf("100: %.1Lf\n", funkcja_pi(100));
  printf("1000: %.2Lf\n", funkcja_pi(1000));
  printf("10_000: %.3Lf\n", funkcja_pi(10000));
  printf("100_000: %.4Lf\n", funkcja_pi(100000));
  printf("1_000_000: %.5Lf\n", funkcja_pi(1000000));
  return 0;
}
```