ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<__strverscmp>:
  endbr64
  cmp    %rsi,%rdi
  je     L50
  movzbl (%rdi),%edx
  movzbl (%rsi),%r9d
  lea    0x1(%rdi),%r8
  lea    0x1(%rsi),%rdi
  lea    -0x30(%rdx),%eax
  mov    %edx,%ecx
  cmp    $0x9,%eax
  setbe  %al
  xor    %esi,%esi
  cmp    $0x30,%dl
  sete   %sil
  movzbl %al,%eax
  add    %esi,%eax
  movzbl %r9b,%esi
  sub    %esi,%edx
  jne    L84
  test   %cl,%cl
  je     L84
  lea    D0(%rip),%r11
  jmp    L75
  nopw   0x0(%rax,%rax,1)
L36:
  test   %cl,%cl
  je     L84
L75:
  movzbl (%r8),%edx
  movslq %eax,%rsi
  add    $0x1,%r8
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%r9d
  lea    -0x30(%rdx),%eax
  mov    %edx,%ecx
  cmp    $0x9,%eax
  setbe  %al
  xor    %r10d,%r10d
  cmp    $0x30,%dl
  sete   %r10b
  movzbl %al,%eax
  add    %r10d,%eax
  or     (%r11,%rsi,4),%eax
  movzbl %r9b,%esi
  sub    %esi,%edx
  je     L36
L84:
  sub    $0x30,%esi
  xor    %ecx,%ecx
  cmp    $0x9,%esi
  setbe  %cl
  xor    %esi,%esi
  cmp    $0x30,%r9b
  sete   %sil
  shl    $0x2,%eax
  add    %esi,%ecx
  or     %ecx,%eax
  lea    D11(%rip),%rcx
  cltq
  mov    (%rcx,%rax,4),%eax
  cmp    $0x2,%eax
  je     L43
  cmp    $0x3,%eax
  jne    L34
  xor    %eax,%eax
  jmp    L47
  nopl   (%rax)
L55:
  add    $0x1,%rax
  cmp    $0x9,%ecx
  ja     L73
L47:
  movzbl (%r8,%rax,1),%esi
  movzbl (%rdi,%rax,1),%ecx
  sub    $0x30,%esi
  sub    $0x30,%ecx
  cmp    $0x9,%esi
  jbe    L55
  cmp    $0x9,%ecx
  mov    $0xffffffff,%eax
  cmovbe %eax,%edx
L43:
  mov    %edx,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L34:
  mov    %eax,%edx
  mov    %edx,%eax
  ret
  nopl   (%rax)
L50:
  xor    %edx,%edx
  mov    %edx,%eax
  ret
  nopl   (%rax)
L73:
  mov    $0x1,%edx
  mov    %edx,%eax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D0
type: string
name: extract_symbolic_data
arguments: data_label: D11
type: string

tool: 

tool: D11: 

assistant: ```c
int __strverscmp(s1, s2) const char *s1;
const char *s2;
{
  const unsigned char *p1 = (const unsigned char *)s1;
  const unsigned char *p2 = (const unsigned char *)s2;
  unsigned char c1, c2;
  int state;
  int diff;
  static const unsigned int next_state[] = {0x0, 0x4, 0xC, 0x0, 0x0, 0x4, 0x4, 0x4, 0x0, 0x8, 0x8, 0x8, 0x0, 0x8, 0xC, 0xC};
  static const int result_type[] = {2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, -1, -1, 2, +1, 3, 3, 2, +1, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, +1, +1, 2, -1, 2, 2, 2, -1, 2, 2, 2};
  if (p1 == p2)
    return 0;
  c1 = *p1++;
  c2 = *p2++;
  state = 0x0 | ((c1 == '0') + (isdigit(c1) != 0));
  while ((diff = c1 - c2) == 0 && c1 != '\0') {
    state = next_state[state];
    c1 = *p1++;
    c2 = *p2++;
    state |= (c1 == '0') + (isdigit(c1) != 0);
  }
  state = result_type[state << 2 | (((c2 == '0') + (isdigit(c2) != 0)))];
  switch (state) {
  case 2:
    return diff;
  case 3:
    while (isdigit(*p1++))
      if (!isdigit(*p2++))
        return 1;
    return isdigit(*p2) ? -1 : diff;
  default:
    return state;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<__strverscmp>:
  endbr64
  cmp    %rsi,%rdi
  je     L101
  movzbl (%rdi),%edx
  movzbl (%rsi),%r9d
  lea    0x1(%rdi),%r8
  lea    0x1(%rsi),%rdi
  lea    -0x30(%rdx),%eax
  mov    %edx,%ecx
  cmp    $0x9,%eax
  setbe  %al
  xor    %esi,%esi
  cmp    $0x30,%dl
  sete   %sil
  movzbl %al,%eax
  add    %esi,%eax
  movzbl %r9b,%esi
  sub    %esi,%edx
  jne    L36
  test   %cl,%cl
  je     L36
  lea    D15(%rip),%r11
  jmp    L84
  nopw   0x0(%rax,%rax,1)
L40:
  test   %cl,%cl
  je     L36
L84:
  movzbl (%r8),%edx
  movslq %eax,%rsi
  add    $0x1,%r8
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%r9d
  lea    -0x30(%rdx),%eax
  mov    %edx,%ecx
  cmp    $0x9,%eax
  setbe  %al
  xor    %r10d,%r10d
  cmp    $0x30,%dl
  sete   %r10b
  movzbl %al,%eax
  add    %r10d,%eax
  or     (%r11,%rsi,4),%eax
  movzbl %r9b,%esi
  sub    %esi,%edx
  je     L40
L36:
  sub    $0x30,%esi
  xor    %ecx,%ecx
  cmp    $0x9,%esi
  setbe  %cl
  xor    %esi,%esi
  cmp    $0x30,%r9b
  sete   %sil
  shl    $0x2,%eax
  add    %esi,%ecx
  or     %ecx,%eax
  lea    D4(%rip),%rcx
  cltq
  mov    (%rcx,%rax,4),%eax
  cmp    $0x2,%eax
  je     L3
  cmp    $0x3,%eax
  jne    L65
  xor    %eax,%eax
  jmp    L83
  nopl   (%rax)
L0:
  add    $0x1,%rax
  cmp    $0x9,%ecx
  ja     L82
L83:
  movzbl (%r8,%rax,1),%esi
  movzbl (%rdi,%rax,1),%ecx
  sub    $0x30,%esi
  sub    $0x30,%ecx
  cmp    $0x9,%esi
  jbe    L0
  cmp    $0x9,%ecx
  mov    $0xffffffff,%eax
  cmovbe %eax,%edx
L3:
  mov    %edx,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L65:
  mov    %eax,%edx
  mov    %edx,%eax
  ret
  nopl   (%rax)
L101:
  xor    %edx,%edx
  mov    %edx,%eax
  ret
  nopl   (%rax)
L82:
  mov    $0x1,%edx
  mov    %edx,%eax
  ret
```

tool_calls: name: unpack_data
arguments: label: D15
type: string
name: unpack_data
arguments: label: D4
type: string

tool: 

tool: D4: 

assistant: ```c
int __strverscmp(s1, s2) const char *s1;
const char *s2;
{
  const unsigned char *p1 = (const unsigned char *)s1;
  const unsigned char *p2 = (const unsigned char *)s2;
  unsigned char c1, c2;
  int state;
  int diff;
  static const unsigned int next_state[] = {0x0, 0x4, 0xC, 0x0, 0x0, 0x4, 0x4, 0x4, 0x0, 0x8, 0x8, 0x8, 0x0, 0x8, 0xC, 0xC};
  static const int result_type[] = {2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, -1, -1, 2, +1, 3, 3, 2, +1, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, +1, +1, 2, -1, 2, 2, 2, -1, 2, 2, 2};
  if (p1 == p2)
    return 0;
  c1 = *p1++;
  c2 = *p2++;
  state = 0x0 | ((c1 == '0') + (isdigit(c1) != 0));
  while ((diff = c1 - c2) == 0 && c1 != '\0') {
    state = next_state[state];
    c1 = *p1++;
    c2 = *p2++;
    state |= (c1 == '0') + (isdigit(c1) != 0);
  }
  state = result_type[state << 2 | (((c2 == '0') + (isdigit(c2) != 0)))];
  switch (state) {
  case 2:
    return diff;
  case 3:
    while (isdigit(*p1++))
      if (!isdigit(*p2++))
        return 1;
    return isdigit(*p2) ? -1 : diff;
  default:
    return state;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<itooct>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L29
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movb   $0x30,(%rax)
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
  jmp    L55
L29:
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L44
L49:
  addl   $0x1,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x3,%eax
  mov    %eax,-0x4(%rbp)
L44:
  cmpl   $0x0,-0x4(%rbp)
  jne    L49
  mov    -0x8(%rbp),%eax
  cltq
  sub    $0x1,%rax
  add    %rax,-0x20(%rbp)
  jmp    L23
L52:
  mov    -0x14(%rbp),%eax
  and    $0x7,%eax
  lea    0x30(%rax),%ecx
  mov    -0x20(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x14(%rbp),%eax
  shr    $0x3,%eax
  mov    %eax,-0x14(%rbp)
L23:
  cmpl   $0x0,-0x14(%rbp)
  jne    L52
  mov    -0x8(%rbp),%eax
  cltq
  add    $0x1,%rax
  add    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
L55:
  pop    %rbp
  ret
```

assistant: ```c
void itooct(unsigned int value, char *str) {
  int length;
  unsigned int tmp;
  if (value == 0) {
    *(str++) = '0';
    *str = '\0';
    return;
  }
  tmp = value;
  length = 0;
  while (tmp > 0) {
    length++;
    tmp /= 0x8;
  }
  str += length - 1;
  while (value > 0) {
    *(str--) = (value % 0x8) + '0';
    value /= 0x8;
  }
  str += length + 1;
  *(str) = '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<itooct>:
  endbr64
  mov    %edi,%eax
  mov    $0x0,%edx
  test   %edi,%edi
  je     L16
L9:
  add    $0x1,%edx
  mov    %eax,%ecx
  shr    $0x3,%eax
  cmp    $0x7,%ecx
  ja     L9
  movslq %edx,%rcx
  lea    -0x1(%rsi,%rcx,1),%rax
L21:
  mov    %rax,%rsi
  sub    $0x1,%rax
  mov    %edi,%edx
  and    $0x7,%edx
  add    $0x30,%edx
  mov    %dl,0x1(%rax)
  mov    %edi,%edx
  shr    $0x3,%edi
  cmp    $0x7,%edx
  ja     L21
  movb   $0x0,(%rsi,%rcx,1)
  ret
L16:
  movb   $0x30,(%rsi)
  movb   $0x0,0x1(%rsi)
  ret
```

assistant: ```c
void itooct(unsigned int value, char *str) {
  int length;
  unsigned int tmp;
  if (value == 0) {
    *(str++) = '0';
    *str = '\0';
    return;
  }
  tmp = value;
  length = 0;
  while (tmp > 0) {
    length++;
    tmp /= 0x8;
  }
  str += length - 1;
  while (value > 0) {
    *(str--) = (value % 0x8) + '0';
    value /= 0x8;
  }
  str += length + 1;
  *(str) = '\0';
}
```
user: What is the c source code of the assembly code below:

```asm
<itooct>:
  endbr64
  mov    %edi,%eax
  xor    %edx,%edx
  test   %edi,%edi
  je     L9
  nopl   0x0(%rax)
L11:
  add    $0x1,%edx
  shr    $0x3,%eax
  jne    L11
  movslq %edx,%rcx
  lea    -0x1(%rsi,%rcx,1),%rax
L8:
  mov    %edi,%edx
  mov    %rax,%rsi
  sub    $0x1,%rax
  and    $0x7,%edx
  add    $0x30,%edx
  shr    $0x3,%edi
  mov    %dl,0x1(%rax)
  jne    L8
  movb   $0x0,(%rsi,%rcx,1)
  ret
  nopl   0x0(%rax)
L9:
  mov    $0x30,%eax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void itooct(unsigned int value, char *str) {
  int length;
  unsigned int tmp;
  if (value == 0) {
    *(str++) = '0';
    *str = '\0';
    return;
  }
  tmp = value;
  length = 0;
  while (tmp > 0) {
    length++;
    tmp /= 0x8;
  }
  str += length - 1;
  while (value > 0) {
    *(str--) = (value % 0x8) + '0';
    value /= 0x8;
  }
  str += length + 1;
  *(str) = '\0';
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<itooct>:
  endbr64
  mov    %edi,%eax
  xor    %edx,%edx
  test   %edi,%edi
  je     L0
  nopl   0x0(%rax)
L6:
  add    $0x1,%edx
  shr    $0x3,%eax
  jne    L6
  movslq %edx,%rcx
  lea    -0x1(%rsi,%rcx,1),%rax
L13:
  mov    %edi,%edx
  mov    %rax,%rsi
  sub    $0x1,%rax
  and    $0x7,%edx
  add    $0x30,%edx
  shr    $0x3,%edi
  mov    %dl,0x1(%rax)
  jne    L13
  movb   $0x0,(%rsi,%rcx,1)
  ret
  nopl   0x0(%rax)
L0:
  mov    $0x30,%eax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void itooct(unsigned int value, char *str) {
  int length;
  unsigned int tmp;
  if (value == 0) {
    *(str++) = '0';
    *str = '\0';
    return;
  }
  tmp = value;
  length = 0;
  while (tmp > 0) {
    length++;
    tmp /= 0x8;
  }
  str += length - 1;
  while (value > 0) {
    *(str--) = (value % 0x8) + '0';
    value /= 0x8;
  }
  str += length + 1;
  *(str) = '\0';
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<GetRange>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  movsd  0x3d0958(%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x1c(%rbp),%xmm1
  movsd  -0x8(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jbe    L0
  mov    -0x1c(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  jmp    L6
L0:
  pxor   %xmm0,%xmm0
L6:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double GetRange(double *sln, int yardage) {
  double size = sln[50001 * 10 + 1];
  if (yardage < size) {
    return sln[10 * yardage];
  } else
    return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<GetRange>:
  endbr64
  pxor   %xmm0,%xmm0
  cvtsi2sd %esi,%xmm0
  movsd  0x3d0958(%rdi),%xmm1
  comisd %xmm0,%xmm1
  jbe    L3
  lea    (%rsi,%rsi,4),%eax
  add    %eax,%eax
  cltq
  movsd  (%rdi,%rax,8),%xmm0
  ret
L3:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
double GetRange(double *sln, int yardage) {
  double size = sln[50001 * 10 + 1];
  if (yardage < size) {
    return sln[10 * yardage];
  } else
    return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<GetRange>:
  endbr64
  pxor   %xmm0,%xmm0
  movsd  0x3d0958(%rdi),%xmm1
  cvtsi2sd %esi,%xmm0
  comisd %xmm0,%xmm1
  jbe    L5
  lea    (%rsi,%rsi,4),%eax
  add    %eax,%eax
  cltq
  movsd  (%rdi,%rax,8),%xmm0
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
double GetRange(double *sln, int yardage) {
  double size = sln[50001 * 10 + 1];
  if (yardage < size) {
    return sln[10 * yardage];
  } else
    return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<GetRange>:
  endbr64
  pxor   %xmm0,%xmm0
  movsd  0x3d0958(%rdi),%xmm1
  cvtsi2sd %esi,%xmm0
  comisd %xmm0,%xmm1
  jbe    L9
  lea    (%rsi,%rsi,4),%eax
  add    %eax,%eax
  cltq
  movsd  (%rdi,%rax,8),%xmm0
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
double GetRange(double *sln, int yardage) {
  double size = sln[50001 * 10 + 1];
  if (yardage < size) {
    return sln[10 * yardage];
  } else
    return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<enemy_destroy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L0
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <destroy@plt>
L0:
  nop
  leave
  ret
```

assistant: ```c
void enemy_destroy(struct Enemy *enemy) {
  if (enemy) {
    destroy(enemy);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<enemy_destroy>:
  endbr64
  test   %rdi,%rdi
  je     L1
  sub    $0x8,%rsp
  call   <destroy@plt>
  add    $0x8,%rsp
  ret
L1:
  ret
```

assistant: ```c
void enemy_destroy(struct Enemy *enemy) {
  if (enemy) {
    destroy(enemy);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<enemy_destroy>:
  endbr64
  test   %rdi,%rdi
  je     L3
  jmp    1050 <destroy@plt>
  xchg   %ax,%ax
L3:
  ret
```

assistant: ```c
void enemy_destroy(struct Enemy *enemy) {
  if (enemy) {
    destroy(enemy);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<enemy_destroy>:
  endbr64
  test   %rdi,%rdi
  je     L2
  jmp    1050 <destroy@plt>
  xchg   %ax,%ax
L2:
  ret
```

assistant: ```c
void enemy_destroy(struct Enemy *enemy) {
  if (enemy) {
    destroy(enemy);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L22
L9:
  addl   $0x1,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L22:
  cmpl   $0x4,-0x4(%rbp)
  jle    L9
  jmp    L25
L18:
  subl   $0x1,-0x8(%rbp)
  subl   $0x1,-0x4(%rbp)
L25:
  cmpl   $0x0,-0x4(%rbp)
  jg     L18
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int a = 0;
  int i;
  for (i = 0; i < 5; i++) {
    a++;
  }
  while (i > 0) {
    a--;
    i--;
  }
  return a;
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
int main() {
  int a = 0;
  int i;
  for (i = 0; i < 5; i++) {
    a++;
  }
  while (i > 0) {
    a--;
    i--;
  }
  return a;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

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

assistant: ```c
int main() {
  int a = 0;
  int i;
  for (i = 0; i < 5; i++) {
    a++;
  }
  while (i > 0) {
    a--;
    i--;
  }
  return a;
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
int main() {
  int a = 0;
  int i;
  for (i = 0; i < 5; i++) {
    a++;
  }
  while (i > 0) {
    a--;
    i--;
  }
  return a;
}
```
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
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  movl   $0x1,-0xc(%rbp)
  jmp    L21
L25:
  mov    -0xc(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x2aaaaaab,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  shl    $0x3,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  test   %edx,%edx
  jne    L35
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
L46:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L46
L35:
  mov    -0xc(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L21:
  mov    -0x10(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jle    L25
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D25
data_type: string
name: read_data
arguments: label: D37
data_type: string
name: read_data
arguments: label: D38
data_type: string
name: read_data
arguments: label: D14
data_type: string
name: read_data
arguments: label: D39
data_type: string

tool: D25: This program prints a table of squares.

tool: D37: Enter the number of entries in table: 

tool: D38: %d

tool: D14: Press ENTER to continue...

tool: D39: %10d	%10d


assistant: ```c
int main(void) {
  int i, n;
  printf("This program prints a table of squares.\n");
  printf("Enter the number of entries in table: ");
  scanf("%d", &n);
  getchar();
  for (i = 1; i <= n; i += 1) {
    if (i % 24 == 0) {
      printf("Press ENTER to continue...\n");
      while (getchar() != '\n') {
      }
    }
    printf("%10d\t%10d\n", i, i * i);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D15(%rip),%rdi
  call   <puts@plt>
  lea    D32(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D11(%rip),%rdi
  call   <scanf@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x1,%ebx
  lea    D44(%rip),%rbp
  lea    D49(%rip),%r12
  cmpl   $0x0,0x4(%rsp)
  jle    L59
L6:
  mov    %ebx,%edx
  imul   %ebx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,0x4(%rsp)
  jl     L59
  movslq %ebx,%rax
  imul   $0x2aaaaaab,%rax,%rax
  sar    $0x22,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,2),%eax
  shl    $0x3,%eax
  cmp    %eax,%ebx
  jne    L6
  mov    %r12,%rdi
  call   <puts@plt>
L31:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L31
  jmp    L6
L59:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D15
type_guessed: string
name: unpack_data
arguments: data_tag: D32
type_guessed: string
name: unpack_data
arguments: data_tag: D11
type_guessed: string
name: unpack_data
arguments: data_tag: D44
type_guessed: string
name: unpack_data
arguments: data_tag: D49
type_guessed: string

tool: D15: This program prints a table of squares.

tool: D32: Enter the number of entries in table: 

tool: D11: %d

tool: D44: %10d	%10d


tool: D49: Press ENTER to continue...

assistant: ```c
int main(void) {
  int i, n;
  printf("This program prints a table of squares.\n");
  printf("Enter the number of entries in table: ");
  scanf("%d", &n);
  getchar();
  for (i = 1; i <= n; i += 1) {
    if (i % 24 == 0) {
      printf("Press ENTER to continue...\n");
      while (getchar() != '\n') {
      }
    }
    printf("%10d\t%10d\n", i, i * i);
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r12
  lea    D47(%rip),%rdi
  lea    D33(%rip),%r12
  push   %rbp
  lea    D20(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D45(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L49
  nopl   0x0(%rax)
L13:
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  imul   %ebx,%edx
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jl     L49
  imul   $0xaaaaaaab,%ebx,%eax
  ror    $0x3,%eax
  cmp    $0xaaaaaaa,%eax
  ja     L13
  mov    %r12,%rdi
  call   <puts@plt>
  nopw   0x0(%rax,%rax,1)
L45:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L45
  jmp    L13
  nopl   (%rax)
L49:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D47
type_infered: string
name: disassemble_data
arguments: data_label: D33
type_infered: string
name: disassemble_data
arguments: data_label: D20
type_infered: string
name: disassemble_data
arguments: data_label: D45
type_infered: string
name: disassemble_data
arguments: data_label: D13
type_infered: string

tool: D47: This program prints a table of squares.

tool: D33: Press ENTER to continue...

tool: D20: %10d	%10d


tool: D45: Enter the number of entries in table: 

tool: D13: %d

assistant: ```c
int main(void) {
  int i, n;
  printf("This program prints a table of squares.\n");
  printf("Enter the number of entries in table: ");
  scanf("%d", &n);
  getchar();
  for (i = 1; i <= n; i += 1) {
    if (i % 24 == 0) {
      printf("Press ENTER to continue...\n");
      while (getchar() != '\n') {
      }
    }
    printf("%10d\t%10d\n", i, i * i);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D3(%rip),%rdi
  lea    D18(%rip),%r12
  push   %rbp
  lea    D46(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D21(%rip),%rdi
  call   <scanf@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L58
  nopl   0x0(%rax)
L37:
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  imul   %ebx,%edx
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jl     L58
  imul   $0xaaaaaaab,%ebx,%eax
  ror    $0x3,%eax
  cmp    $0xaaaaaaa,%eax
  ja     L37
  mov    %r12,%rdi
  call   <puts@plt>
  nopw   0x0(%rax,%rax,1)
L43:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L43
  jmp    L37
  nopl   (%rax)
L58:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D3
type: string
name: read_data
arguments: label: D18
type: string
name: read_data
arguments: label: D46
type: string
name: read_data
arguments: label: D14
type: string
name: read_data
arguments: label: D21
type: string

tool: D3: This program prints a table of squares.

tool: D18: Press ENTER to continue...

tool: D46: %10d	%10d


tool: D14: Enter the number of entries in table: 

tool: D21: %d

assistant: ```c
int main(void) {
  int i, n;
  printf("This program prints a table of squares.\n");
  printf("Enter the number of entries in table: ");
  scanf("%d", &n);
  getchar();
  for (i = 1; i <= n; i += 1) {
    if (i % 24 == 0) {
      printf("Press ENTER to continue...\n");
      while (getchar() != '\n') {
      }
    }
    printf("%10d\t%10d\n", i, i * i);
  }
  return 0;
}
```