ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ft_strsplit>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L50
  mov    %esi,%ebx
  mov    %rdi,(%rsp)
  movsbl %sil,%r13d
  mov    %r13d,%esi
  call   <cnt_wrd@plt>
  mov    %rax,%r14
  lea    0x8(,%rax,8),%edi
  call   <ft_memalloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L80
  mov    $0x0,%ebp
  mov    %rsp,%r15
  test   %r14,%r14
  jne    L21
L83:
  movq   $0x0,(%r12,%r14,8)
L80:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    %r12,%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L60:
  mov    %rax,(%rsp)
  add    $0x1,%rax
  cmp    %bl,-0x1(%rax)
  je     L60
L34:
  mov    %r13d,%esi
  mov    %r15,%rdi
  call   <wrd_cpy@plt>
  mov    %rax,(%r12,%rbp,8)
  add    $0x1,%rbp
  cmp    %rbp,%r14
  je     L83
L21:
  mov    (%rsp),%rdx
  lea    0x1(%rdx),%rax
  cmp    (%rdx),%bl
  je     L60
  jmp    L34
L50:
  mov    %rdi,%r12
  jmp    L80
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char **ft_strsplit(char const *s, char c) {
  char **tab_split;
  char *s1;
  size_t i;
  size_t w;
  if (!s)
    return (NULL);
  s1 = (char *)s;
  i = 0;
  w = cnt_wrd(s1, c);
  if (!(tab_split = ft_memalloc(sizeof(char *) * (w + 1))))
    return (NULL);
  while (i < w) {
    while (*s1 == c)
      ++s1;
    tab_split[i] = wrd_cpy(&s1, c);
    ++i;
  }
  tab_split[i] = 0;
  return (tab_split);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ft_strsplit>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L24
  movsbl %sil,%ebp
  mov    %esi,%ebx
  mov    %rdi,(%rsp)
  mov    %ebp,%esi
  call   <cnt_wrd@plt>
  lea    0x8(,%rax,8),%edi
  mov    %rax,%r13
  call   <ft_memalloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L24
  xor    %r15d,%r15d
  mov    %rsp,%r14
  test   %r13,%r13
  je     L2
  xchg   %ax,%ax
L41:
  mov    (%rsp),%rdx
  lea    0x1(%rdx),%rax
  cmp    (%rdx),%bl
  jne    L75
  nopl   0x0(%rax)
L60:
  mov    %rax,(%rsp)
  add    $0x1,%rax
  cmp    %bl,-0x1(%rax)
  je     L60
L75:
  mov    %ebp,%esi
  mov    %r14,%rdi
  call   <wrd_cpy@plt>
  mov    %rax,(%r12,%r15,8)
  add    $0x1,%r15
  cmp    %r15,%r13
  jne    L41
  lea    (%r12,%r13,8),%rax
L2:
  movq   $0x0,(%rax)
L54:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L64
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L24:
  xor    %r12d,%r12d
  jmp    L54
L64:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char **ft_strsplit(char const *s, char c) {
  char **tab_split;
  char *s1;
  size_t i;
  size_t w;
  if (!s)
    return (NULL);
  s1 = (char *)s;
  i = 0;
  w = cnt_wrd(s1, c);
  if (!(tab_split = ft_memalloc(sizeof(char *) * (w + 1))))
    return (NULL);
  while (i < w) {
    while (*s1 == c)
      ++s1;
    tab_split[i] = wrd_cpy(&s1, c);
    ++i;
  }
  tab_split[i] = 0;
  return (tab_split);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_strsplit>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L69
  movsbl %sil,%ebp
  mov    %esi,%ebx
  mov    %rdi,(%rsp)
  mov    %ebp,%esi
  call   <cnt_wrd@plt>
  lea    0x8(,%rax,8),%edi
  mov    %rax,%r13
  call   <ft_memalloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L69
  xor    %r15d,%r15d
  mov    %rsp,%r14
  test   %r13,%r13
  je     L6
  xchg   %ax,%ax
L38:
  mov    (%rsp),%rdx
  lea    0x1(%rdx),%rax
  cmp    (%rdx),%bl
  jne    L45
  nopl   0x0(%rax)
L56:
  mov    %rax,(%rsp)
  add    $0x1,%rax
  cmp    %bl,-0x1(%rax)
  je     L56
L45:
  mov    %ebp,%esi
  mov    %r14,%rdi
  call   <wrd_cpy@plt>
  mov    %rax,(%r12,%r15,8)
  add    $0x1,%r15
  cmp    %r15,%r13
  jne    L38
  lea    (%r12,%r13,8),%rax
L6:
  movq   $0x0,(%rax)
L41:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L69:
  xor    %r12d,%r12d
  jmp    L41
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char **ft_strsplit(char const *s, char c) {
  char **tab_split;
  char *s1;
  size_t i;
  size_t w;
  if (!s)
    return (NULL);
  s1 = (char *)s;
  i = 0;
  w = cnt_wrd(s1, c);
  if (!(tab_split = ft_memalloc(sizeof(char *) * (w + 1))))
    return (NULL);
  while (i < w) {
    while (*s1 == c)
      ++s1;
    tab_split[i] = wrd_cpy(&s1, c);
    ++i;
  }
  tab_split[i] = 0;
  return (tab_split);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x0,-0x10(%rbp)
  movl   $0x1,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L30
L42:
  shll   $0x2,-0xc(%rbp)
L30:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L42
  jmp    L18
L36:
  mov    -0xc(%rbp),%eax
  lea    0x3(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x2,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x10(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L18
  mov    -0xc(%rbp),%eax
  add    %eax,-0x10(%rbp)
  mov    -0x4(%rbp),%eax
  sub    %eax,-0x8(%rbp)
L18:
  cmpl   $0x1,-0xc(%rbp)
  jne    L36
  mov    -0x10(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int n;
  int p = 0, q = 1, r = n, h;
  while (q <= n)
    q *= 4;
  while (q != 1) {
    n = n;
    q = q / 4;
    h = p + q;
    p = p / 2;
    if (r >= h) {
      p = p + q;
      r = r - h;
    }
  }
  return p;
}
```
user: 这段汇编代码用C语言表示是什么样的?

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

assistant: ```c
int main() {
  int n;
  int p = 0, q = 1, r = n, h;
  while (q <= n)
    q *= 4;
  while (q != 1) {
    n = n;
    q = q / 4;
    h = p + q;
    p = p / 2;
    if (r >= h) {
      p = p + q;
      r = r - h;
    }
  }
  return p;
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
int main() {
  int n;
  int p = 0, q = 1, r = n, h;
  while (q <= n)
    q *= 4;
  while (q != 1) {
    n = n;
    q = q / 4;
    h = p + q;
    p = p / 2;
    if (r >= h) {
      p = p + q;
      r = r - h;
    }
  }
  return p;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

assistant: ```c
int main() {
  int n;
  int p = 0, q = 1, r = n, h;
  while (q <= n)
    q *= 4;
  while (q != 1) {
    n = n;
    q = q / 4;
    h = p + q;
    p = p / 2;
    if (r >= h) {
      p = p + q;
      r = r - h;
    }
  }
  return p;
}
```
user: Translate the below assembly code into C programming language:

```asm
<RandomFunc>:
  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),%rax
  mov    (%rax),%eax
  sub    $0x99f4bb,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  nop
  mov    -0xc(%rbp),%eax
  lea    -0x2bbfcfe6(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy12;
  unsigned short copy13;
  {
    state[0UL] = input[0UL] + 4284877637U;
    local1 = 0UL;
    while (local1 < 0U) {
      if (!(state[0UL] > local1)) {
        state[local1] += state[0UL];
      }
      if (state[0UL] == local1) {
        copy12 = *((unsigned short *)(&state[local1]) + 1);
        *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = copy12;
      } else {
        copy13 = *((unsigned short *)(&state[local1]) + 1);
        *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = copy13;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] - 733990886UL;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  sub    $0x2c59c4a1,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy12;
  unsigned short copy13;
  {
    state[0UL] = input[0UL] + 4284877637U;
    local1 = 0UL;
    while (local1 < 0U) {
      if (!(state[0UL] > local1)) {
        state[local1] += state[0UL];
      }
      if (state[0UL] == local1) {
        copy12 = *((unsigned short *)(&state[local1]) + 1);
        *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = copy12;
      } else {
        copy13 = *((unsigned short *)(&state[local1]) + 1);
        *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = copy13;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] - 733990886UL;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  sub    $0x2c59c4a1,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy12;
  unsigned short copy13;
  {
    state[0UL] = input[0UL] + 4284877637U;
    local1 = 0UL;
    while (local1 < 0U) {
      if (!(state[0UL] > local1)) {
        state[local1] += state[0UL];
      }
      if (state[0UL] == local1) {
        copy12 = *((unsigned short *)(&state[local1]) + 1);
        *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = copy12;
      } else {
        copy13 = *((unsigned short *)(&state[local1]) + 1);
        *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = copy13;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] - 733990886UL;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  sub    $0x2c59c4a1,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy12;
  unsigned short copy13;
  {
    state[0UL] = input[0UL] + 4284877637U;
    local1 = 0UL;
    while (local1 < 0U) {
      if (!(state[0UL] > local1)) {
        state[local1] += state[0UL];
      }
      if (state[0UL] == local1) {
        copy12 = *((unsigned short *)(&state[local1]) + 1);
        *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = copy12;
      } else {
        copy13 = *((unsigned short *)(&state[local1]) + 1);
        *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = copy13;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] - 733990886UL;
  }
}
```
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    $0x100,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x100(%rbp)
  jmp    L41
L20:
  mov    -0x100(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x100(%rbp),%eax
  cltq
  mov    %edx,-0xf0(%rbp,%rax,4)
  addl   $0x1,-0x100(%rbp)
L41:
  cmpl   $0x13,-0x100(%rbp)
  jle    L20
  movl   $0x0,-0xfc(%rbp)
  jmp    L32
L61:
  lea    -0xf0(%rbp),%rax
  mov    $0x14,%esi
  mov    %rax,%rdi
  call   <createList@plt>
  mov    -0xfc(%rbp),%edx
  movslq %edx,%rdx
  mov    %rax,-0xa0(%rbp,%rdx,8)
  addl   $0x1,-0xfc(%rbp)
L32:
  cmpl   $0x11,-0xfc(%rbp)
  jle    L61
  movl   $0x0,-0xf8(%rbp)
  jmp    L55
L67:
  mov    -0xf8(%rbp),%eax
  lea    0x2(%rax),%edx
  mov    -0xf8(%rbp),%eax
  cltq
  mov    -0xa0(%rbp,%rax,8),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <reverseKGroup@plt>
  mov    -0xf8(%rbp),%edx
  movslq %edx,%rdx
  mov    %rax,-0xa0(%rbp,%rdx,8)
  addl   $0x1,-0xf8(%rbp)
L55:
  cmpl   $0x11,-0xf8(%rbp)
  jle    L67
  movl   $0x0,-0xf4(%rbp)
  jmp    L12
L2:
  mov    -0xf4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xf4(%rbp),%eax
  cltq
  mov    -0xa0(%rbp,%rax,8),%rax
  mov    %rax,%rdi
  call   <printList@plt>
  addl   $0x1,-0xf4(%rbp)
L12:
  cmpl   $0x11,-0xf4(%rbp)
  jle    L2
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L84
  call   <__stack_chk_fail@plt>
L84:
  leave
  ret
```

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

tool: D3: No. %d testcase:


assistant: ```c
int main(void) {
  int arr[20];
  struct ListNode *l[18];
  for (int i = 0; i < 20; i++) {
    arr[i] = i + 1;
  }
  for (int i = 0; i < 18; i++) {
    l[i] = createList(arr, 20);
  }
  for (int i = 0; i < 18; i++) {
    l[i] = reverseKGroup(l[i], i + 2);
  }
  for (int i = 0; i < 18; i++) {
    printf("No. %d testcase:\n", i + 1);
    printList(l[i]);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xf8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xe8(%rsp)
  xor    %eax,%eax
  mov    $0x1,%eax
L19:
  mov    %eax,-0x4(%rsp,%rax,4)
  add    $0x1,%rax
  cmp    $0x15,%rax
  jne    L19
  lea    0x50(%rsp),%rbx
  lea    0xe0(%rsp),%r12
  mov    %rbx,%rbp
  mov    %rsp,%r13
L35:
  mov    $0x14,%esi
  mov    %r13,%rdi
  call   <createList@plt>
  mov    %rax,0x0(%rbp)
  add    $0x8,%rbp
  cmp    %r12,%rbp
  jne    L35
  mov    $0x2,%ebp
L38:
  mov    %ebp,%esi
  mov    (%rbx),%rdi
  call   <reverseKGroup@plt>
  mov    %rax,(%rbx)
  add    $0x8,%rbx
  add    $0x1,%ebp
  cmp    %r12,%rbx
  jne    L38
  mov    $0x1,%ebx
  lea    D3(%rip),%rbp
L2:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x48(%rsp,%rbx,8),%rdi
  call   <printList@plt>
  add    $0x1,%rbx
  cmp    $0x13,%rbx
  jne    L2
  mov    0xe8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L46
  mov    $0x0,%eax
  add    $0xf8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L46:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D3
type: string

tool: D3: No. %d testcase:


assistant: ```c
int main(void) {
  int arr[20];
  struct ListNode *l[18];
  for (int i = 0; i < 20; i++) {
    arr[i] = i + 1;
  }
  for (int i = 0; i < 18; i++) {
    l[i] = createList(arr, 20);
  }
  for (int i = 0; i < 18; i++) {
    l[i] = reverseKGroup(l[i], i + 2);
  }
  for (int i = 0; i < 18; i++) {
    printf("No. %d testcase:\n", i + 1);
    printList(l[i]);
  }
  return 0;
}
```
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    $0xf0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xe8(%rsp)
  mov    $0x1,%eax
  mov    %rsp,%r12
  nopl   0x0(%rax)
L46:
  mov    %eax,-0x4(%r12,%rax,4)
  add    $0x1,%rax
  cmp    $0x15,%rax
  jne    L46
  lea    0x50(%rsp),%r13
  lea    0xe0(%rsp),%r14
  mov    %r13,%rbx
  mov    %r13,%rbp
  nopw   0x0(%rax,%rax,1)
L19:
  mov    $0x14,%esi
  mov    %r12,%rdi
  add    $0x8,%rbp
  call   <createList@plt>
  mov    %rax,-0x8(%rbp)
  cmp    %r14,%rbp
  jne    L19
  mov    $0x2,%ebp
  nopw   0x0(%rax,%rax,1)
L25:
  mov    (%rbx),%rdi
  mov    %ebp,%esi
  add    $0x8,%rbx
  add    $0x1,%ebp
  call   <reverseKGroup@plt>
  mov    %rax,-0x8(%rbx)
  cmp    %r14,%rbx
  jne    L25
  mov    $0x1,%ebx
  lea    D5(%rip),%rbp
  cs nopw 0x0(%rax,%rax,1)
L37:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    -0x8(%r13,%rbx,8),%rdi
  add    $0x1,%rbx
  call   <printList@plt>
  cmp    $0x13,%rbx
  jne    L37
  mov    0xe8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  add    $0xf0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D5
type: string

tool: D5: No. %d testcase:


assistant: ```c
int main(void) {
  int arr[20];
  struct ListNode *l[18];
  for (int i = 0; i < 20; i++) {
    arr[i] = i + 1;
  }
  for (int i = 0; i < 18; i++) {
    l[i] = createList(arr, 20);
  }
  for (int i = 0; i < 18; i++) {
    l[i] = reverseKGroup(l[i], i + 2);
  }
  for (int i = 0; i < 18; i++) {
    printf("No. %d testcase:\n", i + 1);
    printList(l[i]);
  }
  return 0;
}
```
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
  sub    $0xf0,%rsp
  movdqa D13(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0xe8(%rsp)
  xor    %eax,%eax
  lea    0x50(%rsp),%r13
  lea    0xe0(%rsp),%r12
  mov    %rsp,%r14
  movaps %xmm0,(%rsp)
  movdqa D7(%rip),%xmm0
  mov    %r13,%rbx
  mov    %r13,%rbp
  movaps %xmm0,0x10(%rsp)
  movdqa D38(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  movdqa D4(%rip),%xmm0
  movaps %xmm0,0x30(%rsp)
  movdqa D57(%rip),%xmm0
  movaps %xmm0,0x40(%rsp)
  nopl   0x0(%rax)
L1:
  mov    $0x14,%esi
  mov    %r14,%rdi
  add    $0x8,%rbp
  call   <createList@plt>
  mov    %rax,-0x8(%rbp)
  cmp    %r12,%rbp
  jne    L1
  mov    $0x2,%ebp
  nop
L15:
  mov    (%rbx),%rdi
  mov    %ebp,%esi
  add    $0x8,%rbx
  add    $0x1,%ebp
  call   <reverseKGroup@plt>
  mov    %rax,-0x8(%rbx)
  cmp    %r12,%rbx
  jne    L15
  mov    $0x1,%ebx
  lea    D35(%rip),%rbp
  cs nopw 0x0(%rax,%rax,1)
L23:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    -0x8(%r13,%rbx,8),%rdi
  add    $0x1,%rbx
  call   <printList@plt>
  cmp    $0x13,%rbx
  jne    L23
  mov    0xe8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0xf0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D13
datatype: byte[16]
name: read_data
arguments: data_label: D7
datatype: byte[16]
name: read_data
arguments: data_label: D38
datatype: byte[16]
name: read_data
arguments: data_label: D4
datatype: byte[16]
name: read_data
arguments: data_label: D57
datatype: byte[16]
name: read_data
arguments: data_label: D35
datatype: string

tool: {"D13": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D7": [0x05,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x08,0x00,0x00,0x00]}

tool: {"D38": [0x09,0x00,0x00,0x00,0x0a,0x00,0x00,0x00,0x0b,0x00,0x00,0x00,0x0c,0x00,0x00,0x00]}

tool: {"D4": [0x0d,0x00,0x00,0x00,0x0e,0x00,0x00,0x00,0x0f,0x00,0x00,0x00,0x10,0x00,0x00,0x00]}

tool: {"D57": [0x11,0x00,0x00,0x00,0x12,0x00,0x00,0x00,0x13,0x00,0x00,0x00,0x14,0x00,0x00,0x00]}

tool: D35: No. %d testcase:


assistant: ```c
int main(void) {
  int arr[20];
  struct ListNode *l[18];
  for (int i = 0; i < 20; i++) {
    arr[i] = i + 1;
  }
  for (int i = 0; i < 18; i++) {
    l[i] = createList(arr, 20);
  }
  for (int i = 0; i < 18; i++) {
    l[i] = reverseKGroup(l[i], i + 2);
  }
  for (int i = 0; i < 18; i++) {
    printf("No. %d testcase:\n", i + 1);
    printList(l[i]);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<__addr_to_num>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  movl   $0x3,-0x14(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L1
L6:
  cmpl   $0x2e,-0x10(%rbp)
  jne    L49
  cmpl   $0xff,-0x18(%rbp)
  jbe    L8
  mov    $0x0,%eax
  jmp    L35
L8:
  shlq   $0x8,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  add    %rax,-0x8(%rbp)
  subl   $0x1,-0x14(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L1
L49:
  mov    -0x18(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  shll   $0x3,-0x18(%rbp)
  mov    -0xc(%rbp),%eax
  add    %eax,-0x18(%rbp)
  mov    -0xc(%rbp),%eax
  add    %eax,-0x18(%rbp)
  subl   $0x30,-0x10(%rbp)
  cmpl   $0x9,-0x10(%rbp)
  jbe    L60
  mov    $0x0,%eax
  jmp    L35
L60:
  mov    -0x10(%rbp),%eax
  add    %eax,-0x18(%rbp)
L1:
  mov    -0x28(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x10(%rbp)
  cmpl   $0x0,-0x10(%rbp)
  jne    L6
  cmpl   $0xff,-0x18(%rbp)
  ja     L20
  cmpl   $0x0,-0x14(%rbp)
  je     L2
L20:
  mov    $0x0,%eax
  jmp    L35
L2:
  shlq   $0x8,-0x8(%rbp)
  mov    -0x18(%rbp),%edx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
L35:
  pop    %rbp
  ret
```

assistant: ```c
unsigned long __addr_to_num(const char *addr) {
  unsigned int c, octet, t;
  unsigned long ipnum;
  int i = 3;
  octet = ipnum = 0;
  while ((c = *addr++)) {
    if (c == '.') {
      if (octet > 255) {
        return 0;
      }
      ipnum <<= 8;
      ipnum += octet;
      i--;
      octet = 0;
    } else {
      t = octet;
      octet <<= 3;
      octet += t;
      octet += t;
      c -= '0';
      if (c > 9) {
        return 0;
      }
      octet += c;
    }
  }
  if ((octet > 255) || (i != 0)) {
    return 0;
  }
  ipnum <<= 8;
  return ipnum + octet;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<__addr_to_num>:
  endbr64
  lea    0x1(%rdi),%rsi
  movsbl (%rdi),%ecx
  test   %ecx,%ecx
  je     L53
  mov    $0x3,%edi
  mov    $0x0,%eax
  mov    $0x0,%edx
  mov    $0x0,%r8d
  jmp    L34
L11:
  cmp    $0xff,%edx
  ja     L62
  shl    $0x8,%rax
  mov    %edx,%edx
  add    %rdx,%rax
  sub    $0x1,%edi
  mov    %r8d,%edx
L12:
  add    $0x1,%rsi
  movsbl -0x1(%rsi),%ecx
  test   %ecx,%ecx
  je     L5
L34:
  cmp    $0x2e,%ecx
  je     L11
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    $0x30,%ecx
  cmp    $0x9,%ecx
  ja     L44
  add    %ecx,%edx
  jmp    L12
L5:
  cmp    $0xff,%edx
  ja     L23
  test   %edi,%edi
  jne    L23
  shl    $0x8,%rax
  mov    %edx,%edx
  add    %rdx,%rax
  ret
L53:
  mov    $0x0,%eax
  ret
L62:
  mov    $0x0,%eax
  ret
L44:
  mov    $0x0,%eax
  ret
L23:
  mov    $0x0,%eax
  ret
```

assistant: ```c
unsigned long __addr_to_num(const char *addr) {
  unsigned int c, octet, t;
  unsigned long ipnum;
  int i = 3;
  octet = ipnum = 0;
  while ((c = *addr++)) {
    if (c == '.') {
      if (octet > 255) {
        return 0;
      }
      ipnum <<= 8;
      ipnum += octet;
      i--;
      octet = 0;
    } else {
      t = octet;
      octet <<= 3;
      octet += t;
      octet += t;
      c -= '0';
      if (c > 9) {
        return 0;
      }
      octet += c;
    }
  }
  if ((octet > 255) || (i != 0)) {
    return 0;
  }
  ipnum <<= 8;
  return ipnum + octet;
}
```