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

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x1,%edi
  jle    L14
  lea    0x10(%rsi),%r13
  lea    -0x2(%rdi),%eax
  mov    0x8(%rsi),%rbp
  mov    $0x5f,%r12d
  lea    0x0(%r13,%rax,8),%r14
  jmp    L28
  nopl   0x0(%rax)
L46:
  movzbl %bl,%edi
  add    $0x1,%rbp
  call   <isalnum@plt>
  test   %eax,%eax
  cmove  %r12d,%ebx
  movsbl %bl,%edi
  call   <putchar@plt>
L28:
  movzbl 0x0(%rbp),%ebx
  test   %bl,%bl
  jne    L46
  cmp    %r14,%r13
  je     L14
  mov    $0x20,%edi
  mov    0x0(%r13),%rbp
  add    $0x8,%r13
  call   <putchar@plt>
  jmp    L28
  nopl   0x0(%rax,%rax,1)
L14:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    D6(%rip),%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  mov    (%rax),%eax
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int i;
  char *p;
  int c;
  for (i = 1; i < argc; i++) {
    p = argv[i];
    if (i != 1)
      putchar(' ');
    while (*p) {
      c = (unsigned char)*p++;
      if (!isalnum(c))
        c = '_';
      putchar(c);
    }
  }
  putchar('\n');
  return EXIT_SUCCESS;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jle    L1
  lea    0x10(%rsi),%r12
  lea    -0x2(%rdi),%eax
  mov    0x8(%rsi),%rbp
  lea    (%r12,%rax,8),%r13
  nopw   0x0(%rax,%rax,1)
L40:
  movzbl 0x0(%rbp),%ebx
  test   %bl,%bl
  je     L19
L29:
  movzbl %bl,%edi
  add    $0x1,%rbp
  call   <isalnum@plt>
  test   %eax,%eax
  je     L0
  movsbl %bl,%edi
  call   <putchar@plt>
  movzbl 0x0(%rbp),%ebx
  test   %bl,%bl
  jne    L29
L19:
  cmp    %r12,%r13
  je     L1
  mov    $0x20,%edi
  mov    (%r12),%rbp
  add    $0x8,%r12
  call   <putchar@plt>
  jmp    L40
  nopl   0x0(%rax)
L0:
  mov    $0x5f,%edi
  call   <putchar@plt>
  jmp    L40
  nopl   0x0(%rax)
L1:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int i;
  char *p;
  int c;
  for (i = 1; i < argc; i++) {
    p = argv[i];
    if (i != 1)
      putchar(' ');
    while (*p) {
      c = (unsigned char)*p++;
      if (!isalnum(c))
        c = '_';
      putchar(c);
    }
  }
  putchar('\n');
  return EXIT_SUCCESS;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x30(%rbp)
  movl   $0x1,-0x2c(%rbp)
  movl   $0x0,-0x28(%rbp)
  movl   $0x1,-0x24(%rbp)
  movl   $0x1,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  movl   $0x1,-0x14(%rbp)
  movl   $0x1,-0x10(%rbp)
  lea    -0x30(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    $0x2,%esi
  mov    %rax,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    $0x3,%esi
  mov    %rax,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    $0x5,%esi
  mov    %rax,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    $0x6,%esi
  mov    %rax,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    $0x7,%esi
  mov    %rax,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    $0x8,%esi
  mov    %rax,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    $0x9,%esi
  mov    %rax,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D51
type: string

tool: D51: %d


assistant: ```c
int main() {
  int arr3[] = {0, 1, 0, 1, 1, 0, 1, 1, 1};
  printf("%d\n", findMaxConsecutiveOnes(arr3, 0));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 1));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 2));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 3));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 4));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 5));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 6));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 7));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 8));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 9));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rsp)
  movl   $0x1,0x4(%rsp)
  movl   $0x0,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  movl   $0x1,0x10(%rsp)
  movl   $0x0,0x14(%rsp)
  movl   $0x1,0x18(%rsp)
  movl   $0x1,0x1c(%rsp)
  movl   $0x1,0x20(%rsp)
  mov    %rsp,%rbp
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  lea    D4(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x2,%esi
  mov    %rbp,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    %rbp,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x5,%esi
  mov    %rbp,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x6,%esi
  mov    %rbp,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x7,%esi
  mov    %rbp,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  mov    %rbp,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x9,%esi
  mov    %rbp,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

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

tool: D4: %d


assistant: ```c
int main() {
  int arr3[] = {0, 1, 0, 1, 1, 0, 1, 1, 1};
  printf("%d\n", findMaxConsecutiveOnes(arr3, 0));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 1));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 2));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 3));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 4));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 5));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 6));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 7));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 8));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 9));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  xor    %esi,%esi
  push   %rbp
  lea    D5(%rip),%rbp
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x100000000,%rax
  mov    %rsp,%r12
  movl   $0x1,0x20(%rsp)
  movq   $0x1,0x10(%rsp)
  mov    %r12,%rdi
  mov    %rax,(%rsp)
  mov    %rax,0x8(%rsp)
  add    $0x1,%rax
  mov    %rax,0x18(%rsp)
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x2,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x4,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x5,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x6,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x7,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x8,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x9,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D5
type_guessed: string

tool: D5: %d


assistant: ```c
int main() {
  int arr3[] = {0, 1, 0, 1, 1, 0, 1, 1, 1};
  printf("%d\n", findMaxConsecutiveOnes(arr3, 0));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 1));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 2));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 3));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 4));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 5));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 6));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 7));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 8));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 9));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r12
  xor    %esi,%esi
  push   %rbp
  lea    D7(%rip),%rbp
  sub    $0x38,%rsp
  movdqa D28(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  movl   $0x1,0x20(%rsp)
  movaps %xmm0,(%rsp)
  movdqa D20(%rip),%xmm0
  mov    %r12,%rdi
  movaps %xmm0,0x10(%rsp)
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x2,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x4,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x5,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x6,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x7,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x8,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x9,%esi
  mov    %r12,%rdi
  call   <findMaxConsecutiveOnes@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D7
type_guessed: string
name: disassemble_data
arguments: data_tag: D28
type_guessed: byte[16]
name: disassemble_data
arguments: data_tag: D20
type_guessed: byte[16]

tool: D7: %d


tool: {"D28": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: {"D20": [0x01,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int arr3[] = {0, 1, 0, 1, 1, 0, 1, 1, 1};
  printf("%d\n", findMaxConsecutiveOnes(arr3, 0));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 1));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 2));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 3));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 4));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 5));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 6));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 7));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 8));
  printf("%d\n", findMaxConsecutiveOnes(arr3, 9));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<hexDigitToValue>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x14(%rbp)
  movb   $0x0,-0x1(%rbp)
  cmpb   $0x2f,-0x14(%rbp)
  jle    L47
  cmpb   $0x39,-0x14(%rbp)
  jg     L47
  movzbl -0x14(%rbp),%eax
  sub    $0x30,%eax
  mov    %al,-0x1(%rbp)
  jmp    L75
L47:
  cmpb   $0x40,-0x14(%rbp)
  jle    L66
  cmpb   $0x46,-0x14(%rbp)
  jg     L66
  movzbl -0x14(%rbp),%eax
  sub    $0x37,%eax
  mov    %al,-0x1(%rbp)
  jmp    L75
L66:
  cmpb   $0x60,-0x14(%rbp)
  jle    L75
  cmpb   $0x66,-0x14(%rbp)
  jg     L75
  movzbl -0x14(%rbp),%eax
  sub    $0x57,%eax
  mov    %al,-0x1(%rbp)
L75:
  movzbl -0x1(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned char hexDigitToValue(char d) {
  char o = 0;
  if (d > 47 && d < 58) {
    o = d - 48;
  } else if (d > 64 && d < 71) {
    o = d - 55;
  } else if (d > 96 && d < 103) {
    o = d - 87;
  }
  return o;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<hexDigitToValue>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L13
  lea    -0x41(%rdi),%eax
  cmp    $0x5,%al
  ja     L2
  lea    -0x37(%rdi),%eax
  ret
L2:
  lea    -0x61(%rdi),%eax
  sub    $0x57,%edi
  cmp    $0x6,%al
  mov    $0x0,%eax
  cmovb  %edi,%eax
L13:
  ret
```

assistant: ```c
unsigned char hexDigitToValue(char d) {
  char o = 0;
  if (d > 47 && d < 58) {
    o = d - 48;
  } else if (d > 64 && d < 71) {
    o = d - 55;
  } else if (d > 96 && d < 103) {
    o = d - 87;
  }
  return o;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<hexDigitToValue>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L18
  lea    -0x41(%rdi),%eax
  cmp    $0x5,%al
  ja     L8
  lea    -0x37(%rdi),%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L8:
  lea    -0x61(%rdi),%eax
  sub    $0x57,%edi
  cmp    $0x6,%al
  mov    $0x0,%eax
  cmovb  %edi,%eax
L18:
  ret
```

assistant: ```c
unsigned char hexDigitToValue(char d) {
  char o = 0;
  if (d > 47 && d < 58) {
    o = d - 48;
  } else if (d > 64 && d < 71) {
    o = d - 55;
  } else if (d > 96 && d < 103) {
    o = d - 87;
  }
  return o;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<hexDigitToValue>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L2
  lea    -0x41(%rdi),%eax
  cmp    $0x5,%al
  ja     L11
  lea    -0x37(%rdi),%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L11:
  lea    -0x61(%rdi),%eax
  sub    $0x57,%edi
  cmp    $0x6,%al
  mov    $0x0,%eax
  cmovb  %edi,%eax
L2:
  ret
```

assistant: ```c
unsigned char hexDigitToValue(char d) {
  char o = 0;
  if (d > 47 && d < 58) {
    o = d - 48;
  } else if (d > 64 && d < 71) {
    o = d - 55;
  } else if (d > 96 && d < 103) {
    o = d - 87;
  }
  return o;
}
```
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    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x48(%rbp),%rdx
  lea    -0x50(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <create_bitset@plt>
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x38(%rbp)
  jmp    L50
L28:
  lea    -0x40(%rbp),%rax
  mov    %rax,%rsi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  jmp    L63
L19:
  mov    -0x28(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <set_bitset@plt>
  mov    -0x30(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x30(%rbp)
  add    %rax,-0x28(%rbp)
L63:
  mov    -0x48(%rbp),%rax
  cmp    %rax,-0x28(%rbp)
  jb     L19
  addq   $0x1,-0x38(%rbp)
L50:
  mov    -0x50(%rbp),%rax
  cmp    %rax,-0x38(%rbp)
  jb     L28
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L71
L22:
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <get_bitset@plt>
  test   %rax,%rax
  je     L61
  addq   $0x1,-0x20(%rbp)
L61:
  addq   $0x1,-0x18(%rbp)
L71:
  mov    -0x48(%rbp),%rax
  cmp    %rax,-0x18(%rbp)
  jb     L22
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <destroy_bitset@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L42
  call   <__stack_chk_fail@plt>
L42:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D17
datatype: string
name: extract_symbolic_data
arguments: tag: D27
datatype: string
name: extract_symbolic_data
arguments: tag: D6
datatype: string

tool: D17: %zu %zu

tool: D27: %zu

tool: D6: %zu


assistant: ```c
int main() {
  size_t infected, count;
  scanf("%zu %zu", &infected, &count);
  void *bitset = create_bitset(count);
  for (size_t i = 0; i < infected; ++i) {
    size_t nr;
    scanf("%zu", &nr);
    size_t delta = nr;
    for (size_t j = nr; j < count; j += delta++) {
      set_bitset(bitset, j, 1);
    }
  }
  size_t infected_count = 0;
  for (size_t i = 0; i < count; ++i) {
    if (get_bitset(bitset, i)) {
      ++infected_count;
    }
  }
  destroy_bitset(bitset);
  printf("%zu\n", infected_count);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%rdi
  call   <create_bitset@plt>
  mov    %rax,%r12
  cmpq   $0x0,(%rsp)
  je     L28
  mov    $0x0,%r13d
  lea    D18(%rip),%r14
L42:
  lea    0x10(%rsp),%rsi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%rbp
  cmp    0x8(%rsp),%rbp
  jae    L44
  mov    %rbp,%rbx
L61:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <set_bitset@plt>
  mov    %rbp,%rax
  add    $0x1,%rbp
  add    %rax,%rbx
  cmp    %rbx,0x8(%rsp)
  ja     L61
L44:
  add    $0x1,%r13
  cmp    %r13,(%rsp)
  ja     L42
L28:
  mov    0x8(%rsp),%rbp
  test   %rbp,%rbp
  je     L16
  mov    $0x0,%ebx
  mov    $0x0,%ebp
L55:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <get_bitset@plt>
  cmp    $0x1,%rax
  sbb    $0xffffffffffffffff,%rbp
  add    $0x1,%rbx
  cmp    %rbx,0x8(%rsp)
  ja     L55
L16:
  mov    %r12,%rdi
  call   <destroy_bitset@plt>
  mov    %rbp,%rsi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L58
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L58:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D14
type_guessed: string
name: read_data
arguments: data_tag: D18
type_guessed: string
name: read_data
arguments: data_tag: D8
type_guessed: string

tool: D14: %zu %zu

tool: D18: %zu

tool: D8: %zu


assistant: ```c
int main() {
  size_t infected, count;
  scanf("%zu %zu", &infected, &count);
  void *bitset = create_bitset(count);
  for (size_t i = 0; i < infected; ++i) {
    size_t nr;
    scanf("%zu", &nr);
    size_t delta = nr;
    for (size_t j = nr; j < count; j += delta++) {
      set_bitset(bitset, j, 1);
    }
  }
  size_t infected_count = 0;
  for (size_t i = 0; i < count; ++i) {
    if (get_bitset(bitset, i)) {
      ++infected_count;
    }
  }
  destroy_bitset(bitset);
  printf("%zu\n", infected_count);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  lea    D3(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rdx
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%rdi
  call   <create_bitset@plt>
  cmpq   $0x0,(%rsp)
  mov    %rax,%r12
  je     L26
  xor    %r13d,%r13d
  lea    0x10(%rsp),%r15
  lea    D27(%rip),%r14
  nopl   0x0(%rax)
L31:
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%rbp
  cmp    0x8(%rsp),%rbp
  jae    L27
  mov    %rbp,%rbx
  nopl   0x0(%rax)
L41:
  mov    %rbx,%rsi
  mov    $0x1,%edx
  mov    %r12,%rdi
  call   <set_bitset@plt>
  mov    %rbp,%rax
  add    $0x1,%rbp
  add    %rax,%rbx
  cmp    %rbx,0x8(%rsp)
  ja     L41
L27:
  add    $0x1,%r13
  cmp    %r13,(%rsp)
  ja     L31
L26:
  mov    0x8(%rsp),%rbp
  test   %rbp,%rbp
  je     L37
  xor    %ebx,%ebx
  xor    %ebp,%ebp
  nopl   0x0(%rax)
L28:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <get_bitset@plt>
  cmp    $0x1,%rax
  sbb    $0xffffffffffffffff,%rbp
  add    $0x1,%rbx
  cmp    %rbx,0x8(%rsp)
  ja     L28
L37:
  mov    %r12,%rdi
  call   <destroy_bitset@plt>
  mov    %rbp,%rsi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D3
data_type: string
name: unpack_data
arguments: data_tag: D27
data_type: string
name: unpack_data
arguments: data_tag: D16
data_type: string

tool: D3: %zu %zu

tool: D27: %zu

tool: D16: %zu


assistant: ```c
int main() {
  size_t infected, count;
  scanf("%zu %zu", &infected, &count);
  void *bitset = create_bitset(count);
  for (size_t i = 0; i < infected; ++i) {
    size_t nr;
    scanf("%zu", &nr);
    size_t delta = nr;
    for (size_t j = nr; j < count; j += delta++) {
      set_bitset(bitset, j, 1);
    }
  }
  size_t infected_count = 0;
  for (size_t i = 0; i < count; ++i) {
    if (get_bitset(bitset, i)) {
      ++infected_count;
    }
  }
  destroy_bitset(bitset);
  printf("%zu\n", infected_count);
}
```