ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fb_move_cursor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x14(%rbp)
  mov    %al,-0x18(%rbp)
  movzbl -0x14(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  movzbl -0x18(%rbp),%eax
  add    %edx,%eax
  mov    %ax,-0x2(%rbp)
  mov    $0xf,%esi
  mov    $0x3d4,%edi
  call   <outb@plt>
  movzwl -0x2(%rbp),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  mov    $0x3d5,%edi
  call   <outb@plt>
  mov    $0xe,%esi
  mov    $0x3d4,%edi
  call   <outb@plt>
  movzwl -0x2(%rbp),%eax
  shr    $0x8,%ax
  movzwl %ax,%eax
  mov    %eax,%esi
  mov    $0x3d5,%edi
  call   <outb@plt>
  nop
  leave
  ret
```

assistant: ```c
void fb_move_cursor(unsigned char row, unsigned char col) {
  unsigned short pos = (row * 80) + col;
  outb(0x3D4, 0x0F);
  outb(0x3D5, pos & 0x00ff);
  outb(0x3D4, 0x0E);
  outb(0x3D5, ((pos >> 8) & 0x00ff));
}
```
user: What is the c source code of the assembly code below?

```asm
<fb_move_cursor>:
  endbr64
  push   %rbx
  movzbl %dil,%edi
  lea    (%rdi,%rdi,4),%edi
  shl    $0x4,%edi
  movzbl %sil,%esi
  lea    (%rdi,%rsi,1),%ebx
  mov    $0xf,%esi
  mov    $0x3d4,%edi
  call   <outb@plt>
  movzbl %bl,%esi
  mov    $0x3d5,%edi
  call   <outb@plt>
  mov    $0xe,%esi
  mov    $0x3d4,%edi
  call   <outb@plt>
  shr    $0x8,%bx
  movzwl %bx,%esi
  mov    $0x3d5,%edi
  call   <outb@plt>
  pop    %rbx
  ret
```

assistant: ```c
void fb_move_cursor(unsigned char row, unsigned char col) {
  unsigned short pos = (row * 80) + col;
  outb(0x3D4, 0x0F);
  outb(0x3D5, pos & 0x00ff);
  outb(0x3D4, 0x0E);
  outb(0x3D5, ((pos >> 8) & 0x00ff));
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<fb_move_cursor>:
  endbr64
  movzbl %dil,%edi
  push   %rbx
  movzbl %sil,%esi
  lea    (%rdi,%rdi,4),%edi
  shl    $0x4,%edi
  lea    (%rdi,%rsi,1),%ebx
  mov    $0xf,%esi
  mov    $0x3d4,%edi
  call   <outb@plt>
  movzbl %bl,%esi
  mov    $0x3d5,%edi
  shr    $0x8,%bx
  call   <outb@plt>
  mov    $0xe,%esi
  mov    $0x3d4,%edi
  call   <outb@plt>
  movzwl %bx,%esi
  mov    $0x3d5,%edi
  pop    %rbx
  jmp    1050 <outb@plt>
```

assistant: ```c
void fb_move_cursor(unsigned char row, unsigned char col) {
  unsigned short pos = (row * 80) + col;
  outb(0x3D4, 0x0F);
  outb(0x3D5, pos & 0x00ff);
  outb(0x3D4, 0x0E);
  outb(0x3D5, ((pos >> 8) & 0x00ff));
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<fb_move_cursor>:
  endbr64
  movzbl %dil,%edi
  push   %rbx
  movzbl %sil,%esi
  lea    (%rdi,%rdi,4),%edi
  shl    $0x4,%edi
  lea    (%rdi,%rsi,1),%ebx
  mov    $0xf,%esi
  mov    $0x3d4,%edi
  call   <outb@plt>
  movzbl %bl,%esi
  mov    $0x3d5,%edi
  shr    $0x8,%bx
  call   <outb@plt>
  mov    $0xe,%esi
  mov    $0x3d4,%edi
  call   <outb@plt>
  movzwl %bx,%esi
  mov    $0x3d5,%edi
  pop    %rbx
  jmp    1050 <outb@plt>
```

assistant: ```c
void fb_move_cursor(unsigned char row, unsigned char col) {
  unsigned short pos = (row * 80) + col;
  outb(0x3D4, 0x0F);
  outb(0x3D5, pos & 0x00ff);
  outb(0x3D4, 0x0E);
  outb(0x3D5, ((pos >> 8) & 0x00ff));
}
```
user: Translate the below assembly code into C programming language:

```asm
<harbol_graph_new_from_vector>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L10
  mov    $0x0,%eax
  jmp    L13
L10:
  mov    $0x0,%eax
  call   <harbol_graph_new@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <harbol_graph_from_vector@plt>
  mov    -0x8(%rbp),%rax
L13:
  leave
  ret
```

assistant: ```c
struct HarbolGraph *harbol_graph_new_from_vector(const struct HarbolVector *const vec) {
  if (!vec)
    return NULL;
  struct HarbolGraph *graph = harbol_graph_new();
  harbol_graph_from_vector(graph, vec);
  return graph;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<harbol_graph_new_from_vector>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L5
  mov    $0x0,%eax
  call   <harbol_graph_new@plt>
  mov    %rax,%rbp
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <harbol_graph_from_vector@plt>
L17:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L5:
  mov    %rdi,%rbp
  jmp    L17
```

assistant: ```c
struct HarbolGraph *harbol_graph_new_from_vector(const struct HarbolVector *const vec) {
  if (!vec)
    return NULL;
  struct HarbolGraph *graph = harbol_graph_new();
  harbol_graph_from_vector(graph, vec);
  return graph;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<harbol_graph_new_from_vector>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L6
  xor    %eax,%eax
  mov    %rdi,%rbp
  call   <harbol_graph_new@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <harbol_graph_from_vector@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L6:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
struct HarbolGraph *harbol_graph_new_from_vector(const struct HarbolVector *const vec) {
  if (!vec)
    return NULL;
  struct HarbolGraph *graph = harbol_graph_new();
  harbol_graph_from_vector(graph, vec);
  return graph;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<harbol_graph_new_from_vector>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L6
  xor    %eax,%eax
  mov    %rdi,%rbp
  call   <harbol_graph_new@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <harbol_graph_from_vector@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L6:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
struct HarbolGraph *harbol_graph_new_from_vector(const struct HarbolVector *const vec) {
  if (!vec)
    return NULL;
  struct HarbolGraph *graph = harbol_graph_new();
  harbol_graph_from_vector(graph, vec);
  return graph;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<OrdCresc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L6
L39:
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  cmp    %edx,-0x4(%rbp)
  jge    L19
  mov    $0x0,%eax
  jmp    L16
L19:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
L6:
  cmpl   $0x0,-0x14(%rbp)
  jne    L39
  mov    $0x1,%eax
L16:
  pop    %rbp
  ret
```

assistant: ```c
int OrdCresc(int a) {
  int max = a % 10;
  a = a / 10;
  while (a != 0) {
    if (a % 10 > max)
      return 0;
    else {
      max = a % 10;
      a = a / 10;
    }
  }
  return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<OrdCresc>:
  endbr64
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,%edx
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  mov    %edi,%eax
  sub    %ecx,%eax
  mov    %edx,%ecx
  test   %edx,%edx
  je     L27
  movslq %edx,%rdi
  imul   $0x66666667,%rdi,%rdi
  sar    $0x22,%rdi
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%edi
  lea    (%rdi,%rdi,4),%esi
  add    %esi,%esi
  mov    %edx,%edi
  sub    %esi,%edi
  cmp    %edi,%eax
  jl     L10
L9:
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  je     L48
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%esi
  sar    $0x1f,%esi
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%esi
  add    %esi,%esi
  sub    %esi,%eax
  cmp    %edi,%eax
  jg     L42
  mov    %eax,%edi
  jmp    L9
L48:
  mov    $0x1,%eax
  ret
L27:
  mov    $0x1,%eax
  ret
L10:
  mov    $0x0,%eax
  ret
L42:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int OrdCresc(int a) {
  int max = a % 10;
  a = a / 10;
  while (a != 0) {
    if (a % 10 > max)
      return 0;
    else {
      max = a % 10;
      a = a / 10;
    }
  }
  return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<OrdCresc>:
  endbr64
  movslq %edi,%rax
  mov    %edi,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  mov    %eax,%ecx
  mov    %edi,%eax
  add    %edx,%edx
  sub    %edx,%eax
  mov    %ecx,%edx
  test   %ecx,%ecx
  je     L38
  movslq %ecx,%rsi
  mov    %ecx,%edi
  imul   $0x66666667,%rsi,%rsi
  sar    $0x1f,%edi
  sar    $0x22,%rsi
  sub    %edi,%esi
  lea    (%rsi,%rsi,4),%edi
  mov    %ecx,%esi
  add    %edi,%edi
  sub    %edi,%esi
  cmp    %esi,%eax
  jge    L0
  jmp    L35
  xchg   %ax,%ax
L4:
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  mov    %edx,%eax
  add    %ecx,%ecx
  sub    %ecx,%eax
  cmp    %esi,%eax
  jg     L35
  mov    %eax,%esi
L0:
  movslq %edx,%rax
  sar    $0x1f,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sub    %edx,%eax
  mov    %eax,%edx
  jne    L4
L38:
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax)
L35:
  xor    %eax,%eax
  ret
```

assistant: ```c
int OrdCresc(int a) {
  int max = a % 10;
  a = a / 10;
  while (a != 0) {
    if (a % 10 > max)
      return 0;
    else {
      max = a % 10;
      a = a / 10;
    }
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<OrdCresc>:
  endbr64
  movslq %edi,%rax
  mov    %edi,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  mov    %eax,%ecx
  mov    %edi,%eax
  add    %edx,%edx
  sub    %edx,%eax
  mov    %ecx,%edx
  test   %ecx,%ecx
  je     L4
  movslq %ecx,%rsi
  mov    %ecx,%edi
  imul   $0x66666667,%rsi,%rsi
  sar    $0x1f,%edi
  sar    $0x22,%rsi
  sub    %edi,%esi
  lea    (%rsi,%rsi,4),%edi
  mov    %ecx,%esi
  add    %edi,%edi
  sub    %edi,%esi
  cmp    %esi,%eax
  jge    L43
  jmp    L37
  xchg   %ax,%ax
L19:
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  mov    %edx,%eax
  add    %ecx,%ecx
  sub    %ecx,%eax
  cmp    %esi,%eax
  jg     L37
  mov    %eax,%esi
L43:
  movslq %edx,%rax
  sar    $0x1f,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sub    %edx,%eax
  mov    %eax,%edx
  jne    L19
L4:
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax)
L37:
  xor    %eax,%eax
  ret
```

assistant: ```c
int OrdCresc(int a) {
  int max = a % 10;
  a = a / 10;
  while (a != 0) {
    if (a % 10 > max)
      return 0;
    else {
      max = a % 10;
      a = a / 10;
    }
  }
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<ft_get_next_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x1,-0x4(%rbp)
  mov    D39(%rip),%rax
  test   %rax,%rax
  jne    L41
  mov    -0x14(%rbp),%eax
  lea    D39(%rip),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <gnl_read_and_save@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L73
L41:
  mov    D39(%rip),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <ft_strchr@plt>
  test   %eax,%eax
  jne    L73
  mov    -0x14(%rbp),%eax
  lea    D39(%rip),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <gnl_read_and_save@plt>
  mov    %eax,-0x4(%rbp)
L73:
  mov    D39(%rip),%rax
  test   %rax,%rax
  jne    L68
  cmpl   $0x0,-0x4(%rbp)
  js     L14
L68:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <gnl_set_line@plt>
  mov    %eax,-0x4(%rbp)
L14:
  cmpl   $0x0,-0x4(%rbp)
  jg     L13
  mov    D39(%rip),%rax
  test   %rax,%rax
  je     L40
  mov    D39(%rip),%rax
  mov    %rax,%rdi
  call   <free@plt>
L40:
  mov    -0x4(%rbp),%eax
  jmp    L60
L13:
  mov    -0x4(%rbp),%eax
L60:
  leave
  ret
```

assistant: ```c
int ft_get_next_line(const int fd, char **line) {
  static char *rbak = NULL;
  int res;
  res = 1;
  if (rbak == NULL)
    res = gnl_read_and_save(fd, &rbak);
  else if (!(ft_strchr(rbak, '\n')))
    res = gnl_read_and_save(fd, &rbak);
  if (rbak != NULL || res >= 0)
    res = gnl_set_line(&rbak, line);
  if (res < 1) {
    if (rbak)
      free(rbak);
    return (res);
  }
  return (res);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ft_get_next_line>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  mov    D2(%rip),%rdi
  test   %rdi,%rdi
  je     L17
  mov    $0xa,%esi
  call   <ft_strchr@plt>
  test   %eax,%eax
  je     L48
L12:
  mov    %rbp,%rsi
  lea    D2(%rip),%rdi
  call   <gnl_set_line@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jle    L85
L83:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L17:
  lea    D2(%rip),%rsi
  mov    %ebx,%edi
  call   <gnl_read_and_save@plt>
  mov    %eax,%ebx
L38:
  cmpq   $0x0,D2(%rip)
  jne    L12
  test   %ebx,%ebx
  js     L83
  jmp    L12
L48:
  lea    D2(%rip),%rsi
  mov    %ebx,%edi
  call   <gnl_read_and_save@plt>
  mov    %eax,%ebx
  jmp    L38
L85:
  mov    D2(%rip),%rdi
  test   %rdi,%rdi
  je     L83
  call   <free@plt>
  jmp    L83
```

assistant: ```c
int ft_get_next_line(const int fd, char **line) {
  static char *rbak = NULL;
  int res;
  res = 1;
  if (rbak == NULL)
    res = gnl_read_and_save(fd, &rbak);
  else if (!(ft_strchr(rbak, '\n')))
    res = gnl_read_and_save(fd, &rbak);
  if (rbak != NULL || res >= 0)
    res = gnl_set_line(&rbak, line);
  if (res < 1) {
    if (rbak)
      free(rbak);
    return (res);
  }
  return (res);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_get_next_line>:
  endbr64
  push   %r13
  push   %r12
  mov    %edi,%r12d
  mov    D10(%rip),%rdi
  push   %rbp
  mov    %rsi,%rbp
  test   %rdi,%rdi
  je     L7
  mov    $0xa,%esi
  lea    D10(%rip),%r13
  call   <ft_strchr@plt>
  test   %eax,%eax
  je     L9
L25:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <gnl_set_line@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jle    L43
L54:
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  mov    %r12d,%edi
  mov    %r13,%rsi
  call   <gnl_read_and_save@plt>
  mov    %eax,%r12d
L50:
  cmpq   $0x0,D10(%rip)
  jne    L25
  test   %r12d,%r12d
  jns    L25
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L7:
  lea    D10(%rip),%r13
  mov    %r12d,%edi
  mov    %r13,%rsi
  call   <gnl_read_and_save@plt>
  mov    %eax,%r12d
  jmp    L50
  nopw   0x0(%rax,%rax,1)
L43:
  mov    D10(%rip),%rdi
  test   %rdi,%rdi
  je     L54
  call   <free@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int ft_get_next_line(const int fd, char **line) {
  static char *rbak = NULL;
  int res;
  res = 1;
  if (rbak == NULL)
    res = gnl_read_and_save(fd, &rbak);
  else if (!(ft_strchr(rbak, '\n')))
    res = gnl_read_and_save(fd, &rbak);
  if (rbak != NULL || res >= 0)
    res = gnl_set_line(&rbak, line);
  if (res < 1) {
    if (rbak)
      free(rbak);
    return (res);
  }
  return (res);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ft_get_next_line>:
  endbr64
  push   %r13
  push   %r12
  mov    %edi,%r12d
  mov    D27(%rip),%rdi
  push   %rbp
  mov    %rsi,%rbp
  test   %rdi,%rdi
  je     L43
  mov    $0xa,%esi
  lea    D27(%rip),%r13
  call   <ft_strchr@plt>
  test   %eax,%eax
  je     L41
L31:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <gnl_set_line@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jle    L60
L19:
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L41:
  mov    %r12d,%edi
  mov    %r13,%rsi
  call   <gnl_read_and_save@plt>
  mov    %eax,%r12d
L55:
  cmpq   $0x0,D27(%rip)
  jne    L31
  test   %r12d,%r12d
  jns    L31
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L43:
  lea    D27(%rip),%r13
  mov    %r12d,%edi
  mov    %r13,%rsi
  call   <gnl_read_and_save@plt>
  mov    %eax,%r12d
  jmp    L55
  nopw   0x0(%rax,%rax,1)
L60:
  mov    D27(%rip),%rdi
  test   %rdi,%rdi
  je     L19
  call   <free@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int ft_get_next_line(const int fd, char **line) {
  static char *rbak = NULL;
  int res;
  res = 1;
  if (rbak == NULL)
    res = gnl_read_and_save(fd, &rbak);
  else if (!(ft_strchr(rbak, '\n')))
    res = gnl_read_and_save(fd, &rbak);
  if (rbak != NULL || res >= 0)
    res = gnl_set_line(&rbak, line);
  if (res < 1) {
    if (rbak)
      free(rbak);
    return (res);
  }
  return (res);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<toto>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x9000(%rsp),%r11
L45:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L45
  sub    $0xdf0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x9dec(%rbp)
  jmp    L19
L59:
  movl   $0x1,-0x9de8(%rbp)
  jmp    L10
L48:
  mov    -0x9dec(%rbp),%edx
  mov    -0x9de8(%rbp),%eax
  add    %edx,%eax
  sub    $0x1,%eax
  cltq
  mov    -0x9de0(%rbp,%rax,4),%eax
  mov    -0x9dec(%rbp),%ecx
  mov    -0x9de8(%rbp),%edx
  add    %edx,%ecx
  lea    0x2(%rax),%edx
  movslq %ecx,%rax
  mov    %edx,-0x9de0(%rbp,%rax,4)
  addl   $0x1,-0x9de8(%rbp)
L10:
  cmpl   $0x63,-0x9de8(%rbp)
  jle    L48
  mov    -0x9dec(%rbp),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x9dec(%rbp),%eax
  add    $0x8,%eax
  cmp    %eax,%edx
  jne    L11
  mov    -0x9dec(%rbp),%eax
  cltq
  imul   $0x194,%rax,%rax
  add    %rbp,%rax
  sub    $0x9c50,%rax
  movl   $0x2,(%rax)
L11:
  movl   $0x1,-0x9de4(%rbp)
  jmp    L6
L32:
  mov    -0x9dec(%rbp),%edx
  mov    -0x9de4(%rbp),%eax
  add    %edx,%eax
  sub    $0x5,%eax
  cltq
  mov    -0x9de0(%rbp,%rax,4),%eax
  mov    -0x9dec(%rbp),%ecx
  mov    -0x9de4(%rbp),%edx
  add    %edx,%ecx
  lea    0x2(%rax),%edx
  movslq %ecx,%rax
  mov    %edx,-0x9de0(%rbp,%rax,4)
  addl   $0x1,-0x9de4(%rbp)
L6:
  cmpl   $0x63,-0x9de4(%rbp)
  jle    L32
  addl   $0x1,-0x9dec(%rbp)
L19:
  cmpl   $0x63,-0x9dec(%rbp)
  jle    L59
  mov    -0x978c(%rbp),%edx
  mov    -0x9ddc(%rbp),%eax
  add    %edx,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

assistant: ```c
int toto() {
  int i, j, k;
  int a[100][100];
  int b[100];
  for (i = 1; i < 100; i++) {
    for (j = 1; j < 100; j++)
      b[i + j] = b[i + j - 1] + 2;
    if (i * 2 == i + 8)
      a[i][i] = 2;
    for (k = 1; k < 100; k++)
      b[i + k] = b[i + k - 5] + 2;
  }
  return a[3][5] + b[1];
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<toto>:
  endbr64
  lea    -0x9000(%rsp),%r11
L29:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L29
  sub    $0xde8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x9dd8(%rsp)
  xor    %eax,%eax
  lea    0x194(%rsp),%rcx
  lea    0x324(%rsp),%r8
  lea    0x190(%rsp),%rsi
  mov    $0x1,%edi
  jmp    L6
L33:
  lea    -0x18c(%rcx),%rax
L17:
  mov    -0x14(%rax),%edx
  add    $0x2,%edx
  mov    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L17
  add    $0x1,%edi
  add    $0x4,%rcx
  add    $0x194,%r8
  add    $0x4,%rsi
  cmp    $0x64,%edi
  je     L12
L6:
  lea    -0x190(%rcx),%rax
L24:
  mov    (%rax),%edx
  add    $0x2,%edx
  mov    %edx,0x4(%rax)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L24
  lea    (%rdi,%rdi,1),%edx
  lea    0x8(%rdi),%eax
  cmp    %eax,%edx
  jne    L33
  movl   $0x2,(%r8)
  jmp    L33
L12:
  mov    0x4(%rsp),%eax
  add    0x654(%rsp),%eax
  mov    0x9dd8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x9de8,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int toto() {
  int i, j, k;
  int a[100][100];
  int b[100];
  for (i = 1; i < 100; i++) {
    for (j = 1; j < 100; j++)
      b[i + j] = b[i + j - 1] + 2;
    if (i * 2 == i + 8)
      a[i][i] = 2;
    for (k = 1; k < 100; k++)
      b[i + k] = b[i + k - 5] + 2;
  }
  return a[3][5] + b[1];
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<toto>:
  endbr64
  lea    -0x9000(%rsp),%r11
L13:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L13
  sub    $0xde8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x9dd8(%rsp)
  xor    %eax,%eax
  lea    0x4b8(%rsp),%rcx
  mov    $0x1,%eax
  jmp    L33
  cs nopw 0x0(%rax,%rax,1)
L12:
  lea    (%rdx,%rdx,1),%esi
  add    $0x9,%eax
  cmp    %eax,%esi
  jne    L7
  movl   $0x2,(%rcx)
L7:
  add    $0x194,%rcx
  mov    %edx,%eax
L33:
  lea    0x1(%rax),%edx
  cmp    $0x63,%eax
  jne    L12
  mov    0x4(%rsp),%eax
  add    0x654(%rsp),%eax
  mov    0x9dd8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L24
  add    $0x9de8,%rsp
  ret
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int toto() {
  int i, j, k;
  int a[100][100];
  int b[100];
  for (i = 1; i < 100; i++) {
    for (j = 1; j < 100; j++)
      b[i + j] = b[i + j - 1] + 2;
    if (i * 2 == i + 8)
      a[i][i] = 2;
    for (k = 1; k < 100; k++)
      b[i + k] = b[i + k - 5] + 2;
  }
  return a[3][5] + b[1];
}
```
user: Translate the below assembly code into C programming language:

```asm
<toto>:
  endbr64
  lea    -0x9000(%rsp),%r11
L37:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L37
  sub    $0xde8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x9dd8(%rsp)
  xor    %eax,%eax
  lea    0x4b8(%rsp),%rdx
  mov    $0x2,%eax
  nopl   0x0(%rax)
L33:
  lea    (%rax,%rax,1),%esi
  lea    0x8(%rax),%ecx
  cmp    %ecx,%esi
  jne    L18
  movl   $0x2,(%rdx)
L18:
  add    $0x1,%eax
  add    $0x194,%rdx
  cmp    $0x64,%eax
  jne    L33
  mov    0x4(%rsp),%eax
  add    0x654(%rsp),%eax
  mov    0x9dd8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L39
  add    $0x9de8,%rsp
  ret
L39:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int toto() {
  int i, j, k;
  int a[100][100];
  int b[100];
  for (i = 1; i < 100; i++) {
    for (j = 1; j < 100; j++)
      b[i + j] = b[i + j - 1] + 2;
    if (i * 2 == i + 8)
      a[i][i] = 2;
    for (k = 1; k < 100; k++)
      b[i + k] = b[i + k - 5] + 2;
  }
  return a[3][5] + b[1];
}
```